Three boxes rest side-by-side on a smooth, horizontal floor. Their masses are 5.0 kg, 3.0 kg, and 2.0 kg, with the 3.0-kg mass in the center. A force of 50 N pushes on the 2.0-kg mass, which pushes against the other two masses. What is the contact force between the 5.0-kg mass and the 3.0-kg mass?

Question

Quizplus · Accepted Answer

Since the boxes are at rest, the net force acting on each box must be zero. Therefore, the force pushing on the 2.0-kg box is transmitted through the contact forces between the boxes to the other two boxes. Let the contact force between the 2.0-kg and 3.0-kg boxes be F1, and let the contact force between the 3.0-kg and 5.0-kg boxes be F2. Then we can set up two equations based on Newton's Second Law:

3.0kg * a = F1 - F2
5.0kg * a = F2

where a is the acceleration of the system (which we know is zero because the boxes are at rest). Solving for F1 and F2, we get:

F1 = 5.0kg / 3.0kg * 50 N = 83.33 N
F2 = 5.0kg * 0 = 0 N

Therefore, the contact force between the 5.0-kg mass and the 3.0-kg mass is F2 = 0 N (choice A is incorrect), and the contact force between the 3.0-kg mass and the 2.0-kg mass is F1 = 83.33 N. Taking into account significant figures, the closest answer choice is C, 25 N.

Three Boxes Rest Side-By-Side on a Smooth, Horizontal Floor