Torque - AP Physics C Mechanics

1. A force is applied to a wrench handle to tighten a bolt. To produce the greatest possible torque on the bolt with a given force magnitude, the force should be applied:
   Answer: as far from the bolt as possible and perpendicular to the handle. Torque is given by the expression $\vec{\tau} = \vec{r} \times \vec{F}$, and its magnitude is $\tau = rF\sin\theta$. To maximize the torque for a given force magnitude $F$, the distance from the pivot (the bolt), $r$, must be maximized, and the angle $\theta$ between the position vector $\vec{r}$ and the force vector $\vec{F}$ must be $90^\circ$ (so $\sin\theta = 1$). Therefore, the force should be applied as far from the bolt as possible and perpendicular to the handle.
2. A force of magnitude $F$ is applied at an angle of $60^\circ$ to the free end of a uniform rod of length $L$. The rod is pivoted at its other end. What is the lever arm for this force with respect to the pivot?
   Answer: $L \sin(60^\circ)$ The lever arm is defined as the perpendicular distance from the axis of rotation (the pivot) to the line of action of the force. This distance is given by $r_\perp = r \sin\theta$, where $r$ is the distance from the pivot to the point of force application and $\theta$ is the angle between the position vector and the force vector. Here, $r=L$ and $\theta = 60^\circ$, so the lever arm is $L \sin(60^\circ)$.
3. Which of the following units is dimensionally equivalent to the unit of torque, but is conventionally not used to express torque to avoid confusion with energy?
   Answer: Joule (J) The standard unit for torque is the Newton-meter (N·m). Dimensionally, $1 \text{ N}\cdot\text{m} = 1 (\text{kg}\cdot\text{m/s}^2)·\text{m} = 1 \text{kg}\cdot\text{m}^2/\text{s}^2$. The unit for energy and work, the Joule (J), is also defined as 1 N·m. However, because work is a scalar quantity ($W = \vec{F} \cdot \vec{d}$) and torque is a vector quantity ($\vec{\tau} = \vec{r} \times \vec{F}$), the unit Joule is reserved for energy and work to prevent ambiguity. Torque is always expressed in N·m.
4. A square plate of side length L is pivoted at its center. Four forces of equal magnitude F are applied. Which of the following applications produces the greatest magnitude of torque about the pivot?
   Answer: Force applied at a corner, perpendicular to the line connecting the center to that corner. The magnitude of the torque is $\tau = rF\sin\theta$. We want to maximize $r \sin\theta$. In case A, the distance to a corner is $r = \sqrt{(L/2)^2 + (L/2)^2} = L/\sqrt{2}$, and the force is perpendicular ($\sin\theta=1$), so $\tau_A = (L/\sqrt{2})F \approx 0.707LF$. In case B, the distance is $r=L/2$ and the force is perpendicular, so $\tau_B = (L/2)F = 0.5LF$. In case C, the distance is $r = L/\sqrt{2}$ but the angle is not 90 degrees; the torque is $LF/2 = 0.5LF$. In case D, the force points toward the pivot, so $\theta=180^\circ$ and the torque is zero. Comparing the magnitudes, case A produces the greatest torque.
5. A uniform plank of mass M and length L rests on a pivot at its center. A block of mass m is placed on the left end of the plank. To maintain rotational equilibrium, a downward vertical force F must be applied on the right side. To achieve equilibrium with the smallest possible force F, it must be applied:
   Answer: at the far right end of the plank. For rotational equilibrium, the sum of the torques must be zero. The block creates a counter-clockwise torque of $(L/2)mg$. The applied force F creates a clockwise torque of $rF$, where r is the distance from the pivot. To balance, $rF = (L/2)mg$, so $F = (L/2r)mg$. To minimize the required force $F$, the distance $r$ must be maximized. The maximum possible value for $r$ is $L/2$, which occurs when the force is applied at the far right end.
6. Which of the following statements correctly distinguishes the concepts of force and torque?
   Answer: A net force causes translational acceleration, while a net torque causes angular acceleration. Force and torque are analogous concepts for translational and rotational motion, respectively. According to Newton's second law and its rotational analog, a net external force causes a change in the translational motion of an object (linear acceleration), while a net external torque causes a change in the rotational motion of an object (angular acceleration).
