Engineering Mechanics part 2
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engineering_mechanics_pt._2_handbook.pdf
Study Unit
Engineering Mechanics, Part 2 By
Andrew Pytel, Ph.D.
Associate Professor, Engineering Mechanics The Pennsylvania State University
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w When you complete this study unit, you’ll be able to
• Describe the characteristics of a body at rest, and the forces that impact on a body
• Describe balanced concurrent forces and the methods necessary to calculate the principles, rules, and formulas that apply to these forces
• Define the balanced nonconcurrent forces and the equations used to determine the reactions to different components
• Discuss the laws and effects of friction
• Identify the conditions that influence sliding friction on a level surface
• Explain the problems that relate to bodies at rest on inclined surfaces
BODIES AT REST 1
Introductory Explanations 1 Balanced Concurrent Forces 13 Balanced Nonconcurrent Forces 38
EQUILIBRIUM INVOLVING FRICTION 51
Characteristics of Friction 51 Bodies on Level Surfaces 58 Bodies on Inclined Surfaces 67
SELF-CHECK ANSWERS 79
PRACTICE PROBLEMS ANSWERS 81
EXAMINATION 83
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C o n t e n t s
C o n t e n t s
1
BODIES AT REST
Introductory Explanations
Equilibrium of Rigid Body at Rest
1 A force has been defined as any action that causes, or tends to cause, a change in the motion of a body. In the
first portion of this text, we shall consider only problems in which a body is acted upon by a system of forces, and it will be assumed that these forces are so related that motion of the body is prevented. In other words, we shall discuss only systems of forces that keep a body at rest. A body is consid- ered to be at rest if it does not move with respect to the earth’s surface. It is said that a body at rest is in equilibrium and that the forces acting on the body form a balanced system. A body moving in a certain way with respect to the earth’s surface also may be in equilibrium under the action of a balanced sys- tem of forces, as will be explained in Engineering Mechanics, Part 4. However, since we shall not consider moving bodies that are in equilibrium in Engineering Mechanics, Part 2, we shall use the expression “at rest” rather than “in equilibrium.”
Engineering Mechanics, Part 2
Engineering Mechanics, Part 22
Although the actual body that is acted upon by a system of forces will be deformed to some extent by the forces, we shall neglect such deformations and assume that every body is rigid. When a rigid body is at rest under the action of a system of concurrent forces, the shape and size of the body need not be considered. It is usually assumed that the lines of action of all the forces in such a system pass through the center of gravity of the body, and that the body is replaced by a single particle located at its center of gravity. In other words, it is assumed that the weight of the entire body is concentrated at the center of gravity of the body. These assumptions will be made in all problems in this text pertaining to balanced sys- tems of concurrent forces. When a rigid body is at rest under the action of a system of nonconcurrent forces, the body must have such a shape and size that the line of action of each force of the system will pass through the body. It may be assumed either that the weight of the entire body is con- centrated at the center of gravity of the body or that the weight of any portion of the body is concentrated at the center of gravity of that portion.
The solution of any problem relating to a body that is at rest under the action of a system of forces is based on the appli- cation of Newton’s first law of motion. According to this law, a particle continues in its state of rest, or of uniform motion along a straight line, unless acted upon by a force. That is, a force acting on a particle at rest tends to move the particle. To prevent movement of a particle under the action of a force, some other force or other forces must also act on the particle.
It is also necessary to keep in mind the principle expressed by Newton’s third law of motion, which may be stated as follows: For every action, there is an equal and opposite reaction.
Newton’s second law of motion is applied only where actual motion is involved, and it will not be used in this text.
Methods of Supporting Body at Rest
2 A body is always subjected to the force known as the weight of the body. This force always acts vertically
downward. To prevent movement of a body under the action of its weight, the body must be supported in some way. Several simple methods of supporting a body at rest are represented
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in Figure 1. The shape and size of each body in this illustra- tion are selected arbitrarily. In Figure 1A, the body is placed on a level floor in a building. In Figure 1B, the body is attached to a steel rod that is fastened to a fixture in the ceiling of a large room. In Figure 1C, the body is attached to a strong wire, which in turn is suspended from a bracket. This bracket consists of a horizontal bar and an inclined bar, which are connected to fixtures in a wall. There are numerous other methods of supporting bodies. Several will be described in problems in this text.
It may be assumed that the floor on which the body in Figure 1A is supported is fixed in position with respect to the earth’s surface. When the body is at rest on the floor, the only two forces that need be considered are the weight of the body and the force exerted on the body by the floor to prevent movement of the body. Such a second force is the “reaction” mentioned in Newton’s third law of motion.
FIGURE 1—Simple Methods of Supporting Bodies
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The body in Figure 1B is attached to a rod. However, so that the weight of the body and the weight of the rod will not cause movement of both the body and the rod, the rod must be fastened to some object that is fixed in position with respect to the earth’s surface. In this case, it may be assumed that the ceiling is fixed and that the rod will not move if it is fastened properly to the fixture in the ceiling. At the connection between the body and the rod, there are the following two forces: the weight of the body, which acts on the rod, and the force (reaction) exerted on this body by the rod to prevent movement of the body. Also, at the connection between the rod and the fixture, there are two other forces. One force is produced by the weights of the body and the rod, and the other force is the reaction exerted on the rod by the fixture to prevent movement of the body and the rod.
When a body is supported in the manner indicated in Figure 1C, it may be assumed that the wall is fixed with respect to the earth’s surface, and four sets of forces must exist to prevent movement of the body, the wire, and the bars. In a case like this, it would usually be permissible to neglect the weights of the wire and the bars.
The four sets of forces that exist in Figure 1C are the follow- ing: First, at the connection of the body to the wire, there are two forces: the weight of the body and the reaction exerted on the body by the wire. Second, at the connection of the wire to the bracket, there are three forces: a force exerted by the wire on both bars, a reaction exerted on the wire by the horizontal bar, and a reaction exerted on the wire by the inclined bar. Third, at the connection of the horizontal bar to the fixture, there are two forces: a force exerted by the bar on the fixture and the reaction exerted on the bar by the fixture. Fourth, at the connection of the inclined bar to the fixture, each of these parts exerts a force on the other part.
External Forces and Internal Forces
3 The forces considered in the preceding article may be classified as external forces and internal forces. Where a
body is supported directly by a fixed object, as the floor in Figure 1A, all forces acting on the body must be external
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forces. Thus, the weight of the body and the reaction of the floor are external forces. If a body is supported by a fixed object with the aid of one or more intermediate links, an external force on any part would be the force representing the weight of the part itself, the force exerted by the supported body directly on the part, or the force exerted by the fixed support directly on the part; whereas any force exerted on or by an intermediate link (other than the force representing the weight of the link) would be considered an internal force.
In Figure 1B, the force exerted by the body on the rod, the weight of the rod itself, and the force exerted by the fixture on the rod are external forces; whereas the force exerted by the rod on the body and the force exerted by the rod on the fixture are internal forces. When a body is supported as in Figure 1C and the weight of the wire and the weights of the bars are neglected, the external forces are the force exerted by the body on the wire, the force exerted by the lower fixture on the horizontal bar, and the force exerted by the upper fix- ture on the inclined bar. Each of the other forces involved is an internal force.
A system of forces considered in the solution of a problem may consist only of external forces, only of internal forces, or of a combination of external and internal forces.
Requirements of Free-Body Diagram
4 To solve a problem involving a body at rest under the action of a system of forces, it is usually desirable to
draw a diagram called a free-body diagram (FBD). In the FBD for a particular problem, it is necessary to represent a body or a portion of a body in some suitable manner and to show all known information about the forces acting on the body or on the portion considered. If the position of the line of action of a force with respect to some fixed point is known, that line of action should be shown in the FBD and identified by the notation used for designating that force. If the magnitude of a force and the direction of the force along its line of action are known, these characteristics should also be included in the FBD. In some cases, the only characteristic of a force that is known is the position of one point on the line of action of the force. This point should be identified in the FBD as part of
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the data, but the line of action of the force may have to be omitted at first. However, when you have to consider a force of this kind, you must remember that such a force exists.
The FBD for a problem need not be drawn accurately to scale. However, it should be a neat sketch on which all the known information about the forces of the system is indicated clearly. If some information about a force, such as the magni- tude of a force or an angle, must be determined in the solution of a problem, the missing value should be denoted in the FBD by a letter in an appropriate place. For instance, an unknown magnitude of a force may be denoted by F or P, or an unknown angle may be denoted by A or H.
Typical Free-Body Diagram
5 To illustrate the general features of free-body diagrams, we shall analyze the systems of forces involved when
bodies are supported by the methods indicated in Figure 1. It is assumed in each case that the body is at rest. In Figure 1A, the body is subjected to a force produced by the weight of the body. This force must act vertically downward. We shall assume that the body weighs 50 lb (pounds). Movement of the body under the action of its weight is prevented because the floor also exerts a force (reaction) on the body. The force rep- resenting this reaction must have the same magnitude as (be equal to) the force representing the weight of the body. Also, for the force representing the reaction to be opposite to the force representing the weight, the reaction must act vertically upward and the two forces must have a common line of action. The FBD for the body in Figure 1A is shown in Figure 2A. The body itself is represented arbitrarily by a small circle at its center of gravity; the weight of the body is represented by a force W = 50 lb acting vertically downward; and the reac- tion of the floor is represented by a force that must act vertically upward and whose magnitude is R.
To analyze a problem based on the conditions in Figure 1B, it is desirable to consider at least the system of forces that con- sists of only the external forces on the rod. In the FBD for this illustration, the rod could be represented by a narrow rectangle, as in Figure 2B. The external forces shown in Figure 2B are a force W1 representing the weight of the body,
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a force W2 representing the weight of the rod, and a force R representing the reaction of the fixture on the rod. Since the forces W1 and W2 must act vertically downward and the reac- tion R must be equal and opposite to the resultant of these two forces, the force R must act vertically upward. The mag- nitudes of the weights W1 and W2 would be known and should be given in the FBD.
Although it probably would not be necessary to draw other free-body diagrams for the conditions in Figure 1B, two such diagrams are drawn in Figure 2C and D. In Figure 2C is shown the FBD for a very short portion of the lower end of the rod in Figure 1B. The only two forces acting on this portion of the rod are the external force W1 in Figure 2C, representing the weight of the body, and the internal force F1, representing the reaction exerted by the rod on the body. In Figure 2D is shown the FBD for a very short portion of the rod at its upper end. The only two forces acting on this portion of the rod are the internal force F2, which represents the sum of the weights of the body and the rod, and the external force R, which represents the reaction of the fixture on the rod.
FIGURE 2—Free-Body Diagrams for Conditions in Figures 1A and B
A. FBD for body in Figure 1A
B. FBD for exter- nal forces in Figure 1B
C. FBD for lower portion of the rod in Figure 1B
D. FBD for upper portion of the rod in Figure 1B
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FBD for a Framework
6 When a body is supported by a framework, as in Figure 1C, it is often advisable to draw several free-body
diagrams. In one of these diagrams, the only forces included are the external forces. In each of the other free-body dia- grams, it is necessary to include both internal forces and external forces or internal forces alone. The several free-body diagrams that may be drawn for the conditions in Figure 1C are shown in Figure 3. The FBD in Figure 3A is that for the three external forces alone. Here, it may be assumed that a single point P represents the entire supporting frame in Figure 1C, which consists of the body, the wire, the bars, and the upper and lower fixtures. The force W in Figure 3A repre- sents the weight of the body; the force R1 represents the reaction exerted on the horizontal bar by the lower fixture; and the force R2 represents the reaction exerted on the inclined bar by the upper fixture. The magnitude of the force W would be known and should be shown in the FBD. To make the diagram in Figure 3A clearer, the lines of action of the reactions are drawn, and the directions of the reactions along their lines of action are indicated. The procedure for determining these characteristics and the magnitudes of the forces R1 and R2 will be described a little later.
In Figure 3B is shown the FBD for a small portion of the lower end of the wire. The only forces acting on this portion are the weight W and the reaction exerted on the body by the wire. In Figure 3C is shown the FBD for a short portion of the horizontal bar near the wall. The only forces acting on this portion are the force F2 exerted by the bar in the fixture in the wall and the reaction R1 exerted on the bar by the fix- ture. In Figure 3D is shown the FBD for a short portion of the inclined bar near the wall. The forces acting on this por- tion are the force F3 exerted by the bar on the fixture in the wall and the reaction R2 exerted on this bar by the fixture. In Figure 3E is shown the FBD for a small portion of the frame at the connection of the wire to the horizontal and inclined bars. The characteristics of the several forces will be discussed later.
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General Procedure in Problem Involving Body at Rest
7 A problem involving a body at rest can always be solved analytically. However, where more than two forces are
involved, it is sometimes convenient to solve such a problem graphically. Whichever method is used in the solution of a problem involving a body at rest, there are four general steps, as follows:
1. Make a sketch indicating the body, the manner in which it is supported, and all the forces that act on it.
2. Draw an FBD for each system of forces that must be considered.
FIGURE 3—Free-Body Diagrams for Conditions in Figure 1C
A. FBD for external forces in Figure 1C B. FBD for lower portion of wire in Figure 1C C. FBD for portion of the horizontal bar in Figure 1C near the wall D. FBD for portion of the inclined bar in Figure 1C near the wall E. FBD for small portion of the frame in Figure 1C at the top of the
wire
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3. Either draw a polygon of forces (which may be a triangle) or write equations expressing the relations between the forces considered in each FBD.
4. Either measure the required quantities in the polygon of forces or compute the required quantities by solving the equations.
