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Pulleys are somewhat tricky, but with some practice, problems including pulleys become simpler. There are two underlying principles:


The tension in a single rope is the same all along the rope, even if it goes over and under pulleys.
If a rope is pulled at a constant rate by a hand, the work done by the hand on the rope is the same as the work done by the rope on some load.


Let's look at some examples.


A person is hanging in air by grabbing the two ends of a rope which is draped around a pulley. If the person is 60 kg, what is the tension in the rope?



First, we DRAW A DIAGRAM (Figure 9-11) showing the forces on the man. The tension on the two sides of the rope is the same, so we can call it T. The person is not accelerating, so the forces add to zero, giving


T + T – mg = 0


T = mg/2


= (60 kg)(10 m/s2)/2


= 300 N


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Figure 9-11



A rope, one end of which is connected to the ceiling, passes through a pulley and then goes up, so that an upward tension is maintained. A mass 30 kg is hung on the pulley. What is the tension in the rope? (See Figure 9-12.)


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Figure 9-12


There are two ways to do this problem. One is to realize that this is essentially the same as Example 1, so that T = 1/2 mg = 150 N. (See Figure 9-13.)


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Figure 9-13

The other way is to imagine pulling up on the rope 1 meter. A bit of study of the diagram will show that the mass will rise 0.5 meters, since 0.5 meters of rope will be pulled from each side of the pulley. The work done by T must be the same as the work done on the mass. The work done by the rope is W1 = FΔxcosφ = T (1 m). The work done on the mass is the change in potential energy W2 = mgΔh = mg(0.5 m). Thus

W1 = W2,
T(1 m) = mg(0.5 m),
T = 1/2 mg = 150 N.



A rope has one end connected to the ceiling. It loops through a pulley with a downward weight of 500 N, goes up to the ceiling where it loops over a second pulley and connects to a mass m. If everything is in equilibrium, what is m? (See Figure 9-14.)


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Figure 9-14


This looks different from the previous problem, but in fact it is essentially the same. The tension T which pulls up on mass m is numerically the same as the tension T pulling up on both sides of the first pulley. So we have


T = 1/2 (500 N) = 250 N


T = mg


So m = 25 kg

Another way to obtain the above equation is to DRAW A DIAGRAM showing the forces on both masses (Figure 9-15).


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Figure 9-15

In this chapter we explored the concept of energy. Whenever you read about a force and a distance through which the force acts, you should think immediately of work W = FΔxcosφ. This will be the key to answering some of the questions, even if no numbers are involved.

If a net force acts on an object, the kinetic energy EK = 1/2 mv2 of the object is changed according to Wtot = ΔEk. The total work gives the size of the energy flow into an object. Another form of energy is gravitational potential energy given by EP = mgh. It is important to keep track of the energy flow because energy is conserved. That is, energy cannot be created from nothing or destroyed, but it can be transferred from one form to another.

The rate at which energy is transformed is called power P = ΔEt.


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