PULLEYS - Newton's Laws - SAT Physics Subject Test

SAT Physics Subject Test

Chapter 3 Newton”s Laws


Pulleys are devices that change the direction of the tension force in the cords that slide over them. For the purposes of this text and the SAT Physics Subject Test, we”ll consider each pulley to be frictionless and massless, which means their masses are so much smaller than the objects attached to the ends of them, that they can be ignored.

In the case of two single masses m1 and m2 that are attached to a pulley and cord, the downward forces are due to the weight (mass and gravity exerted on it) of the masses. The upward forces are due to the tension (T) in the cord.

18. In the diagram above, assume that the tabletop is frictionless. Determine the acceleration of the blocks once they”re released from rest.

Here”s How to Crack It

There are two blocks, so we need to draw two free-body diagrams.

To get the acceleration of each one, we use Newton”s second law, Fnet = ma.

Notice that there are two unknowns, FT and a, but we can eliminate FT by adding the two equations, and then we can solve for a.

A quicker way of solving for the acceleration is to treat the entire system (blocks plus string) as one object. Since we are only concerned with forces acting on the object, we can ignore tension. The string is part of the object. Then we need only consider forces acting in the direction of motion (Mg) and forces opposite the direction of motion (none). Our mass in Newton”s second law becomes M + m, so Fnet = ma becomes Mg = (M + m) a, giving us the same answer for acceleration.

19. Using the diagram from the previous example, assume that m = 2 kg, M = 10 kg, and the coefficient of kinetic friction between the small block and the tabletop is 0.5. What is the acceleration of the blocks?

Here”s How to Crack It

Once again, draw a free-body diagram for each object. Notice that the only difference between these diagrams and the ones in the previous example is the inclusion of friction, Ff, that acts on the block on the table.

As before, we have two equations that contain two unknowns (a and FT).

FTFf = ma (1)

MgFT = Ma (2)

We add the equations (thereby eliminating FT) and solve for a. Notice that, by definition, FfFN and from the free-body diagram for m, we see that FN = mg, so Ff = µmg:

Or, using our shorter metod:

Substituting in the numerical values given for m, M, and µ, we find that a = g (or 7.5 m/s2).

20. In the previous example, calculate the magnitude of the tension in the cord.

Here”s How to Crack It

Since the value of a has been determined, we can use either of the two original equations to calculate FT. Using equation (2), MgFT = Ma (because it”s simpler), we find

As you can see, we would have found the same answer if we”d used equation (1):