The first step requires no special explanation. Suitable sketches will be included in the example problems and practice problems that are given in this text. The general requirements of an FBD have been discussed in Articles 5 and 6. Additional information pertaining to a polygon of forces or to equations is given in the next article. The procedures in the measurement or computation of quantities are similar to the procedures described in Engineering Mechanics, Part 1, for solving problems pertaining to the resultants of systems of forces.
Force Polygon and Equations for Body at Rest
8 When a body is at rest, the forces acting on the body must be related as follows: If vectors representing the
forces are laid off successively in the proper directions and to scale, these vectors will form a polygon. (A triangle is classed as a polygon.) In other words, when the vectors are drawn so that the tail of each vector touches the tip of the preceding vector, the tip of the last vector must touch the initial point or the tail of the first vector.
Another way of expressing the conditions that must exist when a body is at rest is to say that the magnitude of the resultant of the forces acting on the body must be zero. This condition is usually expressed mathematically by the following two statements: 1) The vector sum of the horizontal components of the forces must be zero. 2) The vector sum of the vertical components of the forces must be zero. These two statements can be expressed by the equations.
�Fx = 0 (1) �Fy = 0 (2)
Here, the Greek letter � (sigma) is used to stand for “vector sum.” Thus, �Fx is the vector sum of the horizontal compo- nents of the forces and �Fy is the vector sum of the vertical
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components. For a body to be at rest, both equation 1 and equation 2 must be satisfied. If only one of these equations is satisfied and the other is not, the body will not be at rest.
When equations 1 and 2 are applied in the solution of a problem in this text, we shall always make the following assumptions: Horizontal components acting toward the right are considered as positive quantities, and horizontal components acting toward the left are considered as negative quantities. Vertical components acting upward are considered as positive quantities, and vertical components acting down- ward are considered as negative quantities.
When a body is at rest under the action of a system of con- current forces, the only requirement is either that vectors representing the forces form a polygon or that both �Fx and �Fy equal zero. In a problem involving a body at rest under the action of nonconcurrent forces, the forces must also meet the following requirement: The magnitude of the resultant moment of the system of forces acting on the body, with respect to any point O, must be zero. This requirement can be expressed by the equation
�MO = 0 (3)
in which �MO is the vector sum of the moments of the forces with respect to some reference point O.
When equation 3 is applied in the solution of a problem, we shall assume that clockwise moments are positive quantities and that counterclockwise moments are negative quantities. Equation 3 also must be true for a body at rest under the action of a system of concurrent forces, but it is not usually necessary to use this equation in the solution of a problem involving concurrent forces.
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Self-Check 1 At the end of each section of Engineering Mechanics, Part 2, you’ll be asked to pause and check your understanding of what you’ve just read by completing a “Self-Check” exercise. Answering these questions will help you review what you’ve studied so far. Please complete Self-Check 1 now.
1. Is it possible for a body to be in equilibrium and yet be moving? __________________________________________________________
2. When a body is at rest under the action of a system of concurrent forces, what point in the body is usually selected as the one at which the lines of action of the forces intersect? __________________________________________________________
3. What force always acts on a body? __________________________________________________________
4. If the weight of the rod in Figure 1B, Article 3, must be considered, what are the external forces that act on the rod? __________________________________________________________
5. If the weights of the wire and the bars in Figure 1C, Article 3, can be neglected, what are the external forces in the system? __________________________________________________________
6. In the FBD for a portion of the inclined bar near the wall, Figure 1C, what forces should be included? __________________________________________________________
7. After suitable free-body diagrams have been drawn for solving a problem by the analytic method, what should be the next step in the solution? __________________________________________________________
8. What equations are commonly used to express the statement that the magnitude of the resultant of the forces acting on a body at rest is zero? __________________________________________________________
9. What equation is generally used when it is desirable to consider the resultant moment of the forces acting on a body at rest? __________________________________________________________
Check your answers with those on page 79.
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Balanced Concurrent Forces
Balanced Collinear Forces
9 A system of collinear forces is a special system of con- current forces. Since all forces of a collinear system have
a common line of action, it is sufficient to apply only one equation instead of equations 1 and 2 in Article 8. This single equation may be written in the form
�F = 0
In order to solve a problem involving a system of balanced collinear forces, there can be only one unknown force in the system.
The simplest system of forces that can act on a body at rest consists of only two forces, namely, the weight of the body and a force exerted by a fixed object that supports the body. For the body to be at rest, these two forces must be concur- rent. Also, to satisfy the equation �F = 0, the two forces must act in opposite directions along the common line of action, and they must have the same magnitude. In other words, one force may be considered an action, and the other force is the equal and opposite reaction. Since the characteristics of the first force will be known, the characteristics of the other force can be determined simply by applying Newton’s third law of motion. For instance, when a body rests on a level floor, as in Figure 1A, the FBD for that body would be as shown in Figure 2A. The force W, representing the weight of the body, and the force R, representing the reaction of the floor, must have a common line of action and they must act in opposite directions along this line. Since the force W must act verti- cally downward, the force R must act vertically upward. Finally, since the forces must be equal, the magnitude of the force R must be the same as that of the force W. If W = 50 lb, then R = 50 lb.
Balanced System with More Than Two Forces
10 When a balanced system of collinear forces consists of more than two forces and the characteristics of all
except one are known, the characteristics of the unknown
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force can be determined very easily by applying the equation �F = 0. The procedure may be described as follows: First, make a sketch indicating the conditions and draw a suitable free-body diagram. Next, write the equation �F = 0 in which the magnitudes and the proper signs of the known forces are used, the unknown force is treated as a positive quantity, and the magnitude of this force is denoted by a suitable letter. Then solve this equation to determine the characteristics of the unknown force. The magnitude of the force is determined by ignoring the sign of the result. If the sign of the result is positive, the force must act in the positive direction. If the sign of the result is negative, the force must act in the nega- tive direction.
Example Problem
At intervals throughout this text you will find one or more example problems solved to illustrate clearly the application of a principle, rule, or formula. Read each problem carefully, and study the solution until you understand it thoroughly.
Problem: As indicated in Figure 1B, a body weighing 120 lb is attached to the lower end of a rod that weighs 30 lb, and the upper end of the rod is attached to a suitable fixture in the ceiling of a room. When the body is at rest, what are the magnitude and the direction of the force exerted on the rod by the fixture?
Solution: The conditions are represented in Figure 4A. Here, the body is represented by a sphere, and its weight is repre- sented by a vertical force W1 that acts downward through the center of gravity of the body. The rod is represented by a slender cylinder, and its weight is represented by a vertical force W2 that acts downward through its center of gravity. The center of gravity of the rod will be vertically above the center of gravity of the body, and the two forces W1 and W2 will be collinear. The fixture is represented by a small rectan- gular prism, and the force exerted on the rod by this fixture is represented by the force R. Since the body is at rest, the line of action of the force R must coincide with the lines of action of the forces W1 and W2. The magnitude and direction of the force R have to be determined. However, the positive direction is indicated in Figure 4A.
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The given weights W1 and W2 and the required reaction R are the only external forces acting on the rod. The FBD for these external forces is shown in Figure 4B, in which the rod is represented by a small rectangle.
The equation �F = 0 may be written in the following form:
–120 – 30 + R = 0
R = +150
The magnitude of the reaction R is 150 lb. Since the sign in the computed result is positive, the direction of the reaction must be positive, or upward. Therefore, the required reaction is 150 lb upward, or 150 lb.
#
FIGURE 4—Diagrams for Example Problem in Article 10
A. Characteristics of forces
B. Free-body diagram
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Three Balanced Concurrent Forces
11 Very often a body is acted upon by three concurrent forces that are not collinear. A typical FBD for such
a system of forces is shown in Figure 5A. In this case, it is assumed that the weight of the body is small enough to be neglected. Problems involving bodies at rest under the action of three concurrent forces may be divided into two classes. In one class, the magnitudes and the directions of two of the forces are given, and it is required to determine the magni- tude and the direction of the third force. In the other class the given characteristics are the magnitude and the direction of one force and the positions of the lines of action of the other two forces, and it is required to determine the magni- tudes of those two forces and their directions along their lines of action. To solve a problem of either class, it is neces- sary to consider a triangle of forces whose sides are the vectors representing the three forces composing the system. The unknown characteristics can be determined either graph- ically by laying off the vectors to scale or analytically by computing the proper quantities.
When a problem is to be solved graphically, it is usually advisable to draw the FBD carefully so that the line of action of each force will be exactly at the correct inclination with
FIGURE 5—Three Balanced Concurrent Forces
A. FBD
B. Triangle of forces
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respect to a horizontal or vertical reference line. Each vector in the triangle of forces can then be drawn parallel to the line of action of the corresponding force in the FBD.
Graphical Solution for Three Balanced Concurrent Forces
12 When a problem of the first class in Article 11 is to be solved graphically, the procedure is as follows:
The vectors representing the two given forces are drawn in the proper directions and with the proper lengths so that the tail of the second vector coincides with the tip of the first vec- tor. These two vectors are two sides of the required triangle of forces. The vector representing the unknown force must form the third side of this triangle, and its direction is from the tip of the second vector to the tail of the first vector.
For example, we shall assume that a body is at rest under the action of three concurrent forces F1, F2, and F3, as indi- cated in the FBD in Figure 5A, that the magnitudes and directions of the forces F1 and F2 are known, and that it is required to determine the magnitude and direction of the third force F3. The triangle of forces is shown in Figure 5B. The vector OA is laid off to represent the force F1, and the vector AB is laid off to represent the force F2. Then the vector BO, which completes the triangle, represents the force F3. The characteristics of this force F3 are determined in the manner described in Engineering Mechanics, Part 1, for determining the characteristics of the resultant of two concurrent forces, but the direction of the force F3 along its line of action is opposite to the direction such a resultant would have. After the direction of the force F3 along its line of action has been determined in the triangle of forces, this direction can be indicated in the FBD, as shown in Figure 5A.
When a problem of the second class in Article 11 is to be solved graphically, the first step is to lay off a vector repre- senting the force whose magnitude is known. To complete the triangle of forces, the procedure is as follows: Through the tail of the vector representing the known force, a line of unlimited length is drawn parallel to the line of action of either of the other forces; and through the tip of the vector representing the known force, another line is drawn parallel
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to the line of action of the third force so that it intersects the line that is parallel to the second force. The magnitudes of the forces that were not given are indicated by the lengths of the sides of the triangle that are parallel to the lines of action of those forces. The directions of those forces along their lines of action are determined by pointing the arrowheads so that the tail of each vector coincides with the tip of another vector. This procedure is applied in the following example problems.
Example Problems
Problem 1: A body weighing 250 lb is supported by a bracket in the manner indicated in Figure 1C. The length of the hori- zontal bar is 10 ft (feet), and the distance along the wall between the fixtures is 16 ft. Determine the characteristics of the reaction R1 exerted by the lower fixture on the horizontal bar and those of the reaction R2 exerted by the upper fixture on the inclined bar. The weights of the wire and the two bars may be neglected.
Solution: The positions of the lines of action of the external forces are shown in Figure 6A. It is known that the force W representing the weight of the supported body must act verti- cally downward. The known magnitude of this force is included in the sketch. Also, it is known that the line of action of the reaction R1 must pass through the connection of the horizon- tal bar to the fixture in the wall and must be horizontal. Likewise, it is known that the line of action of the reaction R2 must pass through the connection of the inclined bar to the wall fixture at the end of that bar and must have the same inclination as the bar. It will therefore be seen that the forces representing the weight of the body and the reactions exerted by the fixtures must be concurrent. Since the magnitude of each reaction and its direction along its line of action must be determined, the only information about a reaction that can be included in Figure 6A at first is the notation R1 or R2, which identifies the reaction.
In the FBD for the external forces, the entire frame can be represented by a single point, as P in Figure 6B. The line of action of the force W must be vertical; that of the force R1 must be horizontal; and that of the force R2 must have the same inclination as the inclined bar in the actual bracket. In
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this case, the inclination can be determined by drawing a line through P and another point located as follows: Along a hori- zontal reference line through P, a working point M is located by laying off, to any convenient scale, as 1 in. (inch) = 16 ft, a distance representing 10 ft toward the left; through this point M, a vertical line of unlimited length is drawn upward; and the required point N is located on this vertical line by laying off from M a distance representing 16 ft, to the selected scale. The magnitude and the direction of W along its line of action are shown in the FBD. The only known information about the force R1 or R2 is the position of its line of action.
The triangle of forces, drawn so that 1 in. represents 200 lb, is shown in Figure 6C. The vector OA representing the known force W is drawn vertically. Its length corresponds to the magnitude of the force; and its direction, which is downward, is indicated by an arrowhead. The other two sides of the tri- angle in Figure 6C are located as follows: Through the tail O of the vector OA, a line OB of unlimited length is drawn par- allel to the line of action of the reaction R1 in the FBD, which is horizontal; and through the tip A of the vector OA, a line is drawn parallel to the line of action of the reaction R2 in the FBD so that it intersects the line OB at C. Thus, the lines AC and CO are sides of the triangle of forces.
The directions of the vectors representing the reactions in the triangle of forces are determined by pointing the arrowheads so that the tail of each vector coincides with the tip of another vector. Also, the length of each vector indicates the magnitude of a reaction. The reaction R1 must act horizontally toward the right, and its magnitude is about 160 lb. The reaction R2 must act toward the left and upward, and its magnitude is about 300 lb.
The magnitudes and the directions of the reactions can now be put on the diagrams in Figure 6A and B.
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Problem 2: As indicated in Figure 7A, a body weighing 3500 lb is suspended from a wire sling, which is supported by a bar and an adjustable cable in the position shown. The bar and the cable are fastened to fixtures that are embedded in horizontal ground. The horizontal distance between the fix- tures is 20 ft, the length of the bar is 30 ft, and the horizontal distance from the line of action of the weight of the body to
FIGURE 6—Diagrams for Example Problem 1, Article 12
A. Locations of external forces
B. FBD for external forces
C. Triangle of forces for external forces
Engineering Mechanics, Part 2 21
the first fixture is 15 ft. The weights of the sling, the bar, and the cable may be neglected. Determine the characteristics of the reaction R1 exerted by the first fixture on the bar and of the reaction R2 exerted by the second fixture on the cable.
