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Purpose

Purpose of this lab to get understanding of real motion in which friction plays a role. Get understanding with the coefficient µs and µk.

Pre-lab

There is no pre lab

Solutions to the problems mentioned.

Run the simulation with the 200-kg file cabinet. Try different applied forces until you determine the maximum applied force where the cabinet remains at rest. Record this number, which is also the maximum value of the static friction force.

Above 590 N force the cabinet start moving so the maximum applied force foe which the cabinet remain at rest is 590N.

Use the above result and the formula for maximum static friction to calculate the coefficient of static friction. Show your work.

Fs = µsN

590 = µs * 200*9.8

µs = 0.3

Complete the following table using data directly from the simulation and your knowledge of physics. Be sure to hit the “Clear” button between each trial:

Use the above results and the formula for kinetic friction to determine the coefficient of kinetic friction. Show your work.

Fk = µkN

392 = µk 1960

µk = 0.2

The free-body diagram from the simulation is incomplete as it is missing sources and receivers for each of the forces. It is also missing a specific type for the “applied force”. Draw or describe the type, source, receiver, and direction for each of the four forces.

· Weight of body is acting downward which is equal to mass * gravitational acceleration

· A reactive force of weight is acting in upward direction and is equal to weight in magnitude

· Applied force and force of frictions are equal in magnitude but opposite in direction unless the body start move as soon as body start moving frictional force reduces to a number and remain constant throughout the motion.

Under “More Controls”, change the “Object Position” to a positive number. Set the ramp angle to 5 degrees. Click the play button and note whether or not the block slides down the ramp. Note that the coefficients of friction are given

Mass = 100 kg

µs = 0.5

µk = 0.3

Repeat the previous step multiple times, increasing the angle by several degrees each time. What is the largest angle for which the block will remain at rest? Is this angle consistent with the formula μ​s = tan (Ø)

Ø = 26.5˚

Tan(26.5) = 0.49 which is approximately equal to 0.5

Use the formula μ​ k​ = tanθ to predict the angle at which the block will slide down the ramp at constant velocity. Show your work below.

µk = 0.3

θ = tan-1(0.3)

θ = 16.7˚

Set the angle to that which you calculated in the previous step. Set the position to 8.9 meters.

To give the block a nudge, set F​ Applied​ to -200 N. Then click on play and then pause quick succession.

Set F​ Applied​ to 0 N and click play. What is the net force?

Net force is zero which mean acceleration is zero and velocity id constant.

Repeat the previous three steps with bigger and smaller angles. Record your observations of force and motion.

For angle 10˚

For angle = 20˚

Above diagram shows that as the angle get smaller than 16.7˚ a large initial force is required to move the object with constant velocity. And for angle greater than 16.7 a small force (10N) or a simple push can move the object with constant velocity.

Conclusion

In real time picture a frictional force is present between the body and surface. Motion of body is only possible when you exceed the applied force from the frictional force. A number of measure can be taken to reduce frictional force as it will reduce our input force.