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Two trunks sit side by side on the floor. The large trunk (52kg) is to the left of the smaller trunk (34kg). A person pushes on the larger trunk horizontally toward the right. The coefficient of static friction between the trunks and the floor is 0.35.

a) determine the magnitude of the maximum force the person can exert without moving either trunk. *I got 290N.
b) Calculate the force the larger trunk exerts on the smaller trunk.

Update : c) Would either answer change if the person pushed in the opposite direction on the smaller trunk. Explain your reasoning.

User Homebrand
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1 Answer

5 votes

Answer:

Part a)


F = 290.3 N

Part b)


F = 116.7 N

Part c)

answer for part a) will not change

But here the answer for part b) will change because now the contact force is friction force of large trunk


F = 178.5 N

Step-by-step explanation:

Part a)

The maximum applied force so that both the trunks will not slide will be equal to the total friction force on the two trunks

So we have


F = \mu_1m_1g + \mu_2m_2g


F = 0.35(52 + 34) * 9.81


F = 290.3 N

Part b)

Now the force exerted by large trunk on the smaller trunk is given as


F = F_f


F = \mu m_2 g


F = 0.35 (34)(9.81)


F = 116.7 N

Part c)

If we push smaller trunk from other side then the net force to slide the two trunks will be same

so answer for part a) will not change

But here the answer for part b) will change because now the contact force is friction force of large trunk


F = 0.35 * 52 (9.81)


F = 178.5 N

User Sho Gondo
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9.1k points