A 6 kg mass on a horizontal friction-free air track is accelerated by a string attached to a 115 kg mass hanging vertically from a pulley as shown. Compare the accelerations when the masses are interchanged, that is, for the case where the 115 kg mass dangles over the pulley, and then for the case where the 6 kg mass dangles over the pulley.

Question

anonymous · Answer

A good way to approach these problems is to draw a force diagram, and then find the sum of the forces. In either case, the box resting on the surface is affected by both gravity and normal force, which cancel out. Both boxes are equally affected by tension pulling in opposite directions, so those cancel out. The box hanging from the string is affected by gravity, which is the one remaining force in the system. Since the sum of the forces is equal to the force of gravity on the hanging box, and the sum of the forces is also equal to the sum of the masses times the total acceleration, the acceleration must be equal to the sum of the forces divided by the sum of the masses. In each case, multiply the mass of the hanging box (B) by the acceleration of gravity (9.8 m/s^2) to find the force of gravity of the box (GB), and then divide that number by the sum of the masses of the boxes (MA + MB) to find the acceleration.

In the case where the 115 kg box hangs, I found a = 9.31 m/s^2.
In the case where the 6 kg box hangs, I found a = 0.486 m/s^2

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