Solution: The positions of the lines of action of the external forces are shown in Figure 7A. Here the inclinations of the lines of action of the reactions R1 and R2 must be the same as the inclinations of the bar and the cable. The inclinations
FIGURE 7—Diagrams for Example Problem 2, Article 12
A. Locations of external forces
B. FBD for external forces
C. Triangle of forces for external forces
Engineering Mechanics, Part 222
of the bar and the cable may be determined as follows: Using a suitable scale, locate the two fixtures at points 20 ft apart. Also, locate the point a at the intersection of the bar and the cable by drawing a vertical line 15 ft to the left of the first fix- ture, and also drawing an arc with its center at this fixture and with a radius equal to 30 ft. Then draw lines from the point a to the two fixtures to represent the bar and the cable.
In the FBD for the external forces, which is shown in Figure 7B, the body and the entire frame consisting of the sling, the bar, the cable, and the two fixtures can be repre- sented by a single point P. The lines of action of the external forces W, R1, and R2 should be located accurately, and the magnitude of W and its direction along its line of action also should be indicated.
The triangle of forces in Figure 7C should be started by laying off the vector OA to represent the weight W. In this case, the selected scale is 1 in. = 3000 lb. Then the triangle may be completed by drawing the lines OB and AC parallel to the lines of action of the reactions R1 and R2. The arrowheads are placed on the vectors AC and CO, and the lengths of these vectors are expressed in pounds. It is found that the reaction R1 must act toward the left and upward and that its magni- tude is about 7100 lb. The reaction R2 must act toward the right and downward, and its magnitude is about 4400 lb.
Analytic Method for Three Balanced Concurrent Forces
13 When a problem involving a system composed of three balanced concurrent forces is to be solved ana-
lytically, it is always convenient first to draw a rough FBD and triangle of forces. Since the inclinations of the lines of action of the forces need not be laid off accurately in these diagrams, it is sufficient to make rough sketches indicating the relative positions of the lines of action of the forces in the FBD and the vectors of the forces in the triangle of forces. However, such rough sketches will usually be very helpful in planning and understanding the calculations. The direction of any unknown force along its line of action can be readily
Engineering Mechanics, Part 2 23
determined from the triangle of forces in the manner explained in Article 12, and the missing magnitudes of forces or missing angles can be computed by applying principles of trigonometry.
Computed magnitudes of forces will generally be expressed to three significant figures when the first figure is 2 or more, and to four significant figures when the first figure is 1. Computed angles will be taken to the nearest multiple of 10 min (minutes).
Example Problems
Problem 1: Solve example problem 1 in Article 12 analytically.
Solution: Rough diagrams similar to those shown in Figure 6 would be drawn. In the triangle of forces, the angle between the vectors representing the weight W and the reaction R1 is a right angle. The angle V2 between the line of action of the reaction R2 and the vertical can be found from the relation tan V2 = 10/16 = 0.625. Hence, V2 = 32°00�. The magnitudes of the reactions are
R1 = 250 tan 32°00� = 156 lb
and
The directions of the reactions along their lines of action are determined from the triangle of forces, as explained in exam- ple problem 1 in Article 12.
Problem 2: Find analytically the magnitudes of the reactions R1 and R2 in example problem 2 in Article 12.
Solution: By analyzing rough diagrams similar to those in Figure 7, the required results can be obtained as follows: From the dimensions given in Figure 7A, the angles H1 and H2 can be computed from the relations cos H1
15/30 and tan H2 = . Hence, H1 = 60°00� and H2 = 36°35�.
In the triangle of forces, the magnitude of the force W is 3500 lb; the angle between the vectors for W and R1 is 90° – 60°00� = 30°00�; the angle between the vectors for R1
30 15 20
1sin H +
R2 250 32 00
295= = cos � �
lb
Engineering Mechanics, Part 224
and R2 is 60°00� – 36°35� = 23°25�; and the angle between the vectors for W and R2 is 180° – 30°00� – 23°25� = 126°35�. The magnitudes of the reactions R1 and R2 are
and
Determination of Internal Forces
14 Any one of the procedures described in the preceding articles can also be applied to a balanced system of
forces that includes either internal forces alone or both exter- nal forces and internal forces. For example, the procedure for balanced collinear forces can be applied to each of the free- body diagrams in Figure 2C and D, and Figure 3B, C, and D. The procedure for three concurrent forces can be applied to the FBD in Figure 3E. You should note carefully the direction of an internal force along its line of action. This feature will be given special attention in the following example problems.
Example Problems
Problem 1: In the example problem in Article 10, determine a) the magnitude and the direction of the force exerted on the body by the rod and b) the characteristics of the force exerted on the fixture by the rod.
Solution: a) The FBD for a small portion of the rod in Figure 4A at its lower end would be similar to that in Figure 2C, where W1 represents the weight of the body in Figure 4A and F1 represents the internal force exerted on that body by the rod. It will be seen from the FBD that the forces W1 and F1 must be collinear and that these two forces must be equal and opposite. It is known that the force W1 acts ver- tically downward; and its magnitude, as given in the example problem in Article 10, is 120 lb. Therefore, the force F1 must act vertically upward and its magnitude must be 120 lb.
b) The FBD for a small portion of the rod in Figure 4A at its upper end would be similar to that in Figure 2D. Here, R represents the reaction exerted by the fixture in Figure 4A on
R1 3500 126 35
23 25 7070= =sin
sin �
�
�
� lb
Engineering Mechanics, Part 2 25
the rod, and the force F2 represents the force exerted by the rod on the fixture. As indicated in this FBD, the forces R and F2 must be collinear and these two forces must be equal and opposite. It was found in the example problem in Article 10 that the reaction R acts vertically upward and its magnitude is 150 lb. Therefore, the force F2 must act vertically down- ward and its magnitude must be 150 lb.
The force exerted by the rod on the body is smaller than the force exerted by the rod on the fixture because the weight of the rod itself acts on the fixture but not on the body. You should note that these two forces exerted by the rod act in opposite directions. However, the force exerted by the rod on either the body or the fixture acts away from the object under consideration.
Problem 2: In example problem 1 in Article 12 and example problem 1 in Article 13, determine the characteristics of each internal force.
Solution: The magnitude of the weight of the body, which acts vertically downward, is given in example problem 1 in Article 12 as 250 lb. Also, the characteristics of the reactions were determined analytically in example problem 1 in Article 13.
The internal force F1 exerted by the wire on the body can be determined from an FBD like that in Figure 3B. From this FBD it can be seen that the two forces W and F1 must be collinear and must be equal and opposite. Since W is 250 lb and acts vertically downward, the force F1 must act vertically upward and its magnitude must be 250 lb.
The internal force F2 exerted by the horizontal bar in Figure 1C on the lower fixture can be determined from an FBD like that in Figure 3C. This force F1 and the reaction R1 exerted by the fixture on the bar must be collinear and must be equal and opposite. Since R1 is 156 lb and acts horizontally toward the right, the force F2 must act horizontally toward the left and its magnitude must be 156 lb.
The internal force F3 exerted by the inclined bar in Figure 1C on the upper fixture can be determined from an FBD like that in Figure 3D. The force F3 and the reaction R2 exerted by the fixture on the bar must be collinear and must be equal and opposite. Since R2 is 295 lb. and acts toward the left and
Engineering Mechanics, Part 226
upward along its line of action, the force F3 must act toward the right and downward along its line of action and its mag- nitude must be 295 lb.
The FBD for the point at the intersection of the wire, the horizontal bar, and the inclined bar in Figure 1C would be like that in Figure 3E. The characteristics of the forces F1, F2, and F3 have already been determined. However, the results may be checked by considering a triangle of forces whose sides are vectors representing those three forces. You should note that the direction of each force in Figure 3E along its line of action is opposite to the direction of the same force in one of the other free-body diagrams for the following reason: The force F1 in Figure 3B represents the force exerted by the wire on the body, whereas the force F1 in Figure 3E repre- sents the force exerted by the wire on the bars. Similarly, F2 in Figure 3C represents the force exerted by the horizontal bar on the lower fixture, whereas F3 in Figure 3E represents the force exerted by the horizontal bar on the wire and the inclined bar. F3 in Figure 3D represents the force exerted by the inclined bar on the upper fixture, whereas F3 in Figure 3E represents the force exerted by the inclined bar on the wire and the horizontal bar. In each case, the force exerted by the wire acts away from the object under consideration; the force exerted by the horizontal bar acts toward any other object to which this bar is connected; and the force exerted by the inclined bar acts away from each other object.
If the three vectors representing the three forces in the FBD like that in Figure 3E are drawn accurately, the tip of the third vector should touch the tail of the first vector. If the check is made analytically, the following relationship between the square of the hypotenuse and the sum of the squares of the legs of a right triangle can be applied:
F3 2 = F1
2 + F2 2 or 2952 = 2502 + 1562
Engineering Mechanics, Part 2 27
Practice Problems 1 Practice problems are included in this text to test your ability to apply a rule or a formula. Work each problem carefully and check your answer against the answer given in the Practice Problems Answers section at the end of the text.
Solve each of the following problems both graphically and analytically.
1. As indicated in Figure 8, a body is supported by a wire and two bars that are fastened to two fixtures in a wall. If the body weighs 100 lb and the weights of the wire and the bars are neg- lected, what are the magnitude and the direction of the reaction R1 exerted by the lower fixture on the horizontal bar, and the characteristics of the reaction R2 exerted by the top fix- ture on the inclined bar? Assume that the lines of action of the reactions are continuations of the center lines of the bars. In the graphic solution, use the scale 1 in. = 50 lb.
__________________________________________________________
(Continued)
FIGURE 8—Conditions In Practice Problem 1, Article 14
Engineering Mechanics, Part 228
Practice Problems 1 2. A body weighing 4000 lb is supported by a sling, a bar, and a cable in the manner indicated in
Figure 9. Assuming that the bar and the cable are anchored securely in the ground and that the lines of action of the reactions R1 and R2 are continuations of the center lines of the bar and the cable, determine the magnitude and the direction of each reaction. For the graphic solution, use the scale 1 in. = 2000 lb.
Check your answers with those on page 81.
FIGURE 9—Conditions in Practice Problem 2, Article 14
Engineering Mechanics, Part 2 29
More Than Three Balanced Concurrent Forces
15 It is sometimes necessary to consider a system of balanced concurrent forces that includes more than
three forces. Usually, the known characteristics of the forces are the positions of the lines of action of all the forces and also the magnitudes and the directions along the lines of action of all but two forces. It is then required to determine the unknown characteristics of those two forces. It is nearly always most convenient to solve such a problem analytically by applying equations 1 and 2 in Article 8, which are �Fx = 0 and �Fy = 0. When these equations are applied, it is necessary to combine the magnitude of a force and the sign indicating the direction of the force along its line of action as parts of a single vector quantity. It is then possible to determine such vector quantities for two forces of a balanced system by applying only two equations.
Components of Forces
16 To refresh your memory and to make it easier for you to apply the equations �Fx = 0 and �Fy = 0, we
shall repeat formulas given in Engineering Mechanics, Part 1, which express the relationships between a given force and its rectangular components and also the relationships between the rectangular components of a force and the force itself. If a force whose magnitude is denoted by F makes an angle H with the horizontal, the horizontal component Fx and the ver- tical component Fy can be computed by the formulas
Fx = F cos H (1) Fy = F sin H (2)
It is necessary to consider both the magnitude of a component and its direction, as indicated by the sign + or –, in accor- dance with the assumptions stated in Article 8. When the sign of a component is being determined, the only factor to be considered is whether the component acts along the hori- zontal or the vertical axis in the selected positive direction or in the negative direction. It does not matter whether the line of action of an inclined force is above or below the selected horizontal reference axis or whether this line of action is to the right or to the left of the selected vertical reference axis.
Engineering Mechanics, Part 230
Moreover, it does not matter whether the arrowhead on the line of action of a force points toward or away from the point of intersection of these reference axes. The signs of the two components of a force are determined independently.
When the horizontal and vertical components Fx and Fy of a force are known, the angle H between the line of action of the force and the horizontal can be determined from the relation
(3)
The magnitude F of the force can be determined from either of the following relations:
(4)
or (5)
The signs of the components may be ignored when formula 3, 4, or 5 is applied. However, these signs must be considered for determining the position of the line of action of the force and the direction of the force along its line of action.
Selection of Reference Axes
17 In the preceding article it was assumed that the components of a force will be horizontal and vertical.
However, equations 1 and 2 in Article 8 can be applied if the reference axes are not exactly horizontal and vertical but have any two perpendicular directions. In the solution of a practical problem in which the magnitudes of two forces are unknown, the calculations can usually be simplified if one of the reference axes coincides with the line of action of one of the unknown forces and the other reference axis is perpendi- cular to that line of action. If the directions of the axes are so chosen, the magnitude of one component of the selected unknown force will be equal to the force itself and the magni- tude of the other component of that force will be zero.
F F H
x= cos
tan H F F
y
x
=
Engineering Mechanics, Part 2 31
When the reference axes are not horizontal and vertical, their directions should be indicated clearly in a sketch showing the positions of the lines of action of the forces of the system under consideration, to reduce the possibility of using an incorrect angle in the calculations.
