Practice problems for these concepts can be found at: Newton's Second Law, Fnet = ma Practice Problems for AP Physics B & C
A few words about tension and pulleys . Tension in a rope is the same everywhere in the rope, even if the rope changes direction (such as when it goes around a pulley) or if the tension acts in different directions on different objects. ONE ROPE = ONE TENSION. If there are multiple ropes in a problem, each rope will have its own tension. TWO ROPES = TWO TENSIONS.^{2}
When masses are attached to a pulley, the pulley can only rotate one of two ways. Call one way positive, the other, negative.
We arbitrarily call counterclockwise rotation of the pulley "positive."
^{2} Except for the Physics C corollary, when the pulley is massive—this situation is discussed in Chapter 16.
Step 1: Free–body diagrams.
The tension T is the same for each block—ONE ROPE = ONE TENSION. Also, note that because the blocks are connected, they will have the same acceleration, which we call a.
Step 2: Components.
The vectors already line up with one another. On to step 3.
Step 3: Equations.
Block M: Fnet = Mg – T = Ma
Block m: Fnet = T – mg = ma
Notice how we have been careful to adhere to our convention of which forces act in the positive and negative directions.
Step 4: Solve.
Let's solve for T using the first equation:
T = Mg – Ma.
Plugging this value for T into the second equation, we have
Mg – Ma – mg = ma
Our answer is
Why don't you work this one out for yourself? We have included our solution on a following page.
Solution to Example Problem
Step 1: Free–body diagrams.
Step 2: Components.
Again, our vectors line up nicely, so on to Step 3.
Step 3: Equations.
Before we write any equations, we must be careful about signs: we shall call counterclockwise rotation of the pulley "positive."
For the more massive block, we know that, because it is not flying off the table or tunneling into it, it is in equilibrium in the up–down direction. But it is not in equilibrium in the right–left direction.
F_{net,y} = (F_{N} – Mg) = 0
F_{net,x} = (0 – T ) = Ma
For the less massive block, we only have one direction to concern ourselves with: the up–down direction.
F_{net} = T – mg = ma
We can solve for T from the "F_{net,x}" equation for the more massive block and plug that value into the "F_{net}" equation for the less massive block, giving us
(– Mg) – mg = ma
We rearrange some terms to get
Now we plug in the known values for M and m to find that
To finish the problem, we plug in this value for a into the "F_{net,x}" equation for the more massive block.
More Thoughts on F_{net} = ma
The four example problems in this chapter were all solved using only F_{net} = ma. Problems you might face in the real world—that is, on the AP test—will not always be so straightforward. Here's an example: imagine that this last example problem asked you to find the speed of the blocks after two seconds had elapsed, assuming that the blocks were released from rest. That's a kinematics problem, but to solve it, you have to know the acceleration of the blocks. You would first have to use F_{net} = ma to find the acceleration, and then you could use a kinematics equation to find the final speed. We suggest that you try to solve this problem: it's good practice.
Also, remember the unit of force, the newton, is 1 N = 1 kg.m/s^{2}? Well, now you know why that conversion works: the units of force must be equal to the units of mass multiplied by the units of acceleration.

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