Question

Two blocks of mass m1 = 8.0 kg and m3 = 6.3 kg, connected by a rod of mass m2 = 1.8 kg, are sitting on a very low-friction surface, and you push to the left on the right block (mass m1) with a constant force of (-30 , 0, 0)N. See figure below. (As usual, +x is to the right and +y is up. Ignore effects due to friction between the objects)

What is the force F3 exerted by the rod on the block of mass m3?

Answer 1

The force exerted by the rod on the block of mass m3 is 30 N in the positive x-direction (to the right). Therefore, the force F3 exerted by the rod on the block of mass m3 is 30 N in the positive x-direction (to the right).

Step-by-step explanation:

In this case, the force exerted by the rod on the block of mass m3 is equal in magnitude but opposite in direction to the force exerted by the block of mass m1 on the rod.

From Newton's third law of motion, these forces are equal and opposite.

Therefore, the force F3 exerted by the rod on the block of mass m3 is 30 N in the positive x-direction (to the right).

Answer 2

Since the three bodies are connected properly and the force is exerted at the end of m1, the force is channeled to the other bodies without any losses. If the force on m1 is 30 N, then the force F3 exerted by the rod on the block of mass m3 is also 30 N.