Procedure in Solution
18 In accordance with the foregoing explanations, the procedure for solving a problem involving more than
three balanced concurrent forces may be outlined as follows: Assume that one reference axis for the rectangular compo- nents of the forces is along the line of action of either of the forces whose magnitude is unknown. (In the following expla- nation, we shall call this selected unknown force the first unknown force and call the other unknown force the second unknown force.) Assume that the other reference axis is per- pendicular to the line of action of the first unknown force. Treat these axes as if they were horizontal and vertical, and compute the rectangular components of each of the known forces. Include the components that are parallel to the line of action of the first unknown force in the equation �Fx = 0, and include the components that are perpendicular to this line of action in the equation �Fy = 0. Determine the magnitude and the direction along its line of action of the second unknown force by applying the equation �Fy = 0, since this force will be the only unknown force in the equation. Determine the mag- nitude and the direction along its line of action of the first unknown force by applying the equation �Fx = 0, since the second unknown force will now be known.
Since the directions of the unknown forces along their lines of action will not be established at first, it will be necessary to assume those directions to select suitable signs for the components of these forces. A good method is to assume first that each unknown force acts away from the common point of intersection of the lines of action of the forces. Then, if the sign for a force determined by solving the equation �Fy = 0 or �Fx = 0 comes out positive, the assumed direction is correct. If, on the other hand, it is found that the sign for this force is negative, the correct direction of the force is opposite to the assumed direction. The procedure is illustrated in the follow- ing example problems.
Engineering Mechanics, Part 232
Example Problems
Problem 1: Figure 10 represents a balanced system of con- current forces. The inclinations of the lines of action of the forces are indicated by the given angles. The line of action of the force F1 is vertical, and that of the force F5 is horizontal. The magnitudes of the forces F1, F2, and F3 and the direc- tions of these forces along their lines of action are known and are included in Figure 10. It is required to determine the magnitudes and the directions along their lines of action of the forces F4 and F5.
Solution: The diagram in Figure 10 can serve as the FBD in this case. Since the directions of the forces F4 and F5 along their lines of action are not known at first, the assumed directions are indicated by dashed arrows alongside these lines of action. In this problem, the line of action of the force F5 is horizontal. It will therefore be convenient to locate one reference axis for rectangular components along the line of action of the force F5 as indicated in Figure 10 in which the axis XX� is so located. Then the other reference axis YY� is perpendicular to the line of action of the force F5.
FIGURE 10—Forces in Example Problem 1, Article 18
Engineering Mechanics, Part 2 33
The rectangular components of the known forces F1, F2, and F3 are shown in the accompanying tabulation. Since the numerical values of these components will be added or sub- tracted, each component in this problem is expressed to the nearest multiple of 10 lb. The component of F4 to be used in the equation �Fy = 0 is +F4 sin 48°20� and the component of F5 in that equation is zero.
When the equation �Fy = 0 is applied, we obtain
– 2500 – 3030 – 7600 + F4 sin 48°20� = 0
Hence, F4 =
Since the sign of this result is positive, the assumed direction of F4 along its line of action is correct. So, F4 is 17,580 lb, and this force acts away from the point of intersection of the lines of action of the forces.
When the equation �Fx = 0 is applied, the component of F4 is +17,580 cos 48°20� = +11,690 and the component of F5 is taken as +F5. The resulting equation is
0 – 35,870 + 3800 + 11,690 + F5 = 0
Therefore, F5 = +20,380 lb. Since the sign of this result is positive, the assumed direction of F5 along its line of action is also correct. So F5 is 20,380 lb, which rounds off to 20,400 lb, and this force acts away from the point of intersection of the lines of action of the forces.
Problem 2: The four concurrent forces F1, F2, F3, and F4 in Figure 11A form a balanced system. The lines of action of all the forces are fixed, and the magnitudes and the directions of F1 and F2 are known. Determine the magnitudes and the directions along their lines of action of the forces F3 and F4.
13 130 40 20
17 580, sin
, � �
= + lb
F Fx Fy
F1 0 –2500
F2 –36,000 cos 4°50� = –35,870 –36,000 sin 4°50� = –3030
F3 +8000 cos 63°30� = +3800 –8500 sin 63°30� = –7600
Engineering Mechanics, Part 234
Solution: It does not really matter whether one reference axis for rectangular components is taken along the line of action of F3 or along the line of action of F4. If the line of action of F3 is selected, the positions of the reference axes XX� and YY�
and the angles between the axis XX� and the lines of action or the forces F1, F2, and F4 will be as shown in Figure 11B. The assumed directions of the unknown forces F3 and F4
FIGURE 11—Forces in Example Problem 2, Article 18
A. Locations of actual forces
B. FBD for assumed directions of axes
Engineering Mechanics, Part 2 35
along their lines of action are indicated by dashed arrows alongside those lines of action. The rectangular components of the known forces are as follows:
F1x = +11,000 cos 41°20� = +8260; F1y = –11,000 sin 41°20� = –7265
F2x = +5,000 cos 76°30� = +1170; F2y = –5,000 sin 76°30� = –4860
Also, the component of F4 in the equation �Fy = 0 is taken as –F4 sin 82°40� and the component of F3 in that equation is zero.
The result obtained by applying the equation �Fy = 0 is
–7265 – 4860 – F4 sin 82°40� = 0
Hence, F4 =
Since the sign of this result is negative, the assumed direction of F4 along its line of action is incorrect. So, F4 is 12,225 lb and this force acts toward the point or intersection of the lines of action of the forces.
The component of F4 in the equation �Fx = 0 is –12,225 cos 82°40� = –1560
and the component of F3 is taken as –F3. The equation is
+8260 + 1170 – 1560 – F3 = 0
Therefore, F3 = +7870 lb. Since the sign of this result is posi- tive, the assumed direction of F3 along its line of action is correct. So F3 is 7870 lb and this force acts away from the point of intersection of the lines of action of the forces.
Engineering Mechanics, Part 236
Practice Problems 2 Solve each of the following problems analytically.
1. The five forces F1, F2, F3, F4, and F5 in Figure 12 form a balanced system. The characteristics of the forces F1, F2, and F3 are as indicated; and the lines of action of the forces F4 and F5 are located as shown. For each of the forces F4 and F5, determine the magnitude and the direction along its line of action.
(Continued)
FIGURE 12—Forces ln Practice Problem 1, Article 18
Engineering Mechanics, Part 2 37
Practice Problems 2 2. If the four forces F1, F2, F3, and F4 in Figure 13 form a balanced system, what are the
magnitudes and the directions along their lines of action of the forces F3 and F4?
Check your answers with those on page 81.
FIGURE 13—Forces in Practice Problem 2, Article 18
Engineering Mechanics, Part 238
Balanced Nonconcurrent Forces
Balanced Parallel Forces
19 For a system of parallel forces to be balanced, the following two equations must be satisfied:
�F = 0 (1) �MO = 0 (2)
in which �F denotes the vector sum of the forces and �MO denotes the vector sum of the moments of the forces with respect to any point O.
In almost all practical problems involving bodies at rest under the action of parallel forces, the lines of action of the forces are vertical. In the following explanations, we shall assume that the lines of action of parallel forces are vertical. In order that the vector sum of the forces of a system may be zero, some of the forces must act downward along their lines of action and the other forces must act upward. Also, in order that the vector sum of the moments of the forces may be zero, some of the moments must be clockwise and the other moments must be counterclockwise.
External Forces on Seams
20 In the usual type of problem involving balanced par- allel forces, a relatively long body having a suitable
shape and size is supported near each end and is subjected to forces, or loads, located between the ends. Such a body is commonly called a beam. A typical free-body diagram for a beam is shown in Figure 14. The beam is represented by a single horizontal line, and it is assumed that each force is concentrated along a line.
FIGURE 14—Typical Free- Body Diagram for a Beam
Engineering Mechanics, Part 2 39
Actually, each downward force on a beam is distributed over some measurable area of the beam, and each upward force is distributed over an area of the support. In fact, a beam usually supports a weight that extends over the entire length of the beam as well as other forces that cover relatively small areas. However, when the equations �F = 0 and �MO = 0 are applied, it is assumed that each force is concentrated along a vertical line that passes through the center of gravity of the body that exerts the force on the beam.
A beam may be subjected to any number of forces between its ends. Ordinarily, each of these forces acts downward. There are usually two supports, one near each end, and the forces (commonly called reactions) that are exerted on the beam by the supports act upward. It is possible to support a beam in some other manner, but in this text we shall con- sider only beams that rest on supports at the ends. Internal forces are produced in a beam, but only the external forces will be considered in this text.
Computation of Reactions of Beams Subjected to Vertical Forces
21 When a beam rests on two supports near its ends and is subjected to vertical forces between its ends,
the known information generally includes the following quan- tities: the horizontal distance between the upward forces (or reactions) exerted on the beam by the supports; the location of the line of action of each downward force; and the magni- tude of each downward force. It is easily seen that the line of action of each reaction will be vertical and the direction of each reaction along its line of action will be upward, and it is required to determine the magnitude of each reaction. Although the magnitudes of the reactions can be determined by a graphic method, it is almost always more convenient to compute the reactions.
The procedure usually adopted for computing the magnitude of either reaction of a beam may be outlined as follows: Any point on the line of action of the other reaction is selected as the center of moments, and the equation �MO= 0 is applied. In this equation, the moment of the reaction whose line of action passes through the center of moments is zero; the
Engineering Mechanics, Part 240
moment of each of the given downward forces can easily be determined; and the moment of the reaction whose magni- tude is to be computed is expressed as that magnitude times the horizontal distance between the lines of action of the reactions. If the sign of the moment of this reaction is deter- mined by assuming that the reaction acts upward and if the moments of the downward forces are given the correct signs, then the sign of the result will always be positive.
After the magnitudes of the two reactions have been computed by applying the equation �MO = 0 twice with two different centers of moments, these values should be checked by applying the relation �F = 0. In other words, the sum of the magnitudes of the two reactions should be equal to the sum of the magnitudes of the given downward forces.
Example Problem
Problem: As indicated in Figure 15, a horizontal beam rests on two supports at A and B, which are so located that the horizontal distance between the lines of action of the reac- tions exerted by the supports is 16 ft. The beam is subjected to three forces P1, P2, and P3 that act vertically downward and have the magnitudes and positions indicated. Determine the magnitudes of the reactions R1 and R2.
FIGURE 15—Beam in Example Problem, Article 21
Engineering Mechanics, Part 2 41
Solution: To determine the magnitude of the reaction R1, the center of moments should be at B on the line of action of the reaction R2. Then, the moment of R2 is zero; the moment of the reaction R1 will be clockwise, or positive; and the moment of each downward force will be counterclockwise, or negative. The equation �MB = 0 becomes
+ R1 � 16 – 5000 � (16 – 3) – 1000 � (16 – 8) – 3600 � (16 – 12) = 0
or
+ R1 � 16 – 65,000 – 8000 – 14,400 = 0
Hence,
When the center of moments is at A on the line of action of the reaction R1, the moment of R1 will be zero, the moment of each downward force will be clockwise, and the moment of the reaction R2 will be counterclockwise. The equation �MA = 0 for this center of moments is
+ 5000 � 3 + 1000 � 8 + 3600 � 12 – R2 � 16 = 0
Therefore,
As a check, the sum of the magnitudes of the downward forces is 5000 + 1000 + 3600 = 9600 lb, and the sum of the magnitudes of the reaction is 5460 + 4140 = 9600 lb.
Balanced Nonparallel, Nonconcurrent Forces
22 A body at rest is sometimes subjected to a system of forces whose lines of action are not concurrent and
are not parallel. The body may have any shape, and the lines of action of the forces may have any directions. A fairly com- mon type of problem involving a body at rest under the action of balanced nonparallel, nonconcurrent forces is represented in Figure 16.
R2 66 200
16 4140= + = +, lb
Engineering Mechanics, Part 242
Here a horizontal beam is supported at its ends at A and B and is subjected to forces whose lines of action are not parallel. Usually a beam that is subjected to a system of non- parallel forces is supported in such a way that one reaction will be vertical. If the reaction at the left-hand end A of the beam in Figure 16 is vertical, the lines of action of the reac- tions would be as shown for R1 and R2. If the reaction at the right-hand end B is vertical, the lines of action of the reactions would be those for R1� and R2�. Sometimes it is assumed that the line of action of each reaction is parallel to the line of action of the resultant of the downward forces.
Since the vertical components of all three forces P1, P2, and P3 in Figure 16 act downward, the vertical component of each reaction will act upward. However, the magnitudes of the reactions and the actual position of the line of action of an inclined reaction will depend on the characteristics and loca- tions of the forces P1, P2, and P3.
In another type of problem involving a body at rest under the action of balanced nonparallel, nonconcurrent forces, the conditions are similar to those shown in Figure 17. In this case, it is assumed that the body is composed of three parts, AB, BC, and CD, which are rigidly connected to one another. It is assumed that the line of action of the reaction R2 must be vertical. The magnitudes of the reactions R1 and R2 and the other characteristics of the reaction R1 will depend on the dimensions of the parts of the body and on the characteris- tics and locations of the forces P1 and P2. The direction of the vertical component of R1 may be upward for some conditions and downward for other conditions, and the direction of the
FIGURE 16—Beam Subjected to Nonparallel Forces
Engineering Mechanics, Part 2 43
horizontal component of R1 may be toward the right for some conditions and toward the left for other conditions. The line of action of R1 is therefore represented by a dashed line, and no arrowhead is shown on that line of action.
It is usually advisable to replace each inclined force (including an inclined reaction) by its horizontal and vertical compo- nents at the point of intersection of the line of action of the force and the body.
