SATII物理专题分析Pulleys
SAT考试是美国高中生进入美国大学的标准入学考试，包括SAT1和SAT2。SAT2是专项测验，有数学、物理、化学、生物、外语等，大部分为选择题，是可以选科的。今天我们来学习SATII物理专题分析Pulleys，希望对大家的复习有所帮助。
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 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 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?
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:
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 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
, 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 t