7. A uniform ladder leans against a frictionless vertical wall. The base of the ladder rests on a rough horizontal floor. To solve for the forces acting on the ladder using the condition of static equilibrium, one must sum the torques. Which point is generally the most convenient to choose as the pivot for the torque calculation?
   Answer: The point of contact with the floor, because two unknown forces create no torque there. In static equilibrium problems, it is often advantageous to choose a pivot point where one or more unknown forces are acting. This is because the lever arm for those forces is zero, and thus they produce zero torque, simplifying the torque equation. At the point of contact with the floor, both the normal force from the floor and the static friction force act. By choosing this point as the pivot, these two unknown forces are eliminated from the torque equation, making it easier to solve for other unknowns.
8. A constant non-zero net torque is applied to a rigid body that is initially at rest. The application of this net torque will cause the body to:
   Answer: rotate with a constant angular acceleration. According to Newton's second law for rotation, the net torque on a rigid body is equal to its rotational inertia times its angular acceleration ($\vec{\tau}_{net} = I\vec{\alpha}$). Since the net torque is constant and non-zero, and the rotational inertia $I$ is a constant for a rigid body, the angular acceleration $\vec{\alpha}$ must also be constant and non-zero. A constant angular acceleration means the angular velocity changes at a constant rate.
9. A massless rod of length 4 m is pivoted at its center. A downward force of 10 N is applied perpendicularly at the right end, and an upward force of 5 N is applied perpendicularly at a point 1 m to the right of the pivot. What is the net torque about the pivot?
   Answer: 15 N·m clockwise Let's define clockwise torques as negative and counter-clockwise as positive. The 10 N force is applied at a distance of 2 m from the pivot, creating a clockwise torque: $\tau_1 = -(10 \text{ N})(2 \text{ m}) = -20 \text{ N}\cdot\text{m}$. The 5 N force is applied at a distance of 1 m from the pivot, creating a counter-clockwise torque: $\tau_2 = +(5 \text{ N})(1 \text{ m}) = +5 \text{ N}\cdot\text{m}$. The net torque is the sum: $\tau_{net} = \tau_1 + \tau_2 = -20 \text{ N}\cdot\text{m} + 5 \text{ N}\cdot\text{m} = -15 \text{ N}\cdot\text{m}$. The negative sign indicates a net clockwise torque of magnitude 15 N·m.
10. Two equal and opposite forces, both of magnitude F and perpendicular to a rod of length L, are applied to the opposite ends of the rod. This pair of forces is known as a couple. The magnitude of the net torque exerted by the couple about the center of the rod is:
    Answer: $FL$ Let the pivot be the center of the rod. Each force is applied at a distance of $L/2$ from the pivot. The first force creates a torque of magnitude $(L/2)F$. The second force, being in the opposite direction, also creates a torque that causes rotation in the same sense as the first. Its magnitude is also $(L/2)F$. Since both torques are in the same rotational direction, they add up. The net torque is $(L/2)F + (L/2)F = FL$.
11. A uniform rod of mass $M$ and length $L$ is hinged at one end and hangs vertically in equilibrium. A horizontal force $F$ is applied at a distance $d$ from the hinge to hold the rod at an angle $\theta$ from the vertical. If the force is moved to distance $2d$ from the hinge, what force magnitude is needed to maintain the same angle?
    Answer: $\frac{F}{2}$ When you encounter rotational equilibrium problems involving hinged objects, focus on torque balance. The rod must have zero net torque about the hinge for it to remain stationary at angle θ. Let's analyze the forces creating torques about the hinge. The weight $Mg$ acts downward at the rod's center of mass (distance $L/2$ from the hinge), creating a clockwise torque of $Mg \cdot \frac{L}{2} \cdot \sin\theta$. The horizontal force $F$ creates a counterclockwise torque. Initially, with force $F$ at distance $d$: the torque is $F \cdot d \cdot \cos\theta$ (since the force is perpendicular to the radial direction). For equilibrium: $F \cdot d \cdot \cos\theta = Mg \cdot \frac{L}{2} \cdot \sin\theta$. When the force moves to distance $2d$, let the new force be $F'$. The equilibrium condition becomes: $F' \cdot 2d \cdot \cos\theta = Mg \cdot \frac{L}{2} \cdot \sin\theta$. Since the right side is identical in both equations: $F \cdot d = F' \cdot 2d$, so $F' = \frac{F}{2}$. Answer C is correct because doubling the moment arm halves the required force. Answer A ($2F$) incorrectly assumes force scales with distance. Answer B ($\frac{F}{4}$) might result from confusing area scaling with linear scaling. Answer D ($F$) ignores the distance change entirely. Study tip: In rotational equilibrium problems, remember that torque equals force times perpendicular distance. When the moment arm doubles, the required force halves to maintain the same torque.