When it is known that the line of action of a reaction is verti- cal, the magnitude of that reaction and its direction along its line of action can be determined by assuming that the center of moments is on the line of action of the other reaction and applying the equation �MO = 0. Since the moment of this other reaction will then be zero, the magnitude of the vertical reaction will be the only unknown quantity in the equation thus obtained. If the line of action of a reaction is inclined, the characteristics of the vertical component of that reaction can be determined in the manner just described for a vertical reaction. After the vertical components of both reactions have been computed by applying the equation �MO = 0 twice, the results should be checked by applying the equation �Fy = 0.
FIGURE 17—Body Subjected to Nonparallel Forces
Engineering Mechanics, Part 244
When it is known that the line of action of one reaction is vertical, the characteristics of the horizontal component of the other reaction can be determined by applying the equation �Fx = 0. If both reactions are inclined, the characteristics of the horizontal components of both reactions or the character- istics of the reactions themselves can be determined by using information that will be known. For instance, if the lines of action of the reactions are to be parallel to the line of action of the resultant of the given forces, the angle that this line of action makes with a horizontal reference line can be deter- mined. Then the magnitude of the horizontal component of each reaction can be found by dividing the vertical compo- nent of that reaction by the tangent of that angle; and the magnitude of each reaction can be found by dividing the vertical component of the reaction by the sine of that angle.
Example Problems
Problem 1: In Figure 16, the magnitudes of the downward forces are P1 = 6000 lb, P2 = 1000 lb, and P3 = 4000 lb. Determine the characteristics of the reactions R1 and R2 for each of the following conditions: a) The line of action of R1 is vertical; b) the line of action of R2 is vertical; c) the lines of action of R1 and R2 are parallel to the line of action of the resultant of the forces P1, P2, and P3.
Solution: a) The first step is to compute the horizontal and vertical components of the forces P1 and P3, as follows:
P1x = +6,000 cos 60° = +3000; P1y = –6,000 sin 60° = –5200
P3x = –4,000 cos 75° = –1040; P3y = –4,000 sin 75° = –3860
When the line of action of the reaction R1 is vertical, the magnitude and the direction along its line of action can be determined by applying the equation �MB = 0 and taking the center of moments at the point B at which the line of action of R2 intersects the beam. It will be assumed that R1 acts upward. Since the moments of the horizontal components of the forces P1 and P3 are zero and the moment of the reaction R2 also is zero, the equation is
Engineering Mechanics, Part 2 45
�MB = 0 = +R1 � 24 – 5200 � (24 – 7) – 1000 � 12 – 3860 � 8 Hence, R1 = +5470 lb, or 5470 lb
#
The vertical component of the reaction R2 can be computed by taking moments with respect to the point A at which the line of action of R1 intersects the beam. If it is assumed that this vertical component acts upward, the result obtained by applying the equation �MA = 0 is
+ 5200 � 7 + 1000 � (7 + 5) + 3860 � 16 – R2y � 24 = 0 Therefore, R2y = +4590 lb, or 4590 lb.
#
As a check, the sum of the vertical components of the down- ward forces is 5200 + 1000 + 3860 = 10,060 lb, and the sum of the magnitudes of the vertical components of the reactions is 5470 + 4590 = 10,060 lb.
The horizontal component of R2 can be found by applying the equation �Fx = 0. If it is assumed that this component acts toward the right, the equation is
�Fx = 0 = +3000 – 1040 + R2x
The result obtained by solving this equation is R2x = –1960 lb. Hence, the horizontal component of R2 is 1960 lb toward the left, or 1960 lb.
!
The characteristics of R2 can be determined by applying formulas 3 and 5 in Article 16. Thus,
and H = 66°50�
Since this reaction acts toward the left and upward along its line of action, R2 = 5000 lb.
66°50�
R2 4590 66 50
5000= = sin � �
lb
tan H = 4590 1960
Engineering Mechanics, Part 246
b) In this case, the procedure is basically similar to that in part (a). The first step would be to compute the components of the forces P1 and P3. The next step would be to determine the magnitude and the direction along its line of action of the vertical reaction R2. The solution would be completed by determining the vertical component of the inclined reaction R1, the horizontal component of that reaction, and the magni- tude and direction of the reaction itself. After the components of P1 and P3 have been computed, the remainder of the solu- tion may be indicated as follows:
�MA = 0 = +5200 � 7 + 1000 � 12 + 3860 � 16 – R2y � 24
R2y = +4590 lb, or 4590 lb #
�MB = 0 = +R1y � 24 – 5200 � 17 – 1000 � 12 – 3860 � 8
R1y = +5470 lb, or 5470 lb #
�Fx = 0 = +3000 – 1040 + R1x
R1x = –1960 lb, or 1960 lb !
and H = 70°20�
R1 = 5810 lb 70°20�
c) Since both reactions are inclined, the vertical component of each reaction is computed by applying the equation �MO = 0. After the components of P1 and P3 have been determined, the procedure is as follows:
�MB = 0 = +R1y � 24 – 5200 � 17 – 1000 � 12 – 3860 � 8
R1y = +5470 lb, or 5470 lb #
�MA = 0 = +5200 � 7 + 1000 � 12 + 3860 � 16 – R2y � 24
R2y = +4590 lb, or 4590 lb #
R1 5470 70 20
5810= = sin � �
lb
tan H + 5470 1960
Engineering Mechanics, Part 2 47
The angle that the resultant of the forces P1, P2, and P3 makes with a horizontal reference line can be determined from the following relation:
Then
Thus, R1 = 5570 lb and R2 = 4670 lb 79°00� 79°00�
Problem 2: In Figure 17 the magnitude of the force P1 is 500 lb and the magnitude of the force P2 is 350 lb. Assume that the reaction R2 is vertical and determine a) the magnitude of R2 and b) the magnitude of the reaction R1 and the angle that the line of action of R1 makes with a horizontal reference line.
Solution: a) The horizontal and vertical components of the forces P1 and P2 are as follows:
P1x = – 500 cos 67° = – 195; P1y = – 500 sin 67° = – 460 P2x = + 350 cos 50° = + 225; P2y = – 350 sin 50° = – 268
It is convenient to imagine that the force P1 is replaced by its components at the point C and that the force P2 is replaced by its components at the point D. To determine the magnitude and the direction of the reaction R2, the equation �MA = 0 should be applied with the center of moments at the point A. It is known that this point is on the line of action of the reac- tion R1, and the moment of that reaction with respect to the point A will be zero. If it is assumed that the reaction R2 acts upward along its line of action, the equation is
�MA = 0 = – 195 � 3 + 460 � 4 + 225 � 3 + 268 � (4 + 5) – R2 � 4 = 0
Hence,
R2 = +1086 lb, or 1086 lb #
and
and R x2 4590 79 00
890= − = − tan � �
lb
R x1 5470 79 00
1070= − = − tan � �
lb
Engineering Mechanics, Part 248
b) If it is assumed that the vertical component of R1 acts upward and moments are taken with respect to the point B on the line of action of the reaction R2, the moment of the reaction R2 and the moment of the vertical component of the force P1 will be zero and the equation �MB = 0 becomes
– 195 � 3 + 225 � 3 + 268 � 5 + R1y � 4 = 0
Therefore, R1y = – 358 lb, or 358 lb $
As a check, the sum of the magnitudes of the vertical compo- nents of the downward forces (including the reaction R1) is 460 + 268 + 358 = 1086 lb, which is equal to the vertical component of the upward force R2. Hence, �Fy = 0.
The horizontal component of the reaction R1 can be found by applying the equation �Fx = 0. If it is assumed that this com- ponent acts toward the right, the equation is
– 195 + 115 + R1x = 0
Since it is found that R1x = –30 lb, the magnitude of the hori- zontal component of R1 is 30 lb and this component must act toward the left. Hence, R1x = 30 lb.
!
The direction and the magnitude of the reaction R1 may be determined from its components as follows:
and H = 85°10�
Therefore, R1 = 360 lb 85°10�
Engineering Mechanics, Part 2 49
Practice Problems 3 1. Figure 18 shows a horizontal beam that rests on two supports at A and B and carries two
vertical loads P1 and P2. Determine the magnitudes of the reactions R1 and R2.
2. Figure 19 represents a horizontal beam that is supported at A and B and is subjected to the forces P1, P2, and P3. Assume that the reaction R1 is vertical, and determine a) the magnitude of R1 and b) the magnitude and the direction of the reaction R2.
(Continued)
FIGURE 18—Conditions in Practice Problem 1, Article 22
FIGURE 19—Conditions in Practice Problem 2, Article 22
Engineering Mechanics, Part 250
Practice Problems 3 3. Assume that the lines of action of the reactions R1 and R2 for the beam in problem 2 are par-
allel to the line of action of the resultant of the forces P1, P2, and P3. Determine a) the angle that the line of action of this resultant makes with a horizontal reference line, b) the magni- tude of the reaction R1, and c) the magnitude of the reaction R2.
4. Figure 20 represents a body that is formed by connecting the three parts AB, BC, and CD together rigidly. If the body is subjected to the forces P1 and P2 and the line of action of the reaction R1 is vertical, what would be a) the magnitude and the direction along its line of action of the reaction R1 and b) the characteristics of the reaction R2?
Check your answers with those on page 81.
FIGURE 20—Conditions in Practice Problem 4, Article 22
Engineering Mechanics, Part 2 51
EQUILIBRIUM INVOLVING FRICTION
Characteristics of Friction
Sliding Friction and Rolling Friction
23 No surface of an actual body is absolutely smooth. Even though a certain surface may appear to be per-
fectly smooth, it is actually rough to some extent. Of course, some surfaces obviously are very rough. In the types of prob- lems involving bodies at rest that have been discussed so far in this text, it was not necessary to consider the smoothness of the surfaces of the bodies. We shall now deal with prob- lems in which the smoothness of a surface is an important factor.
When two bodies are in contact and one body moves or tends to move with respect to the other body, a force is developed that resists the movement. Such a force is called a friction force. For instance, if you tried to slide a heavy box across a level floor having a relatively rough concrete surface, you would have to apply a fairly large horizontal force to the box in order to move it. Naturally, the magnitude of a friction force depends on the roughness of the surfaces of the two bodies that are in contact. If you tried to slide the same box across a level floor made of relatively smooth wood, a smaller force would be sufficient to move the box.
In most problems involving friction forces, a flat surface of one body is in contact with a flat surface of another body, and one body slides or tends to slide along the other. The friction in such a case is called sliding friction. Whenever there is sliding friction, two surfaces rub against each other and resistance to movement is created because of the rough- ness of the surfaces.
Another type of movement occurs when one body with a curved surface rolls along a surface of another body. For example, the wheels of an automobile roll along a highway, or the wheels of a train roll along the rails of a track. Resistance
Engineering Mechanics, Part 252
to movement by rolling is developed because the weight on a wheel causes the wheel to flatten out to some extent and also causes the wheel to sink slightly into the supporting surface. As a result, a small ridge is formed in front of the wheel. The action that resists rolling is called rolling friction. Rolling fric- tion will not be discussed in this text.
Relation of Friction Force to Applied Force
24 A friction force exists only when a body is subjected to some other force that moves the body or tends to
move it. The direction of a friction force must always be oppo- site to the direction of such an applied force. To explain sliding friction, we shall assume that a horizontal force tends to slide a body along a level (horizontal) surface and that the shape of the body is a rectangular prism. Figure 21A repre- sents a body that is supported by a level surface and is acted upon by a vertical force representing the weight of the body, by a horizontal force P, and by other forces that are not shown. These other forces must include the following: a vertical force, which is exerted by the supporting surface on the body and acts upward to prevent the body from moving downward under the action of the force W; and a horizontal friction force, which is exerted by the supporting surface on the body and acts toward the left to resist movement of the body toward the right under the action of the force P. These two forces are indicated in Figure 21B and are designated as N and F.
The force P may be applied at any point on the body. The friction force F must be applied along the lower surface of the body, but its line of action is shown a little below that surface for convenience. Actually, the force W is the resultant of a very large number of parallel forces, each of which represents the weight of a portion of the body. This force W must pass through the center of gravity of the body. The force N is also the resultant of a very large number of parallel forces, each of which represents the force exerted on the body by a portion of the supporting surface. The position of the line of action of the force N will be discussed a little later.
Engineering Mechanics, Part 2 53
The magnitude of a friction force can never exceed the magni- tude of the applied force. When a force that tends to slide a body is less than the force required to move the body, the friction force is exactly equal to the applied force. As the applied force is increased (but before movement occurs), the friction force also becomes greater. However, the friction force cannot increase beyond a certain limiting value. In a particu- lar case, this limit depends on the roughness of the surfaces and on other factors. If the applied force exceeds the limiting value of the friction force, the body will move in the direction in which the applied force acts. If the body in Figure 21 is at rest, the magnitude of the friction force F must be exactly equal to the magnitude of the applied force P. If the magnitude of the force P becomes greater than the maximum possible magnitude of the friction force, the body will start to move toward the right.
FIGURE 21—Body at Rest on Level Surface
Engineering Mechanics, Part 254
Laws of Friction
25 The most important facts relating to friction, which have been established by experience, may be stated
as follows:
1. A friction force is greater when the surfaces in contact are rough than when these surfaces are smooth. Thus, friction force can be reduced by polishing a surface or providing lubrication between the surfaces. However, only friction without lubrication, or dry friction, will be considered in this text.
2. A friction force is greater just before a body starts to move than after the body has been put in motion.
3. A friction force always acts parallel to the surfaces that are in contact and in a direction opposite to the direction in which movement occurs or tends to occur.
4. The magnitude of the maximum friction force that can be developed is directly proportional to the magnitude of the total perpendicular, or normal, force between the two surfaces in contact. For instance, we shall suppose that the maximum friction force is 10 lb when the normal force is 50 lb. Then if the normal pressure is increased to 100 lb, or made twice as great, the maximum friction force would be 2 � 10 = 20 lb.
