A very long, mass-less rope hangs over a mass-less, friction-less pulley. There
ID: 2210097 • Letter: A
Question
A very long, mass-less rope hangs over a mass-less, friction-less pulley. There is a mokey on one end of the rope of mass mM and on the other there is a bunch of bananas of mass mB. The monkey and the bananas are both initially at rest. When the monkey sees the bananas it begins to climb the rope with constant acceleration aM relative to the ground. Analyze the motion of the bananas. Under what conditions can the monkey reach the bananas? As the rope is very long, you need not worry about what happens if the bananas or monkey reach the height of the pulley.Explanation / Answer
Before we do anything else, let’s understand what forces are acting on the two masses in this problem. The free–body diagrams are shown in Fig. 4.10. The monkey holds onto the rope so it exerts an upward force of magnitude T, where T is the tension in the rope. Gravity pulls down on the monkey with a force of magnitude mg, where m is the mass of the monkey. These are all the forces. Note that they will not cancel since the problem talks about the monkey having an acceleration and so the net force on the monkey will not be zero. The forces acting on the box are also shown. Gravity pulls downward on the box with a force of magnitude Mg, M being the mass of the box. The rope pulls upward with a force T, If the box is resting on the ground, the ground will be pushing upward with some force Fground. (Here, the ground cannot pull downward.) However when the box is not touching the ground then Fground will be zero. In the first part of the problem, the monkey is moving along the rope. It is not stuck to any point of the rope, so there is no obvious relation between the acceleration of the monkey and the acceleration of the box. Suppose we let ay,monkey be the vertical acceleration of the monkey and ay,box be the vertical acceleration of the box. Then from our free–body diagrams, Newton’s Second Law gives the acceleration of the monkey: T - mg = may,monkey When the box is on the ground its acceleration is zero and then T + Fgr = mg. But when the box is off then ground then we have: T - Mg = Mabox (Box off the ground) In the first part of the problem we are solving for the condition that the monkey climbs just barely fast enough for the box to be lifted off the ground. If so, then the ground would exert no force but the net force on the box would be so small as to be virtually zero; the box has a very, very tiny acceleration upwards. From this we know: T - Mg = 0 =) T = Mg and substituting this result into the first equation gives Mg - mg = mamonkey