Two trunks sit side by side on the floor. The large trunk (52 kg) is to the left of the smaller trunk (34 kg). 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.

b) Calculate the force the larger trunk exerts on the smaller trunk.

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

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 ;

Answered by aristocles | 2024-06-12

(a)The maximum force the person can exert without moving either trunk is approximately 295 N, (b) the force the larger trunk exerts on the smaller trunk is about 117 N. (c) Pushing in the opposite direction on the smaller trunk would not change these results.
Let's analyze the problem step-by-step. We have two trunks with the following masses:

Large trunk: 52 kg
Small trunk: 34 kg

The coefficient of static friction ( μ ₛ) between the trunks and the floor is 0.35. To find the maximum force the person can exert without moving either trunk, we need to calculate the combined normal force (N) and the maximum static friction force (fₛ max).
a) Maximum Force (fₛ max)
Total mass (m) = 52 kg + 34 kg = 86 kg
Gravitational acceleration (g) = 9.8 m/s²
Total normal force (N) = m * g = 86 kg * 9.8 m/s² = 842.8 N
Maximum static friction (fₛ max) = μₛ * N = 0.35 * 842.8 N ≈ 294.98 N
So, the maximum force the person can exert without moving either trunk is approximately 295 N.
b) Force Exerted on the Smaller Trunk
Since the trunks are side by side and the larger trunk is being pushed, the force between them is equal to the maximum static friction force required to prevent the smaller trunk from moving.
Force on smaller trunk = (mass of smaller trunk) * g * μₛ = 34 kg * 9.8 m/s² * 0.35 ≈ 116.62 N
Therefore, the larger trunk exerts a force of approximately 117 N on the smaller trunk.
c) Effect of Pushing in the Opposite Direction
If the person were to push on the smaller trunk in the opposite direction, the results would not change. The combined mass and the coefficient of static friction remain the same, thus the forces required to initiate movement remain unchanged.
Therefore, the maximum force the person can exert without moving either trunk is approximately 295 N, and the force the larger trunk exerts on the smaller trunk is about 117 N

Answered by VeraMindy | 2024-06-27

The maximum force the person can exert on the trunks without moving them is approximately 290.3 N. The force exerted by the large trunk on the smaller trunk is approximately 116.7 N when pushed from the left. If pushing from the right, the force would change to about 178.5 N due to altered friction dynamics.
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Answered by aristocles | 2024-10-12