5. The magnitude of a friction force is independent of the area of contact between the surfaces. That is, if the body in Figure 21 were placed on the surface in the position indicated in Figure 21C, the magnitude of the friction force F would be the same as the magnitude of the force F for the conditions in Figure 21B.
Effects of Friction
26 One effect of friction is to offer resistance to move- ment when it is desired to slide a body along a flat
surface. This effect is often considered objectionable. However, the effect of friction in offering resistance to movement is utilized to advantage in numerous ways. In fact, if there were no friction, many common actions would be impossible. For instance, friction between a person’s shoes and a floor or a
Engineering Mechanics, Part 2 55
street pavement makes it possible for the person to walk. Friction between the brakes and the wheels of an automobile enables the driver to stop the automobile quickly. Friction is very useful in many other ways. In some cases, it is desirable to reduce the resistance that would be offered to movement by friction. For example, a common practice is to make the surfaces of moving parts in a machine as smooth as possible at a reasonable cost and also to provide lubrication between such surfaces.
In this text, we shall consider primarily problems in which it is required to determine the magnitude of an applied force that would overcome the effect of friction and start motion under certain conditions, or to determine the magnitude of a friction force that would prevent motion under other conditions.
Coefficient of Sliding Friction
27 As stated in Article 25, it has been observed that the magnitude of the friction force that is produced
between two surfaces in contact is directly proportional to the magnitude of the total normal force between the surfaces. In other words, when two bodies made of any materials are in contact, there is a constant ratio between the friction force that can be produced to resist sliding of one body on the other and the total normal force between the surfaces in con- tact. This constant ratio depends on the roughness of the surfaces and is called the coefficient of sliding friction for the materials used in the bodies whose surfaces are in contact. These materials may be either alike or different. Since the various portions of a relatively large surface usually have somewhat different roughnesses, it is not practicable to determine the coefficient of friction for two materials with a high degree of precision. As an example, for the materials given in Article 25 in connection with fact 4, the coefficient of friction would be 10/50 = 0.20.
It has also been observed, as stated in Article 25, that the friction force produced between two surfaces before sliding starts is greater than the friction force produced while sliding occurs. There are therefore two different coefficients of fric- tion for specified materials. The value before motion starts may be called the coefficient of static friction, and the value
Engineering Mechanics, Part 256
while motion occurs may be called the coefficient of kinetic friction. However, the difference between these two values for two given materials and specified conditions of the surfaces in contact is comparatively small. Since it is not possible to determine either value accurately, the same value of the coef- ficient of friction is generally used for either bodies at rest or bodies in motion. In this text, the coefficient of sliding friction will be denoted by the Greek letter � (mu). A few values of this coefficient that are generally used for dry friction are as follows: metal on metal, 0.15; concrete on concrete, 0.65; and hard wood on hard wood, 0.5.
The relation involving the total normal force between two surfaces in contact, the coefficient of sliding friction, and the maximum magnitude of the friction force that can be devel- oped between the surfaces may be expressed by the following equation:
Fm = �N
in which Fm = maximum friction force � = coefficient of friction N = normal force
Types of Problems Involving Friction
28 In each problem considered in this text, it will be assumed that a body is at rest under the action of a
system of four forces, which may be described as follows: 1) the weight of the body; 2) a single applied force that tends to slide the body under consideration along a plane surface of another body; 3) a friction force that resists sliding and acts parallel to the plane surface of contact between the two bodies; and 4) a force that is exerted on the body under con- sideration by the other body and acts perpendicular to the plane surface of contact.
In some problems the surface of contact will be level, and in other problems the surface of contact will be inclined. Also, in some problems the applied force will be parallel to the sur- face of contact, and in other problems the applied force will be inclined to that surface. Moreover, in most problems the magnitude of the friction force will be equal to the maximum value that can be developed, but in a few problems the mag- nitude of the friction force will be less than the maximum possible value.
Engineering Mechanics, Part 2 57
Self-Check 2 1. When two bodies are in contact and a force tends to slide one body along a surface of the
other body, what condition causes resistance to motion?
__________________________________________________________
2. When a wheel of an automobile rolls along a highway, describe the action at the surface of contact that tends to prevent rotation of the wheel.
__________________________________________________________
3. A horizontal force tends to slide a body along a level surface. If the magnitude of this applied force is increased but the body does not move, would the magnitude of the friction force be the same for all values of the applied force? Explain.
__________________________________________________________
4. Would the maximum possible magnitude of the friction force for a smooth surface be greater or less than the maximum possible value for a rough surface? Explain.
__________________________________________________________
5. The area of one plane surface of a certain body is 30 sq ft (square feet) and the area of another plane surface of this body is 20 sq ft. When the first surface of the body was placed in contact with a level surface and a horizontal force was applied to the body, the magnitude of the maximum friction force that was developed was 60 lb. What would be the magnitude of the maximum friction force that can be developed when the second surface of the body is in contact with this level surface? Explain.
__________________________________________________________
6. When the total normal pressure between two bodies in contact is 30 lb, the maximum friction force is 15 lb. What would be the maximum friction force when the normal pressure is 10 lb?
__________________________________________________________
7. Name two ways in which the friction force between two surfaces can be reduced.
__________________________________________________________
8. Is the coefficient of static friction greater than or less than the coefficient of kinetic friction? Explain.
__________________________________________________________
Check your answers with those on page 79.
Engineering Mechanics, Part 258
Bodies on Level Surfaces
Applied Force Horizontal
29 When a body rests on a level surface and no hori- zontal force is applied to the body, the only forces
acting on the body would be the weight of the body and a vertical force exerted on the body by the supporting surface. Since the weight of the body and the other force are an action and a reaction, the two forces must have equal magnitudes and must act in opposite directions along a common line of action.
If a horizontal force is applied to a body that is at rest on a level surface, a friction force also is developed. As previously explained, the line of action of this friction force must lie in the surface of contact and its direction along its line of action must be opposite to the direction of the applied horizontal force. Furthermore, since the body is at rest, the magnitude of the friction force must be equal to the magnitude of the applied force. However, another action must be considered. As seen in Figure 21B and C, the applied horizontal force P and the friction force F form a couple, because their lines of action are parallel, their magnitudes are equal, and they act in opposite directions. These two forces therefore tend to cause rotation of the body. Rotation is prevented in the fol- lowing way.
Since the only two vertical forces acting on the body are the weight of the body and the upward force exerted on the body by the supporting surface, the equation �Fy = 0 can be satis- fied only if the lines of action of these forces are parallel, their magnitudes are equal, and they act in opposite directions. Therefore, these two forces must also form a couple. The body will be at rest if the magnitude of the moment of this couple is equal to the magnitude of the moment of the couple formed by the horizontal forces and if the two couples tend to cause rotation in opposite directions.
For the conditions represented in Figure 21B, the moment of the couple formed by the horizontal forces P and F would be Ph1 (P times h1, clockwise). The moment of the couple formed�
Engineering Mechanics, Part 2 59
by the vertical forces W and N would have to be counter- clockwise. Hence, the line of action of the force N would have to lie to the right of the line of action of the force W, as shown. The magnitude of this second couple, which is Wd1, must be equal to Ph1. Therefore, the distance d1 can be computed from the relation
(1)
For the conditions in Figure 21C, the moment of the couple formed by the horizontal forces would be Ph2. So the line of action of the force N would have to lie to the right of the line of action of the force W. Also, since Wd2 must be equal to Ph2, the distance d2 can be computed from the relation
(2)
You should note the following similarity and the following dif- ference between Figure 21B and C, in which the same body is shown on the same surface in two positions. Regardless of the area of the body that is in contact with the supporting surface, the magnitude of the friction force F will be the same. Also, the magnitude of the force N will always be equal to that of the force W. However, the position of the line of action of the force N depends on the magnitude and the direction of the couple formed by the forces P and F.
Example Problem
Problem: Figure 22A shows a body supported by a level sur- face. The dimensions of the body in the plane of the paper are 6 ft and 4 ft, as indicated, and the body weighs 200 lb. (The dimension of the body at right angles to the plane of the paper need not be considered in this problem.) The coefficient of friction is 0.5. a) If the body is subjected to a horizontal force P, whose magnitude is 40 lb and whose line of action is 2 ft above the supporting surface, will the force cause the body to slide along the supporting surface? b) What will be the magnitude of the smallest horizontal force that would cause sliding? c) When the horizontal force is located as shown in Figure 22A and its magnitude is equal to the maximum
d Ph W2
2=
d Ph W1
1=
�
Engineering Mechanics, Part 260
possible value of the friction force, what would be the dis- tance between the line of action of the weight of the body and the line of action of the resultant upward force exerted on the body by the supporting surface?
Solution: a) Figure 22B shows a free-body diagram, in which are included the upward force N exerted on the body by the surface and the friction force F. Since the force P acts toward the left, the friction force F must act toward the right. Since the moment of the couple formed by the forces P and F is counterclockwise, the moment of the couple formed by the forces W and N must be clockwise and the line of action of N must lie to the left of the line of action of W.
The maximum friction force that can be developed is
Fm = �N = 0.5 � 200 = 100 lb
When P = 40 lb, the magnitude of the friction force actually needed to prevent sliding would be 40 lb; so the body would not slide under the action of such a force.
b) Since the magnitude of the maximum friction force that can be developed is 100 lb, the magnitude of the smallest horizontal force that would cause sliding would be 100 lb.
FIGURE 22—Conditions in Example Problem, Article 29
A. Locations of applied forces
B. Free-body diagram
Engineering Mechanics, Part 2 61
c) When P = 100 lb, W = 200 lb, and the moment arm of the couple formed by the forces P and F is 2 ft, the moment arm d of the couple formed by the forces W and N would be, by either formula 1 or 2,
Possibility of Overturning
30 The procedure described in the preceding article for locating the line of action of the upward vertical
force N can be applied only when the moment of the couple formed by the horizontal forces P and F is small enough. If the distance h2 in Figure 21C were relatively great and the dimension of the body in the plane of the paper were compar- atively short, it would be found that the computed value of the distance d2 would be greater than the distance between the line of action of the force W and the edge a of the body. As a result, the body would overturn about that edge before the body would slide along the supporting surface. For exam- ple, if you place an ordinary building brick on a fairly rough horizontal surface in an upright position (that is, resting on one of its small ends) and you attempt to slide the brick along the supporting surface by applying a horizontal force at the center of gravity of the brick, the brick will topple over before it starts to slide.
In the problems relating to sliding friction in this text, the dimensions of the object are so selected that the body will not topple over. However, in an actual problem it may be neces- sary to consider the possibility that the body will overturn before it slides.
Applied Force Inclined
31 When the force applied to a body on a level surface is inclined, it is convenient to replace this applied
force by its horizontal and vertical components. The horizon- tal component tends to cause sliding of the body along the surface and also tends to cause rotation of the body. The vertical component has an effect on the magnitude of the upward force exerted on the body by the supporting surface
Engineering Mechanics, Part 262
and also on the position of the line of action of that force. Typical conditions are considered in the following example problems.
Example Problems
Problem 1: Figure 23A shows a body that is supported on a level surface and is subjected to an applied force P. The dimensions and the weight of the body and the characteristics of the applied force are indicated. Determine a) the magni- tude of the friction force that would be required to prevent sliding of the body along the surface; b) the magnitude of the total vertical force N that would be exerted on the body by the supporting surface when this friction force is developed; and c) the horizontal distance d from the line of action of the force W to the line of action of the force N.
Solution: a) The FBD is shown in Figure 23B. The horizontal and vertical components of the applied inclined force are:
Px = +100 cos 30° = 86.6 lb and Py = –100 sin 30° = 50.0 lb " $
Since the only horizontal forces acting on the body are Px and F, and since F must act toward the left,
�Fx = 0 = +86.6 – F
FIGURE 23—Conditions in Example Problem 1, Article 31
A. Locations of applied forces
B. Free-Body diagram
Engineering Mechanics, Part 2 63
Hence, the minimum magnitude of the friction force that will prevent sliding is F, or 86.6 lb.
b) From the equation �Fy = 0,
– W – Py + N = 0
or
– 200 – 50 + N = 0 Hence, N = 250 lb.
#
c) If the center of moments is taken at the point O at which the line of action of the force W intersects the supporting surface, the moments of the forces W and F are zero. If we assume that the line of action of N lies to the right of the line of action of W, we can write the following equation:
MO = 0 = +86.6 � 4 – 50 � 3 – 250d
Hence, d = +0.79 ft, and the line of action of N would be 0.79 ft to the right of the line of action of W.
Problem 2: As shown in Figure 24A, a body is supported on a level surface and is subjected to an applied force P. The known information is given in the illustration. Determine a) the magnitude of the friction force that would be required to prevent sliding of the body; b) the magnitude of the total vertical force N that would be exerted on the body by the supporting surface when this force is developed; and c) the horizontal distance d between the lines of action of the forces W and N.
Solution: a) The FBD is shown in Figure 24B. The horizontal and vertical components of the applied force P are
Px = –50 cos 45° = 35.4 lb and Py = +50 sin 45° = 35.4 lb ! #
The minimum magnitude of the friction force that will pre- vent sliding can be found by applying the equation �Fx = 0. Since this friction force must act toward the right, the equa- tion is
– 35.4 + F = 0
Hence, the required friction force would be 35.4 lb "
Engineering Mechanics, Part 264
b) If it is assumed that the force N will act upward, the rela- tion obtained by applying the equation �Fy = 0 would be as follows:
– 200 + 35.4 + N = 0
Hence, N = 164.6 lb. #
c) If the center of moments is taken at the point O at which the line of action of the force W intersects the supporting sur- face and we assume that the line of action of the force N lies to the left of the line of action of W, we can write the following equation:
�MO = 0 = – 35.4 � 3 + 35.4 � 2.5 + 164.6d
Hence, d = + 0.11 ft, and the line of action of N would be 0.11 ft to the left of the line of action of W.