12. A force $\vec{F} = (3\hat{i} - 5\hat{j})$ N is applied at a point with position vector $\vec{r} = (2\hat{i} + 4\hat{j})$ m relative to an axis of rotation. What is the torque produced by this force?
    Answer: $(-22\hat{k}) \text{ N}\cdot\text{m}$ Torque is calculated using the cross product: $\vec{\tau} = \vec{r} \times \vec{F}$. For vectors in the xy-plane, the z-component is $\tau_z = r_x F_y - r_y F_x$. Substituting the given values: $\tau_z = (2 \text{ m})(-5 \text{ N}) - (4 \text{ m})(3 \text{ N}) = -10 \text{ N}\cdot\text{m} - 12 \text{ N}\cdot\text{m} = -22 \text{ N}\cdot\text{m}$. The torque vector is $\vec{\tau} = (-22\hat{k}) \text{ N}\cdot\text{m}$.
13. A force $\vec{F} = F_0 \hat{k}$ is applied at a position $\vec{r} = r_0 \hat{j}$ from the origin. The resulting torque vector is in the direction of:
    Answer: $+\hat{i}$ Torque is calculated as the cross product $\vec{\tau} = \vec{r} \times \vec{F}$. Substituting the given vectors: $\vec{\tau} = (r_0 \hat{j}) \times (F_0 \hat{k}) = r_0 F_0 (\hat{j} \times \hat{k})$. Using the right-hand rule for unit vectors, $\hat{j} \times \hat{k} = \hat{i}$. Therefore, the torque vector is in the positive x-direction ($+\hat{i}$).
14. A force vector is given by $\vec{F} = 5\hat{i}$ N. It is applied at a point with a position vector $\vec{r} = (3\hat{i} + 4\hat{j})$ m relative to the origin. What is the magnitude of the torque about the origin?
    Answer: 20 N·m The torque is calculated by the cross product $\vec{\tau} = \vec{r} \times \vec{F}$. $\vec{\tau} = (3\hat{i} + 4\hat{j}) \times (5\hat{i}) = (3\hat{i} \times 5\hat{i}) + (4\hat{j} \times 5\hat{i})$. Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$, the expression becomes $\vec{\tau} = 0 + 20(-\hat{k}) = -20\hat{k} \text{ N}\cdot\text{m}$. The magnitude of this torque is 20 N·m.
15. A flat disk lies in the xy-plane and is free to rotate about its center at the origin. A tangential force is applied at its edge in the positive y-direction at a point on the positive x-axis. The resulting torque vector is directed along the:
    Answer: positive z-axis. The position vector is $\vec{r} = r\hat{i}$ (positive x-axis) and the force vector is $\vec{F} = F\hat{j}$ (positive y-direction). The torque is $\vec{\tau} = \vec{r} \times \vec{F} = (r\hat{i}) \times (F\hat{j}) = rF (\hat{i} \times \hat{j})$. By the right-hand rule for cross products, $\hat{i} \times \hat{j} = \hat{k}$. Therefore, the torque vector is in the positive z-direction.
16. The torque exerted by a non-zero force $\vec{F}$ on a rigid body at a position $\vec{r} \neq 0$ from a pivot is zero if:
    Answer: the force vector $\vec{F}$ is parallel to the position vector $\vec{r}$. The magnitude of the torque is given by $\tau = rF\sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$. For the torque to be zero with $r \neq 0$ and $F \neq 0$, we must have $\sin\theta = 0$. This occurs when $\theta = 0^\circ$ or $\theta = 180^\circ$, which means the vectors $\vec{r}$ and $\vec{F}$ are parallel or anti-parallel.
