Free-Body Diagrams and Equilibrium Practice Problems for AP Physics B & C
Review the following concepts if necessary:
- Free-Body Diagram and Equilibrium for AP Physics B & C
- Normal Force and Tension for AP Physics B & C
- Friction for AP Physics B & C
- Inclined Planes and Torque for AP Physics B & C
- Free-Body Diagrams and Equilibrium: Of Special Interest to Physics C Students
- A 50-g mass is hung by string as shown in the picture above. The left-hand string is horizontal; the angled string measures 30° to the horizontal. What is the tension in the angled string?
- A 6000-kg bus sits on a 30° incline. A crane attempts to lift the bus off of the plane. The crane pulls perpendicular to the plane, as shown in the diagram. How much force must the crane apply so that the bus is suspended just above the surface? [cos 30° = 0.87, sin 30° = 0.50]
- 52,000 N
- 30,000 N
- 6000 N
- 5200 N
- 300 N
- Give two examples of a situation in which the normal force on an object is less than the object's weight. Then give an example of a situation in which there is NO normal force on an object.
- A 150-N box sits motionless on an inclined plane, as shown above. What is the angle of the incline?
- A 50-g meterstick is to be suspended by a single string. A 100-g ball hangs from the left-hand edge of the meterstick. Where should the string be attached so that the meterstick hangs in equilibrium?
- at the left-hand edge
- 40 cm from left-hand edge
- 30 cm from right-hand edge
- 17 cm from left-hand edge
- at the midpoint of the meterstick
- When a block rests on an inclined plane, the normal force on the block is less than the block's weight, as discussed in the answer to #2. Another example in which the normal force is less than an object's weight occurs when you pull a toy wagon.
- This free-body diagram should be very familiar to you by now.
Call the tension in the angled rope T2. In the y-direction, we have T2, y = T2(sin 30°) acting up, and mg acting down. Set "up" forces equal to "down" forces and solve for tension: T2 = mg/(sin 30°). Don't forget to use the mass in KILOgrams, i.e., 0.050 kg. The tension thus is (0.050 kg)(10 N/kg)/(0.5) = 1.0 N. This is reasonable because the tension is about the same order of magnitude as the weight of the mass.
A—Because the force of the crane, Fc, acts perpendicular to the plane, the parallel-to-the-plane direction is irrelevant. So all we need to do is set Fc equal to mg(cos 30°) = (6000 kg)(10 N/kg)(.87) and plug in. Fc = 52,000 N. This is a reasonable answer because it is less than—but on the same order of magnitude as—the weight of the bus.
In any situation where an object does not rest on a surface (for example, when something floats in space), there is no normal force.
The box is in equilibrium, so Ff must equal mg (sinθ), and FN must equal mg(cosθ).
- μ · FN = μ · mg(cosθ) = mg(sinθ).
Plugging in the values given in the problem we find that μ = 17°. This answer seems reasonable because we'd expect the incline to be fairly shallow.
Because the stick is in equilibrium, the clockwise torques equal the counterclockwise torques: (1 N)(x) = (0.5 N)(50 – x). So x = something in the neighborhood of 25/1.5 17 cm. This answer is less than 50 cm, and is closer to the edge with the heavy mass, so it makes sense.
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