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# Inclined Planes and Torque for AP Physics B & C

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By McGraw-Hill Professional
Updated on Feb 10, 2011

Practice problems for these concepts can be found at:

Free-Body Diagrams and Equilibrium Practice Problems for AP Physics B & C

### Inclined Planes

These could be the most popular physics problems around. You've probably seen way too many of these already in your physics class, so we'll just give you a few tips on approaching them.

In Figure 10.4 we have a block of mass m resting on a plane elevated an angle θ above the horizontal. The plane is not frictionless. We've drawn a free-body diagram of the forces acting on the block in Figure 10.5a.

Ff is directed parallel to the surface of the plane, and FN is, by definition, directed perpendicular to the plane. It would be a pain to break these two forces into x- and y-components, so instead we will break the "weight" vector into components that "line up" with Ff and FN, as shown in Figure 10.5b.

This rule always works, as long as the angle of the plane is measured from the horizontal.

### Torque

Torque occurs when a force is applied to an object, and that force can cause the object to rotate.

In other words, the torque exerted on an object equals the force exerted on that object (F) multiplied by the distance between where the force is applied and the fulcrum (d) as long as the force acts perpendicular to the object.

Figure 10.6 shows what we mean:

The unit of torque is the newton meter.

Torque problems on the AP Physics B exam tend to be pretty straightforward. To solve them, set counterclockwise torques equal to clockwise torques.

Here's an example.

Step 1: Free-body diagram.

We'll use point A as the fulcrum to start with. Why? In a static equilbrium situation, since the bridge isn't actually rotating, any point on the bridge could serve as a fulcrum. But we have two unknown forces here, the forces of the supports A and B. We choose the location of one of these supports as the fulcrum, because now that support provides zero torque—the distance from the fulcrum becomes zero! Now all we have to do is solve for the force of support B.

The diagram below isn't a true "free-body diagram," because it includes both distance and forces, but it is useful for a torque problem. Bob's weight acts downward right where he stands. The bridge's weight is taken into account with a force vector acting at the bridge's center of mass; that is, 40 m to the right of pillar A.

Step 2: Vector components.

We don't have to worry about vector components here. (We would have if the forces had not acted perpendicular to the bridge. Such a situation is much more likely to appear on the C than the B exam—see the Physics C section below.)

Step 3: Equations.

Torquenet = counterclockwise – clockwise = 0.
(FB)(80 m) – [(100 kg.10 N/kg)(20 m) + (10,000 N)(40 m)] = 0.

Step 4: Solve. FB = 5300 N

This is reasonable because pillar B is supporting less than half of the 11,000 N weight of the bridge and Bob. Because Bob is closer to pillar A, and otherwise the bridge is symmetric, A should bear the majority of the weight.

Practice problems for these concepts can be found at:

Free-Body Diagrams and Equilibrium Practice Problems for AP Physics

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