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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.

Adetermine the magnitude of the maximum force the person can exert without moving either trunk. *I got 290N.
BCalculate 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.

Verified Answer :

AUsing the equation, \(F_{r} = \mu \times N\)
Where Fr is friction in N, mu is the coefficient of friction and N is our weight.

For the big trunk we can work out the resistive force with:
\(F_r = 0.35 \times (52 \times 9.8) \\ \therefore F_r = 178.36N\)
The 52 times 9.8 is weight, which is it's mass times gravity.

We can do the same for the little trunk:
\(F_r = 0.35 \times (43\times9.8) \\ \therefore F_r = 147.49N\)

Now we can add together the 2 forces:
\(178.36 + 147.49 = 325.85N\)

This is the total amount of friction that both logs could withstand before they would start to move.

B) This is rather more simple than A. Since the big trunk resists 178.36N of the 325.85N of the total pushing force, it means that the big trunk is exerting a force the same amount of force as the force that the small trunk is resisting. This means the big trunk is exerting 147.49N.

C) The force that is being exerted on the small trunk from the big trunk if you pushed from the other side would be different. This is because the big trunk is pushing against the small trunk, in order for it not to move. This means it's still exerting a force on the small trunk.

Hope this is easy to understand, I'm not too sure how advanced your knowledge of forces are. 

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

\(F = 178.5 N\)

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

\(F = \mu_1m_1g + \mu_2m_2g\)

\(F = 0.35(52 + 34) \times 9.81\)

\(F = 290.3 N\)

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

\(F = 0.35 \times 52 (9.81)\)

\(F = 178.5 N\)

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