Pulleys are simple machines that consist of a rope that slides around a disk, called a block. Their main function is to change the direction of the tension force in a rope. The pulley systems that appear on SAT II Physics almost always consist of idealized, massless and frictionless pulleys, and idealized ropes that are massless and that don’t stretch. These somewhat unrealistic parameters mean that:

The one exception to this rule is the occasional problem you might find regarding the torque applied to a pulley block. In such a problem, you will have to take the pulley’s mass into account. We’ll deal with this special case in Chapter 7, when we look at torque.

We use pulleys to lift objects because they reduce the amount of force we need to exert. For example, say that you are applying force F to the mass in the figure above. How does F compare to the force you would have to exert in the absence of a pulley?

To lift mass m at a constant velocity without a pulley, you would have to apply a force equal to the mass’s weight, or a force of mg upward. Using a pulley, the mass must still be lifted with a force of mg upward, but this force is distributed between the tension of the rope attached to the ceiling, T, and the tension of the rope gripped in your hand, F.

Because there are two ropes pulling the block, and hence the mass, upward, there are two equal upward forces, F and T. We know that the sum of these forces is equal to the gravitational force pulling the mass down, so F + T = 2F = mg or F = mg/2. Therefore, you need to pull with only one half the force you would have to use to lift mass m if there were no pulley.

The figure above represents a pulley system where masses m and M are connected by a rope over a massless and frictionless pulley. Note that M > m and both masses are at the same height above the ground. The system is initially held at rest, and is then released. We will learn to calculate the acceleration of the masses, the velocity of mass m when it moves a distance h, and the work done by the tension force on mass m as it moves a distance h.

Before we start calculating values for acceleration, velocity, and work, let’s go through the three steps for problem solving:

Note that the tension force, T, on each of the blocks is of the same magnitude. In any nonstretching rope (the only kind of rope you’ll encounter on SAT II Physics), the tension, as well as the velocity and acceleration, is the same at every point. Now, after preparing ourselves to understand the problem, we can begin answering some questions.

Because the acceleration of the rope is of the same magnitude at every point in the rope, the acceleration of the two masses will also be of equal magnitude. If we label the acceleration of mass m as a, then the acceleration of mass M is –a. Using Newton’s Second Law we find:

By subtracting the first equation from the second, we find (M – m)g = (M + m)a or a = (M – m)g/(M + m). Because M – m > 0, a is positive and mass m accelerates upward as anticipated. This result gives us a general formula for the acceleration of any pulley system with unequal masses, M and m. Remember, the acceleration is positive for m and negative for M, since m is moving up and M is going down.

We could solve this problem by plugging numbers into the kinematics equations, but as you can see, the formula for the acceleration of the pulleys is a bit unwieldy, so the kinematics equations may not be the best approach. Instead, we can tackle this problem in terms of energy. Because the masses in the pulley system are moving up and down, their movement corresponds with a change in gravitational potential energy. Because mechanical energy, E, is conserved, we know that any change in the potential energy, U, of the system will be accompanied by an equal but opposite change in the kinetic energy, KE, of the system.

Remember that since the system begins at rest, . As the masses move, mass M loses Mgh joules of potential energy, whereas mass m gains mgh joules of potential energy. Applying the law of conservation of mechanical energy, we find:

Mass m is moving in the positive y direction.

We admit it: the above formula is pretty scary to look at. But since SAT II Physics doesn’t allow calculators, you almost certainly will not have to calculate precise numbers for a mass’s velocity. It’s less important that you have this exact formula memorized, and more important that you understand the principle by which it was derived. You may find a question that involves a derivation of this or some related formula, so it’s good to have at least a rough understanding of the relationship between mass, displacement, and velocity in a pulley system.

Since the tension force, T, is in the same direction as the displacement, h, we know that the work done is equal to hT. But what is the magnitude of the tension force? We know that the sum of forces acting on m is T – mg which is equal to ma. Therefore, T = m(g – a). From the solution to question 1, we know that a = g(M – m)/(M + m), so substituting in for a, we get:

Now imagine that masses m and M are in the following arrangement:

Let’s assume that mass M has already begun to slide along the table, and its movement is opposed by the force of kinetic friction, , where is the coefficient of kinetic friction, and N is the normal force acting between the mass and the table. If the mention of friction and normal forces frightens you, you might want to flip back to Chapter 3 and do a little reviewing.

So let’s approach this problem with our handy three-step problem-solving method:

Given this information, can you calculate the acceleration of the masses? If you think analytically and don’t panic, you can. Since they are attached by a rope, we know that both masses have the same velocity, and hence the same acceleration, a. We also know the net force acting on both masses: the net force acting on mass M is , and the net force acting on mass m is T – mg. We can then apply Newton’s Second Law to both of the masses, giving us two equations involving a:

Adding the two equations, we find . Solving for a, we get:

Since m is moving downward, a must be negative. Therefore, .

It is highly unlikely that SAT II Physics will ask a question that involves remembering and then plugging numbers into an equation like this one. Remember: SAT II Physics places far less emphasis on math than your high school physics class. The test writers don’t want to test your ability to recall a formula or do some simple math. Rather, they want to determine whether you understand the formulas you’ve memorized. Here are some examples of the kinds of questions you might be asked regarding the pulley system in the free-body diagram above:

These examples are perhaps less demanding than a question that expects you to derive or recall a complex formula and then plug numbers into it, but they are still difficult questions. In fact, they are about as difficult as mechanics questions on SAT II Physics will get.