FIGURE 24—Conditions in Example Problem 2, Article 31
A. Locations of applied forces
B. Free-body diagram
Engineering Mechanics, Part 2 65
Practice Problems 4 1. A body that weighs 80 lb and is supported by a level surface is subjected to a horizontal force
P that is located as indicated in Figure 25. It is found that the body will start to slide when the magnitude of the force P is 20 lb. What is the value of the coefficient of friction?
2. When the magnitude of the force P in problem 1 is 20 lb but the body has not started to slide, what would be the horizontal distance between the line of action of the force W and the line of action of the total vertical force N exerted on the body by the supporting surface?
(Continued)
FIGURE 25—Conditions in Practice Problem 1, Article 31
Engineering Mechanics, Part 266
Practice Problems 4 3. Figure 26 shows a body that is supported by a level surface and is subjected to an inclined
force. The known information is given in the illustration. Compute a) the magnitude of the minimum friction force that would be required to prevent sliding of the body along the sur- face; b) the magnitude of the total vertical force N that would be exerted on the body by the supporting surface just before sliding would start; and c) the horizontal distance between the lines of action of the forces W and N.
4. For the conditions indicated in Figure 27, determine the same quantities that are specified in problem 3.
Check your answers with those on page 82.
FIGURE 27—Conditions in Practice Problem 4, Article 31
FIGURE 26—Conditions ln Practice Problem 3, Article 31
Engineering Mechanics, Part 2 67
Bodies on Inclined Surfaces
Types of Problems Involving Inclined Surfaces
32 When a body is supported by an inclined surface, a component of the weight of the body always acts
parallel to the surface in a downward direction and tends to cause the body to slide down the surface. If the angle between a horizontal plane and the surface is relatively small or if the surface is rough enough, the magnitude of the friction force that can be developed may be sufficient to prevent sliding of a body under the action of its own weight. In some cases, however, the only way to prevent a body from sliding down- ward on an inclined surface under the action of its own weight is by applying a force that has a component that acts parallel to the inclined surface in an upward direction.
Problems in which it is necessary to consider sliding of bodies down inclined surfaces can be divided into three types. In one type, it is only necessary to determine whether the mag- nitude of the maximum possible friction force will be sufficient to prevent sliding. In another type, it is required to determine the magnitude of the minimum force that must be applied in a certain specified direction in order to prevent sliding or to permit sliding at a uniform speed. In a third type, it is desired to cause a body to slide upward along an inclined surface and it is necessary to determine the magnitude of the minimum force that must be applied in a certain specified direction in order to start the body moving upward.
Friction Force Sufficient to Prevent Sliding
33 Figure 28 shows a body supported by an inclined plane surface that makes a relatively small angle A
with a horizontal reference plane. The weight of the body is represented by the vertical force W, which passes through the center of gravity of the body at the point O. This force can be replaced by its rectangular components that act parallel and perpendicular to the inclined surface. These components are
Engineering Mechanics, Part 268
represented by the forces X and Y, which also pass through the center of gravity of the body. The magnitudes of X and Y are
X = W sin A Y = W cos A
The component X tends to cause the body to slide downward along the inclined surface. However, it may be assumed that sliding will be prevented by the friction force F that will be developed. Also, the component Y pushes the body against the inclined surface, and the inclined surface exerts a force N on the body. For the body to be in equilibrium under the action of the forces W (or X and Y), F, and N, the lines of action of these forces must be concurrent. In Figure 28 the lines of action of the forces F and N pass through the point at which the line of action of the weight W intersects the inclined surface.
In this case it is convenient to analyze the system of forces by taking reference axes parallel and perpendicular to the inclined surface. A pair of axes having these inclinations are represented above the body in Figure 28 and are designated as X�X� and Y�Y�. For the axis X�X� parallel to the surface, the only forces that have components are X and F. Therefore, if it is assumed that sliding does not occur, we can write the equation
�Fx� = 0 = – X + F
FIGURE 28—Body at Rest on Inclined Plane Surface
Engineering Mechanics, Part 2 69
It follows that the magnitude of F is equal to the magnitude of X and that the friction force F acts upward along its line of action. For the axis Y�Y� perpendicular to the inclined sur- face, the only forces that have components are Y and N. So we can write the equation
�Fy� = 0 = – Y + N
It is seen that the magnitude of N is equal to the magnitude of Y and that the force N acts upward along its line of action.
If the magnitude of X or F is less than the maximum friction force that can be developed, or less than Fm = �N, the body will not slide down the inclined surface under the action of its own weight. Here � denotes the coefficient of sliding fric- tion. If X is greater than �N, the body will slide unless some other force is applied to the body to prevent it from sliding.
Example Problem
Problem: A body that weighs 40 lb is placed in contact with a plane surface, which is inclined to the horizontal at an angle equal to 15°. If the coefficient of friction between the body and the inclined surface is 0.3, will the body remain at rest on the surface or will it slide down the surface?
Solution: In this problem the magnitudes of the components of the weight of the body that are parallel and perpendicular to the inclined surface are
X = 400 sin 15 ° = 10.4 lb and Y = 40 cos 15° = 38.6 lb
Also, � = 0.3, and the magnitude of the maximum friction force that can be developed would be
Fm = �N = 0.3 � 38.6 = 11.6 lb. Since X is less than Fm, the body will not slide under the action of its own weight.
Angle of Repose
34 We shall now consider a special condition that may exist when a body is placed on an inclined plane
surface. If the angle between this surface and a horizontal reference plane is small, the body will remain at rest on the surface. If the angle is large, the body will slide down the plane under the action of its own weight. For certain rough-
Engineering Mechanics, Part 270
nesses of the surfaces in contact, there is some angle for which the body will be at rest but just ready to start to slide. In other words, the body would be at rest for that angle but would slide for a slightly greater angle. The greatest angle between a horizontal reference plane and a certain inclined surface for which a particular body would remain at rest on the surface is called the angle of repose for the materials of that body and that surface. If the actual angle of inclination for a particular case is not greater than the angle of repose, the body will remain at rest on the surface under the action of its own weight. If the actual angle of inclination is greater than the angle of repose, the body will slide under the action of its own weight.
The value of the angle of repose for certain materials can be determined as follows: Figure 29 shows a body that is placed on an inclined plane surface for which the angle Z between the horizontal and the surface is equal to the angle of repose. For this condition, the magnitude of the actual friction force must be equal to the magnitude of the force Fm = �N, where � is the coefficient of friction for the surfaces in contact. The mag- nitudes of the components of the weight W of the body that are parallel and perpendicular to the inclined surface are
X = W sin Z and Y = W cos Z
For the assumed conditions, the magnitude of X must be the same as the magnitude of the maximum friction force that can be developed, or X = Fm = ��. Also, since N = Y,
W sin Z = �W cos Z
Therefore, = � or tan Z = �
FIGURE 29—Angle of Repose
Engineering Mechanics, Part 2 71
In other words, the value of the angle of repose for a particu- lar body in contact with a certain inclined surface must be such that the tangent of this angle is equal to the coefficient of sliding friction for the surfaces in contact.
You should note that the angle of repose is not affected by the weight of the body.
Example Problem
Problem: What would be the angle of repose for the surfaces considered in the example problem in Article 33?
Solution: In this case, the coefficient of sliding friction is 0.3. The angle for which the tangent is equal to 0.3 is 16°40�. Hence, the angle of repose is 16°40�.
Force Required to Prevent Sliding on Inclined Surface
35 In some problems a body is to be supported by a plane surface that is inclined to a horizontal refer-
ence plane at an angle greater than the angle of repose. To prevent sliding of the body under the action of its own weight, it is necessary to apply a force that has a component that acts parallel to the inclined surface and upward along its line of action. The line of action of such an applied force may be parallel to the inclined surface that supports the body, or it may make an angle with this surface. Regardless of the position of the line of action of the applied force, all the forces acting on the body must form a balanced system.
Typical conditions are represented in Figure 30. Here, the angle B is greater than the angle of repose for the surfaces in contact, and the line of action of the applied force P that holds the body in equilibrium is parallel to the inclined sur- face and passes through the center of gravity of the body. Since the body tends to slide down the surface, the friction force F must act upward along the surface to resist sliding. However, the maximum possible magnitude of the friction force will be less than the magnitude of the component X of the weight of the body. The forces can be analyzed most
Engineering Mechanics, Part 272
easily if the reference axes are parallel and perpendicular to the inclined surface, as indicated by the directions of the axes X�X� and Y�Y� in Figure 30.
Procedures for determining the magnitude of an applied force that will prevent sliding of a body on an inclined surface are described in the following example problems.
Example Problems
Problem 1: As indicated in Figure 30, a body is to be sup- ported by an inclined surface. The angle B at which the surface is inclined to a horizontal reference plane is 40°, the body weighs 140 lb, and the coefficient of friction between the body and the supporting surface is 0.4. If sliding is to be pre- vented by a force P whose line of action is parallel to the inclined surface, what would be the least magnitude of this force?
Solution: The force that tends to cause the body to slide is the component X of the weight of the body, which acts parallel to the inclined surface. The magnitude of this component is X = W sin B = 140 sin 40° = 90.0 lb. Also, since the force P that is to be applied will be parallel to the inclined surface, the normal pressure N between the surfaces in contact is equal to the component Y of the weight of the body, which acts perpendicular to the inclined surface. The magnitude of
FIGURE 30—Force Applied to Prevent Sliding of Body (Force Parallel to Supporting Surface)
Engineering Mechanics, Part 2 73
this component is Y = 140 cos 40° = 107.2 lb. The greatest magnitude of the friction force that can be developed to resist sliding is Fm = �N = 0.4 � 107.2 = 42.9 lb.
Since 42.9 is less than 90.0, the body will slide down unless sliding is prevented by a force P that acts upward.
When we consider the forces that act parallel to the inclined surface, or parallel to the axis X�X� in Figure 30, we can write the following equation:
�Fx� = 0 = – X + Fm + P = – 90.0 + 42.9 + P or P = 90.0 – 42.9 = 47.1 lb
Hence, the required magnitude of P is at least 47.1 lb.
You should note that friction is helpful under the given con- ditions. Without the resistance furnished by friction, the required magnitude of P would be 90 lb.
Problem 2: Suppose that the body in example problem 1 is to be supported by the same inclined surface but with the conditions represented in Figure 31, where the line of action of the force P� that is applied to prevent sliding of the body is not parallel to the supporting inclined surface. If the angle M is 30°, what would be the least magnitude of the force P�?
FIGURE 31—Force Applied to Prevent Sliding of Body (Force Not Parallel to the Supporting Surface)
Engineering Mechanics, Part 274
Solution: In this problem, the force that tends to cause the body to slide is the component X of the weight of the body. As in example problem 1, its magnitude is X = 140 sin 40° = 90.0 lb. However, since the line of action of the force P� is not parallel to the supporting surface, the normal pressure N is not equal to the component Y of the weight of the body. When we consider forces perpendicular to the inclined sur- face or parallel to the reference axis Y�Y� in Figure 31, we must include the component of the force P� in that direction. The magnitude of this component is P�y� = P� sin 30° = 0.500 P�. The magnitude of the component Y of the weight of the body is Y = 140 cos 40° = 107.2 lb. The equation for forces parallel to the axis Y�Y� then becomes:
�Fy� = 0 = – Y + P�y� + N = – 107.2 + 0.5P� + N
Hence, N = 107.2 – 0.5P�
The greatest magnitude of the friction force F that can be developed to resist sliding is
Fm = �N = 0.4 � (107.2 – 0.5P�) = 42.9 – 0.2P�
When the force P� is applied, sliding is resisted only by the component of this force that acts parallel to the supporting surface or parallel to the axis X�X�. The magnitude of this component is P�x� = P� cos 30° = 0.866 P�. When we consider the forces parallel to the axis X�X�, we can write the following equation:
�Fx� = 0 = – X + Fm + P�x�
= – 90.0 + 42.9 – 0.2P� + 0.866P�
= – 47.1 + 0.666P�
or P� = 70.7 lb
Hence, the required magnitude of P� is at least 70.7 lb.
You should note that the force P� must be greater than the force P computed in example problem 1 for the following reasons: The component of P� that acts perpendicular to the supporting surface reduces the maximum friction force that can be developed; and only the component of P� that is paral- lel to the supporting surface is effective in preventing sliding.
Engineering Mechanics, Part 2 75
Force Required to Move Body upward on Inclined Surface
36 It may be required to determine the force that will cause a body actually to slide upward along an
inclined plane surface. In a problem of this type, the proce- dure is the same regardless of whether the angle of inclination to the horizontal is less than, equal to, or greater than the angle of repose. Obviously, the applied force must have a component that is parallel to the inclined surface and acts upward along its line of action. Also, since the body tends to move upward along the surface, the friction force (which tends to resist movement) will act downward. The applied force may or may not be parallel to the supporting surface. Typical con- ditions are represented in Figure 32, where the applied force P is parallel to the inclined supporting surface. The forces that resist the upward slide of the body along the inclined surface are the component X of the weight that acts parallel to this surface and the friction force F that would be developed under the particular conditions. The magnitudes of these forces would be X = W sin A and F = �N = �Y = �W cos A.
When the applied force P is parallel to the inclined surface and when we consider forces that are parallel to that surface or parallel to the reference axis X�X� in Figure 32, we can write the following equation:
�Fx� = 0 = – X – F + P = – W sin A – � W cos A + P
Hence, P = W (sin A + � cos A) (1) FIGURE 32—Force Required to Slide Body upward on Inclined Surface (Force Parallel to Surface)
Engineering Mechanics, Part 276
If the applied force is not parallel to the inclined supporting surface, it is necessary to consider the components of the applied force in the manner described in example problem 2 in Article 35. For instance, the position of the line of action of the applied force may be as indicated for the force P� in Figure 33. In this case, the magnitudes of the components X and Y of the weight of the body are X = W sin A and Y = W cos A.
