7 9 7
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–49.
The uniform crate has a mass of 150 kg. If the coefficient of
static friction between the crate and the floor is ms = 0.2,
determine whether the 85-kg man can move the crate. The
coefficient of static friction between his shoes and the floor
is m′s = 0.4. Assume the man only exerts a horizontal force
on the crate.
Ans:
He is able to move the crate.
SOLUTION
Equations of Equilibrium. Referring to the FBD of the crate shown in Fig. a,
S+ ΣFx = 0; P - FC = 0 (1)
+ c ΣFy = 0; NC - 150(9.81) = 0 NC = 1471.5 N
a + ΣMO = 0; 150(9.81)x - P(1.6) = 0 (2)
Also, from the FBD of the man, Fig. b,
+ c ΣFy = 0; Nm - 85(9.81) = 0 Nm = 833.85 N
S+ ΣFx = 0; Fm - P = 0 (3)
Friction. Assuming that the crate slips before tipping. Then
FC = ms NC = 0.2(1471.5) = 294.3 N
Solving Eqs. (1) to (3) using this result,
Fm = P = 294.3 N x = 0.32 m
Since x < 0.6 m, the crate indeed slips before tipping as assumed. Also
since Fm 6 (Fm)max = m′s Nm = 0.4(833.85) = 333.54 N, the man will not slip.
Therefore, he is able to move the crate.
2.4 m
1.2 m
1.6 m
8 1 0
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–62.
If P = 250 N, determine the required minimum compression
in the spring so that the wedge will not move to the right.
Neglect the weight of A and B. The coefficient of static
friction for all contacting surfaces is Neglect
friction at the rollers.
SOLUTION
Free-Body Diagram: The spring force acting on the cylinder is .
Since it is required that the wedge is on the verge to slide to the right, the frictional
force must act to the left on the top and bottom surfaces of the wedge and their
magnitude can be determined using friction formula.
Equations of Equilibrium: Referring to the FBD of the cylinder, Fig. a,
Referring to the FBD of the wedge shown in Fig. b,
Ans.x = 0.01830 m = 18.3 mm
= 0- 316.233(103)x4sin 10°
250 - 5.25(103)x - 0.35316.233(103)x4cos 10°©Fx = 0;:+
N2 = 16.233(103)x
N2 cos 10° - 0.35N2 sin 10° - 15(103)x = 0+ c ©Fy = 0;
Thus, (Ff)1 = 0.35315(103)x4 = 5.25(103)x
N1 = 15(103)xN1 - 15(103)x = 0+ c ©Fy = 0;
(Ff)2 = 0.35N2(Ff)1 = mN1 = 0.35N1
Fsp = kx = 15(103)x
ms = 0.35. k ! 15 kN/m
A
P
B
10"
Ans:
x = 18.3 mm
8 1 0
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–62.
If P = 250 N, determine the required minimum compression
in the spring so that the wedge will not move to the right.
Neglect the weight of A and B. The coefficient of static
friction for all contacting surfaces is Neglect
friction at the rollers.
SOLUTION
Free-Body Diagram: The spring force acting on the cylinder is .
Since it is required that the wedge is on the verge to slide to the right, the frictional
force must act to the left on the top and bottom surfaces of the wedge and their
magnitude can be determined using friction formula.
Equations of Equilibrium: Referring to the FBD of the cylinder, Fig. a,
Referring to the FBD of the wedge shown in Fig. b,
Ans.x = 0.01830 m = 18.3 mm
= 0- 316.233(103)x4sin 10°
250 - 5.25(103)x - 0.35316.233(103)x4cos 10°©Fx = 0;:+
N2 = 16.233(103)x
N2 cos 10° - 0.35N2 sin 10° - 15(103)x = 0+ c ©Fy = 0;
Thus, (Ff)1 = 0.35315(103)x4 = 5.25(103)x
N1 = 15(103)xN1 - 15(103)x = 0+ c ©Fy = 0;
(Ff)2 = 0.35N2(Ff)1 = mN1 = 0.35N1
Fsp = kx = 15(103)x
ms = 0.35. k ! 15 kN/m
A
P
B
10"
Ans:
x = 18.3 mm
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