1. The problem statement, all variables and given/known data We begin by considering the forces exerted on a ring that is connected to a spring and that is free to slide along a rod TOP VIEW: The diagram is top view Diagram Description (sorry don't have a pic): A pulse traveling towards a free end (has NOT reached the boundary yet), moving to the right. Similar Pic Assume the ring is massless and that the rod is frictionless i.) What is the net force on the ring? (Hint: Consider what happens to the net force on an object as its mass approaches zero) ii) Does the net force on a massless ring depend on the acceleration of the ring? Explain iii) What is the magnitude of the gravitational force exerted on the ring? (Recall the ring is massless.) iv) Does the force exerted on the ring by the rod have a component that is parallel to the rod? Explain (Hint: Recall the assumptions made above) 2. Relevant equations F = ma F = kx (Hooke's Law) 3. The attempt at a solution Solved? i) Since the ring is initially at equilibrium along the rod, we assume the net force is equal to zero. Horizontal displacement is caused by the spiring force and thus the net force once the pulse reaches the boundary, the ring accelerates upward by the spring force exerted by the spring ii) No. The acceleration of spring is entirely due to the spring force exerted on the ring by the spring. Therefore the net force is entirely due to the spring force and does not depend on the acceleration of the ring. iii) Since the ring is at equilibrium before the pulse reaches the boundary, we can say the contact force between the rod and the ring has a equal magnitude to the gravitational force. Thus gravity has a magnitude equal to the normal force (contact force) of the rod and the ring (since ring is considered massless). iv) No. The force exerted by the rod on the ring is a normal force that is vertical and thus orthogonal to the rod. Next they ask for a free body diagram when the ring is farthest from the equilibrium point For this, I assume that the vertical forces are equal in magnitude and opposite in direction. For the horizontal force, its a spring force OPPOSITE in direction of motion. Does this seem right.