AP Physics 1: Algebra-Based Practice Quiz: Torque

Correct Answer: B

The provided content explicitly states that 'Torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force.' A parallel force component produces no torque.

Correct Answer: D

The content provides a direct definition: 'The lever arm is the perpendicular distance from the axis of rotation to the line of action of the exerted force.' This is distinct from the distance 'r' to the point of application unless the force is applied at a 90-degree angle.

Correct Answer: C

In the standard torque equation $\tau = rF\sin\theta$, 'r' is the magnitude of the position vector, which is the straight-line distance from the axis of rotation to the point where the force is applied. The lever arm is represented by the term $r\sin\theta$.

Correct Answer: C

The magnitude of torque is given by $\tau = rF\sin\theta$. The sine function has a maximum value of 1 when the angle $\theta$ is 90°. Therefore, applying the force perpendicular to the position vector maximizes the torque.

Correct Answer: C

The content states, 'Force diagrams also depict the location at which those forces are exerted relative to the axis of rotation.' This is in addition to showing the forces themselves, which is the primary purpose of any force diagram.

Correct Answer: C

Torque is calculated by $\tau = rF\sin\theta$. If the force is parallel to the position vector, the angle $\theta$ between them is 0° or 180°. Since $\sin(0°) = 0$ and $\sin(180°) = 0$, the resulting torque is zero. This aligns with the principle that only the perpendicular component of the force creates torque.

Correct Answer: C

Using trigonometry, if $\theta$ is the angle between the force vector F and the position vector r, then $F\sin\theta$ is the magnitude of the component of F that is perpendicular to r. The equation can be thought of as $\tau = r \times (F_{\perp})$.

Correct Answer: C

The lever arm is the perpendicular distance from the axis to the line of action of the force. By rearranging the equation as $\tau = F(r\sin\theta)$, the term $r\sin\theta$ represents this perpendicular distance. This is an alternative but equivalent way to calculate torque.

Correct Answer: C

Using the formula $\tau = rF\sin\theta$. Here, r = 2 m, F = 10 N, and since the force is perpendicular, $\theta = 90°$. We know that $\sin(90°) = 1$. Therefore, $\tau = (2 \text{ m})(10 \text{ N})(1) = 20 \text{ N·m}$.

Correct Answer: B

The magnitude of torque is directly proportional to the distance 'r' as shown in the equation $\tau = rF\sin\theta$. If 'r' is doubled and F and $\sin\theta$ are constant, the torque $\tau$ will also be doubled.

Correct Answer: C

The provided content explicitly mentions the identification and description of torques exerted on a 'rigid system'.

Correct Answer: C

Using the formula $\tau = rF\sin\theta$. We are given r = 0.5 m, F = 20 N, and $\theta = 30°$. The sine of 30° is 0.5. Therefore, $\tau = (0.5 \text{ m})(20 \text{ N})(\sin 30°) = (0.5)(20)(0.5) = 5 \text{ N·m}$.

Correct Answer: B

This is a fundamental concept from the provided text: 'Torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force.' A force can be applied without creating torque if it is directed through the axis of rotation or parallel to the position vector.

Correct Answer: C

Torque is given by $\tau = rF\sin\theta$. If we want to keep $\tau$ constant while doubling the force F to 2F (assuming the angle is the same), the distance r must be halved. $\tau = (r/2)(2F)\sin\theta = rF\sin\theta$.

Correct Answer: B

The provided content states, 'Torques can be described using force diagrams. Force diagrams also depict the location at which those forces are exerted relative to the axis of rotation.' This visual tool is essential for analyzing torques.