


Pulleys
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 rope slides without any resistance over the pulley, so that the pulley changes the direction of the tension force without changing its magnitude.
 You can apply the law of conservation of energy to the system without worrying about the energy of the rope and pulley.
 You don’t have to factor in the mass of the pulley or rope when calculating the effect of a force exerted on an object attached to a pulley system.
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.
The Purpose of Pulleys
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.
Standard Pulley Problem
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:
 Ask yourself how the system will move: From experience, we know that the heavy mass, M, will fall, lifting the smaller mass, m. Because the masses are connected, we know that the velocity of mass m is equal in magnitude to the velocity of mass M, but opposite in direction. Likewise, the acceleration of mass m is equal in magnitude to the acceleration of mass M, but opposite in direction.
 Choose a coordinate system: Some diagrams on SAT II Physics will provide a coordinate system for you. If they don’t, choose one that will simplify your calculations. In this case, let’s follow the standard convention of saying that up is the positive y direction and down is the negative y direction.
 Draw freebody diagrams: We know that this pulley system will accelerate when released, so we shouldn’t expect the net forces acting on the bodies in the system to be zero. Your freebody diagram should end up looking something like the figure below.
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.

1. What is the acceleration of mass M?
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.
2. What is the velocity of mass m after
it travels a distance h?
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.
3. What is the work done by the force of tension
in lifting mass m a distance h?
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:
A Pulley on a Table
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 threestep
problemsolving method:
 Ask yourself how the system will move: First, we know that mass m is falling and dragging mass M off the table. The force of kinetic friction opposes the motion of mass M. We also know, since both masses are connected by a nonstretching rope, that the two masses must have the same velocity and the same acceleration.
 Choose a coordinate system: For the purposes of this problem, it will be easier if we set our coordinate system relative to the rope rather than to the table. If we say that the xaxis runs parallel to the rope, this means the xaxis will be the updown axis for mass m and the leftright axis for mass M. Further, we can say that gravity pulls in the negative x direction. The yaxis, then, is perpendicular to the rope, and the positive y direction is away from the table.
 Draw freebody diagrams: The above description of the coordinate system may be a bit confusing. That’s why a diagram can often be a lifesaver.
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, .
How Complex Formulas Will Be Tested on SAT II Physics
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
freebody diagram above:
 Which of the following five formulas represents the acceleration of the pulley system? You would then be given five different mathematical formulas, one of which is the correct formula. The test writers would not expect you to have memorized the correct formula, but they would expect you to be able to derive it.
 Which of the following is a way of maximizing the system’s acceleration? You would then be given options like “maximize M and m and minimize ,” or “maximize and m and minimize M.” With such a question, you don’t even need to know the correct formula, but you do need to understand how the pulley system works. The downward motion is due to the gravitational force on m and is opposed by the force of friction on M, so we would maximize the downward acceleration by maximizing m and minimizing M and
 If the system does not move, which of the following must be true? You would then be given a number of formulas relating M, m, and . The idea behind such a question is that the system does not move if the downward force on m is less than or equal to the force of friction on M, so .
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.