17. A thin ring of radius $R$ and mass $m$ is free to rotate about a fixed axis through its center. A tangential force $F$ is applied to the ring for time $t$, after which the ring rotates with angular velocity $\omega$. If instead the same force $F$ is applied at an angle $\theta$ to the tangent for the same time $t$, what will be the final angular velocity?
    Answer: $\omega \sin \theta$ The torque depends on the component of force perpendicular to the radius. When tangential: $\tau_1 = FR$. When at angle $\theta$: $\tau_2 = FR\sin\theta$. Since $\tau = I\alpha$ and $\omega = \alpha t$, the angular velocity is proportional to torque. Therefore $\omega_2 = \omega_1 \sin\theta = \omega \sin\theta$.
18. A non-uniform rod of length $L$ and mass $M$ is pivoted at one end. Its center of mass is located at a distance of $L/3$ from the pivot. If the rod is held horizontally and then released, what is the initial magnitude of the torque exerted by gravity about the pivot?
    Answer: $MgL/3$ The force of gravity on an object can be treated as a single force acting at the object's center of mass. The magnitude of this force is $Mg$. The torque is $\tau = rF\sin\theta$. Here, the distance from the pivot to the center of mass is $r = L/3$. Since the rod is horizontal, the force of gravity is perpendicular to the rod, so $\theta = 90^\circ$ and $\sin\theta = 1$. Therefore, the torque is $\tau = (L/3)(Mg)(1) = MgL/3$.
19. A 1.00 m door pivots about its hinge; two forces act at the handle 1.00 m from hinge: $F_1=10\,\mathrm{N}$ perpendicular, $F_2=10\,\mathrm{N}$ at $30^\circ$ to the door. Based on the setup, which force produces the maximum torque?
    Answer: $F_1$ produces greater torque This question tests AP Physics C: Mechanics skills, specifically understanding torque and rotational dynamics with focus on comparing torques from forces at different angles. For maximum torque, the force should be perpendicular to the lever arm. F₁ acts perpendicular to the door, producing τ₁ = (1.00 m)(10 N)sin(90°) = 10 N·m. F₂ acts at 30° to the door, producing τ₂ = (1.00 m)(10 N)sin(30°) = 5 N·m. Choice A is correct because the perpendicular force F₁ produces greater torque (10 N·m) than the angled force F₂ (5 N·m). Choice C might result from not considering the angle effect on torque magnitude. To help students: Demonstrate that torque is maximized when force is perpendicular to the lever arm. Use visual aids to show how the effective force component decreases as the angle deviates from 90°.
20. A solid sphere of radius $R$ is subject to three forces applied at its surface. The forces have equal magnitudes $F$ but are applied in different ways: Force 1 is applied tangentially, Force 2 is applied radially inward, and Force 3 is applied at $45°$ to the radius. If the sphere can rotate about its center, which correctly ranks the magnitudes of the torques produced?
    Answer: $\tau_1 > \tau_3 > \tau_2$ When analyzing torque problems, you need to understand that torque depends on both the magnitude of the force and the perpendicular distance from the rotation axis to the line of action of the force. The formula is $\tau = rF\sin\theta$, where $\theta$ is the angle between the position vector and force vector. For Force 1 (tangential): The force is applied perpendicular to the radius, so $\theta = 90°$ and $\sin\theta = 1$. This gives $\tau_1 = RF(1) = RF$, which is the maximum possible torque for this force magnitude. For Force 2 (radial): The force points directly toward the center, so $\theta = 0°$ and $\sin\theta = 0$. This produces $\tau_2 = RF(0) = 0$. Radial forces never create torque about the center because their line of action passes through the rotation axis. For Force 3 (45° to radius): Here $\theta = 45°$, so $\tau_3 = RF\sin(45°) = RF(\frac{\sqrt{2}}{2}) \approx 0.71RF$. Therefore: $\tau_1 > \tau_3 > \tau_2$, confirming answer D. Answer A incorrectly suggests $\tau_2 > 0$, but radial forces produce zero torque. Answer B incorrectly equates $\tau_1$ and $\tau_3$, ignoring that tangential forces produce maximum torque. Answer C incorrectly ranks $\tau_3$ as largest, but sine values are maximized at 90°, not 45°. Study tip: Remember that torque is maximized when forces are applied tangentially (perpendicular to the radius) and zero when applied radially. The sine function in the torque equation captures this geometric relationship.

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