The components of the applied force P� that act parallel to the reference axes Y�Y� and X�X� in Figure 33 are P�y� = P� sin M and P�x� = P� cos M. When we consider the forces parallel to the axis Y�Y�, we can write the following equation:
�Fy� = 0 = – Y + P�y� + N Hence, N = Y – P�y� = W cos A – P� sin M
When we consider the forces parallel to the axis X�X�, the equation is
�Fx� = 0 = – X – F + P�x�
= – W sin A – � (W cos A – P� sin M) + P� cos M
When this equation is solved for P�, the result is
(2)P W A A M M
� = + +
(sin cos ) cos sin
μ μ
FIGURE 33—Force Required to Slide Body upward on Inclined Surface (Force Not Parallel to Surface)
Engineering Mechanics, Part 2 77
Another possible position of the line of action of the applied force is indicated in Figure 34, where the body is being pushed by a horizontal force P��. The magnitudes of the components of the weight of the body that are parallel and perpendicular to the inclined supporting surface are X = W sin A and Y = W cos A. Also, the magnitudes of the components of the applied force P�� that are parallel to the reference axes Y�Y�
and X�X� are P��y� = P�� sin A and P��x� = P�� cos A. When we consider the forces parallel to the axis Y�Y�, we can write the following equation:
�Fy� = 0 = – Y – P��y� + N Hence, N = Y + P��y� = W cos A – P�� sin A
When we consider the forces parallel to the axis X�X�, the equation is
�Fx� = 0 = – X – F + P��x�
= – W sin A – �(W cos A + P�� sin A) + P�� cos A
When this equation is solved for P��, the result is
(3)P W A A A M
� = + −
(sin cos ) cos sin
μ μ
FIGURE 34—Force Required to Slide Body upward on Inclined Surface (Force Horizontal)
Engineering Mechanics, Part 278
Practice Problems 5 1. A body that weighs 50 lb is placed in contact with an inclined plane surface. The coefficient of
friction between the surfaces in contact is 0.25. If the angle between the inclined surface and a horizontal reference plane is 28°, will the body remain at rest on the inclined surface or will it slide down the surface?
2. What would be the angle of repose for the body and the material used for the surface in problem 1?
3. If the body in problem 1 is to be prevented from sliding by an applied force whose line of action is parallel to the inclined supporting surface, what is the least permissible magnitude of the force?
4. If the conditions are similar to those indicated in Figure 31 and the body in problem 1 is prevented from sliding by a force P� whose line of action is located so that the angle M is 20°, what would be the least permissible magnitude of the force P�?
5. If the body in problem 1 is to be slid upward along the inclined surface by a force whose line of action is parallel to the inclined surface, what would be the least permissible magnitude of the force?
6. If the body in problem 1 is to be slid upward along the inclined surface by a force whose line of action is located as indicated in Figure 33 and the angle M is 20°, what would be the least permissible magnitude of the force?
7. If the body in problem 1 is to be pushed upward along the inclined surface by a horizontal force as indicated in Figure 34, what would be the least permissible magnitude of the force?
Check your answers with those on page 82.
Self-Check 1 1. Yes Article 1
2. The center of gravity of the body Article 1
3. The weight of the body Article 2
4. The weight of the body, the weight of the rod itself, and the reaction of the fixture in the ceiling Article 3,
Figure 1
5. The weight of the supported body, the force exerted by the lower fixture on the horizontal bar, and the force exerted by the upper fixture on the inclined bar Article 3,
Figure 1
6. The horizontal and vertical components of the force exerted by the upper fixture on the inclined bar and the force exerted on the upper fixture by the inclined bar Article 6,
Figure 3
7. Equations should be written to express the relations between the forces. Article 7
8. �Fx = 0 and �Fy = 0 Article 8
9. �M = 0 Article 8
Self-Check 2 1. Friction force Article 23
2. The wheel flattens out slightly and sinks into the supporting surface to a slight extent. Article 23
3. No. The friction force would become greater as the applied force is increased. Article 24
79
A n s w e r s
A n s w e r s
4. Less, because the friction force is greater when the surfaces in contact are rough Article 25
5. 60 lb. The friction force is not affected by the area of the surfaces in contact. Article 25
6. Since 10/30 = 1/3, it follows that F/15 = 1/3 and F = 5 lb Article 25
7. Polishing the surfaces or providing lubrication between them Article 25
8. Greater, because the force resisting movement of a body is greatest just before the body starts to slide Article 27
Self-Check Answers80
Practice Problems 1 1. R1 = 67 lb toward the left; R2 = 120 lb toward the right
and upward
2. R1 = 5507 lb toward the right and upward; R2 = 1917 lb toward the left and downward
Practice Problems 2 1. F4 = 40,540 lb toward intersection point; F5 = 17,800 lb
away from intersection point
2. F3 = 4240 lb toward intersection point; F4 = 4830 lb toward intersection point
Practice Problems 3 1. R1 = 5900 lb; R2 = 6100 lb
2. a) R1 = 3500 lb #
b) R2 = 4110 lb 50°40�
3. a) 68°40�
b) 3760 lb c) 3420 lb
4. a) 1853 lb. #
b) 520 lb 77°10�
81
A n s w e r s
A n s w e r s
Practice Problems 4 1. 0.25
2. 0.38 ft
3. a) 141 lb; b) 349 lb; c) 1.10 ft ! #
4. a) 129 lb; b) 453 lb; c) 0.50 ft " #
Practice Problems 5 1. The body will slide.
2. 14°00�
3. 12.5 lb
4. 14.6 lb
5. 34.5 lb
6. 33.6 lb
7. 45 lb
Practice Problems Answers82
83
1. As indicated in Examination Figure 1, a wooden box containing several objects is supported on a horizontal surface. The total weight of the box and its contents is 150 lb. The horizontal force P required to start the box sliding along this surface is 60 lb. If additional objects weighing 50 lb were placed in the box, the magnitude of the horizontal force required to slide the box along the supporting surface would be
A. 60 lb. C. 80 lb. B. 70 lb. D. 90 lb.
E x a m in a tio
n E x a m in a tio
n Engineering Mechanics, Part 2
When you feel confident that you have mastered the material in this study unit, go to http://www.takeexamsonline.com and submit your answers online. If you don’t have access to the Internet, you can phone in or mail in your exam. Submit your answers for this examination as soon as you complete it. Do not wait until another examination is ready.
Questions 1–20: Select the one best answer to each question.
EXAMINATION NUMBER
28603701 Whichever method you use in submitting your exam
answers to the school, you must use the number above.
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Examination Figure 1
Examination84
2. A body that weighs 50 lb is placed in contact with an inclined plane surface. The coefficient of friction between the surfaces is 0.25 and the angle between the inclined surface and the horizontal reference plane is 28°. In order to keep the body from sliding, by using an applied force whose line of action is parallel to the inclined supporting surface, what is the least permissible magnitude of force?
A. 6.3 lb C. 14 lb B. 12.5 lb D. 50 lb
3. A sign is suspended from a supporting frame in the manner indicated in Examination Figure 2. The frame consists of two vertical posts and a horizontal beam, with the sign suspended from this beam by means of two vertical bars. The total weight of the sign is 400 lb and the weight of each of the bars is 50 lb. If each of these bars carries one- half of the total weight of the sign, the force exerted by the left bar on the fixture is
A. 150 lb. C. 225 lb. B. 200 lb. D. 250 lb.
Examination Figure 2
Examination 85
4. Examination Figure 3 gives the dimensions of the parts of a bracket that supports a body with the aid of a vertical cable. The bracket is composed of a horizontal bar and an inclined bar, and these bars are rigidly fastened to the wall of the building by means of upper and lower fixtures. The weight of the body is 3000 lb, and the weights of the cable and the bars may be neglected. The magnitude of the reaction R1 exerted by the lower fixture on the horizontal bar is
A. 1500 lb. C. 2700 lb. B. 2100 lb. D. 3350 lb.
Examination Figure 3
5. What equation is generally used when it is desirable to consider the resultant moment of the forces acting on a body at rest?
A. �F1 = 0 C. �M0 = 0 B. �F0 = 1 D. �Z0 = 1
Examination86
6. Examination Figure 4 represents a balanced system consisting of four concurrent forces F1, F2, F3, and F4. The positions of the lines of action of the forces are as shown. The line of action of F1 is vertical, and that of F4 is horizontal. The diagram also includes the magnitudes of the forces F1 and F2 and the directions of these two forces along their lines of action. The magnitude of the force F3 and its direction along its line of action are
A. 19,780 lb away from P. C. 30,200 lb away from P. B. 19,780 lb toward P. D. 30,200 lb toward P.
Examination Figure 4
7. In Examination Figure 4, the magnitude of the force F4 and its direction along its line of action are
A. 24,400 lb toward P. C. 28,800 lb toward P. B. 24,400 lb away from P. D. 28,800 lb away from P.
Examination 87
8. As indicated in Examination Figure 5, a body having the shape of a rectangular prism is supported on a horizontal surface. The weight W of the body is 150 lb, and the body is acted upon by a horizontal force P whose magnitude and location are as shown. If the magnitude of the friction force is just sufficient to prevent sliding of the body along the supporting surface, the horizontal distance between the line of action of the weight of the body and the line of action of the total vertical force N exerted on the body by the supporting surface would be
A. 1.00 ft. C. 1.50 ft. B. 1.20 ft. D. 2.00 ft.
Examination Figure 5
9. The condition that must exist when a body is at rest is that the magnitude of the resultant of forces acting on the body must be
A. 0. C. dependent upon the incline. B. 20 percent. D. external.
10. When solving a practical problem where the magnitudes of two forces are unknown, your calculations can be simplified if one of the reference axes coincides with the line of action of one of the unknown forces and the other reference axis is
A. equal to zero. B. horizontal to that line of action. C. perpendicular to that line of action. D. vertical to that line of action.
Examination88
11. The horizontal beam represented in Examination Figure 6 carries three loads P1, P2, and P3, which act vertically downward and have the magnitudes and positions indicated. The beam rests on two supports at A and B, which are so located that the horizontal distance between the lines of action of the reactions R1 and R2 is 24 ft. The magnitude of the reaction R1 is
A. 14,330 lb. C. 15,220 lb. B. 14,780 lb. D. 15,670 lb.
Examination Figure 6
Examination 89
12. As indicated in Examination Figure 7, a body that weighs 200 lb is to be supported on a plane surface, which is inclined so that the angle between a horizontal reference plane and the surface is 35°. If the coefficient of friction for the materials used for the body and the surface is 0.3 and the body is to be prevented from sliding down the surface by a force P whose line of action is parallel to the surface, the least magnitude of the force P (rounded off to the nearest pound) should be
A. 49 lb. C. 82 lb. B. 66 lb. D. 115 lb.
Examination Figure 7
13. What will be the result when the applied horizontal force and the friction force form a couple, the lines of action are parallel, and the magnitudes are equal?
A. They will act in the same direction. C. They will create a mutual force. B. They will cause the body to rotate. D. They will create friction.
14. A body that weighs 50 lb is placed in contact with an inclined plane surface. The coefficient of friction between the surfaces in contact is 0.25. If the angle between the inclined surface and a horizontal reference plane is 28°, and if the body is to be prevented from sliding by an applied force whose line of action is parallel to the inclined supporting surface, what is the least permissible magnitude of the force?
A. 1.25 lb C. 125 lb B. 12.5 lb D. 0.125 lb
Examination90
15. As indicated in Examination Figure 8, a body that weighs 80 lb is to be pushed upward along a plane surface that is inclined so that the angle between a horizontal reference line and the surface is 30°. If the coefficient of friction for the materials of the body and the inclined surface is 0.2, the magnitude of the smallest horizontal force P that would start the body moving (rounded off to the nearest pound) would be
A. 34 lb. C. 56 lb. B. 52 lb. D. 70 lb.
Examination Figure 8
16. Assume that the conditions are the same as in Examination Figure 8 and question 15, except that the line of action of the force applied to push the body upward along the inclined surface is to be parallel to that surface instead of horizontal. The magnitude of the smallest inclined force that would start the body moving would be
A. 54 lb. C. 64 lb. B. 58 lb. D. 70 lb.
Examination 91
17. Examination Figure 9 represents a beam that is supported at A and B. The characteris- tics of the loads P1 and P2 that are applied to the beam and the positions of their lines of action are as indicated. If it is assumed that the line of action of the reaction R2 at B will be vertical, the magnitude of that reaction would be
A. 3480 lb. C. 3760 lb. B. 3620 lb. D. 3900 lb.
Examination Figure 9
18. For the beam in Examination Figure 9, the magnitude of the inclined reaction R1 at A would be
A. 3610 lb. C. 4000 lb. B. 3800 lb. D. 4200 lb.
Examination92
19. As indicated in Examination Figure 10, a body that is composed of the three parts AB, BC, and CD (which are rigidly connected to one another) is supported at the points A and B and is subjected to two loads P1 and P2 whose characteristics and locations are as shown. If it is assumed that the line of action of the reaction R1 at B is vertical, the magnitude of that reaction would be
A. 498 lb. C. 658 lb. B. 578 lb. D. 738 lb.
Examination Figure 10
20. A body that weighs 50 lb is placed in contact with an inclined plane surface. The coefficient of friction between the surfaces in contact is 0.25. If the angle between the inclined surface and a horizontal reference plane is 28°, what would be the angle of repose for the body and the material used for the surface?
A. 14° 00� C. 20° 00�
B. 17° 00� D. 30° 00�
