- 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 > mand 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 subject test 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 free-body 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 free-body 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 subject test 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?
2. What is the velocity of mass m after it travels a distance h?
3. What is the work done by the force of tension in lifting mass m a distance h?
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:
for mass M:T- Mg = -Ma
for mass m:T - mg = ma
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.
?KE = -?U
Remember that since the system begins at rest,KEinitial = 0 . 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 subject test 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, Fƒr = µN, µ 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:
- 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 x-axis runs parallel to the rope, this means
the x-axis will be the up-down axis for mass
m and the left-right axis for mass
M. Further, we can say that gravity
pulls in the negative x direction. The y-axis, then, is
perpendicular to the rope, and the positive y direction is away from the
table.
- Draw free-body 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 µMG -T ,
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:
for mass M: µMG -T = Ma
for mass m: mg sin? -T = ma
Adding the two equations, we find µMg - mg = (M + m)a. Solving for a, we get:
Since m is moving downward, a must be negative. Therefore, µM < m .
How Complex Formulas Will Be Tested on subject test Physics
It is highly unlikely that subject test Physics will ask a question that involves
remembering and then plugging numbers into an equation like this one. Remember:
subject test 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:
- 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
µ,”
or “maximize
µ
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
µ
- 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
µ.
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
mg = ?Mg.
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 subject test Physics will get.
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