AP Physics C Mechanics Quiz: Torque
20 questions · exam conditions
0:00
TorqueQuestion 1 of 20

A force F applied perpendicularly at a distance r from a pivot creates a torque of magnitude τ\tau. If the force is doubled to 2F2F and the distance from the pivot is halved to r/2r/2, the magnitude of the new torque is:

τ/4\tau / 4
τ/2\tau / 2
τ\tau
2τ2\tau
← Back to quizzes

AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Torque

Practice Torque in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Torque, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A force F applied perpendicularly at a distance r from a pivot creates a torque of magnitude τ\tau. If the force is doubled to 2F2F and the distance from the pivot is halved to r/2r/2, the magnitude of the new torque is:

  1. τ/4\tau / 4
  2. τ/2\tau / 2
  3. τ\tau (correct answer)
  4. 2τ2\tau
Explanation: The original torque magnitude is given by τ=rF\tau = rF, since the force is applied perpendicularly. The new torque, τ′\tau', is calculated with the new force F′=2FF' = 2F and new distance r′=r/2r' = r/2. Thus, τ′=r′F′=(r/2)(2F)=rF\tau' = r'F' = (r/2)(2F) = rF. The new torque is equal to the original torque, τ′=τ\tau' = \tau.

Question 2

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?

  1. The center of mass of the ladder, because the gravitational force creates no torque there.
  2. The point of contact with the wall, because the wall's normal force creates no torque there.
  3. The point of contact with the floor, because two unknown forces create no torque there. (correct answer)
  4. The top end of the ladder, as it simplifies the geometry of the force from the wall.
Explanation: 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.

Question 3

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:

  1. positive x-axis.
  2. positive y-axis.
  3. positive z-axis. (correct answer)
  4. negative z-axis.
Explanation: The position vector is r⃗=ri^\vec{r} = r\hat{i} (positive x-axis) and the force vector is F⃗=Fj^\vec{F} = F\hat{j} (positive y-direction). The torque is τ⃗=r⃗×F⃗=(ri^)×(Fj^)=rF(i^×j^)\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, i^×j^=k^\hat{i} \times \hat{j} = \hat{k}. Therefore, the torque vector is in the positive z-direction.

Question 4

A rigid body is subject to two forces, F⃗1\vec{F}_1 at position r⃗1\vec{r}_1 and F⃗2\vec{F}_2 at position r⃗2\vec{r}_2. The net torque about the origin is τ⃗O\vec{\tau}_O. If the pivot point is moved to a new position r⃗P\vec{r}_P, the new net torque τ⃗P\vec{\tau}_P is related to τ⃗O\vec{\tau}_O by:

  1. τ⃗P=τ⃗O\vec{\tau}_P = \vec{\tau}_O if and only if the net force is zero.
  2. τ⃗P=τ⃗O−r⃗P×(F⃗1+F⃗2)\vec{\tau}_P = \vec{\tau}_O - \vec{r}_P \times (\vec{F}_1 + \vec{F}_2) (correct answer)
  3. τ⃗P=τ⃗O+r⃗P×(F⃗1+F⃗2)\vec{\tau}_P = \vec{\tau}_O + \vec{r}_P \times (\vec{F}_1 + \vec{F}_2)
  4. τ⃗P\vec{\tau}_P is independent of r⃗P\vec{r}_P and always equals τ⃗O\vec{\tau}_O.
Explanation: The original torque is τ⃗O=r⃗1×F⃗1+r⃗2×F⃗2\vec{\tau}_O = \vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2. The new position vectors relative to r⃗P\vec{r}_P are r⃗1′=r⃗1−r⃗P\vec{r}'_1 = \vec{r}_1 - \vec{r}_P and r⃗2′=r⃗2−r⃗P\vec{r}'_2 = \vec{r}_2 - \vec{r}_P. The new torque is τ⃗P=(r⃗1−r⃗P)×F⃗1+(r⃗2−r⃗P)×F⃗2=(r⃗1×F⃗1+r⃗2×F⃗2)−(r⃗P×F⃗1+r⃗P×F⃗2)=τ⃗O−r⃗P×(F⃗1+F⃗2)\vec{\tau}_P = (\vec{r}_1 - \vec{r}_P) \times \vec{F}_1 + (\vec{r}_2 - \vec{r}_P) \times \vec{F}_2 = (\vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2) - (\vec{r}_P \times \vec{F}_1 + \vec{r}_P \times \vec{F}_2) = \vec{\tau}_O - \vec{r}_P \times (\vec{F}_1 + \vec{F}_2). The torque is independent of the pivot only when the net force (F⃗1+F⃗2)(\vec{F}_1 + \vec{F}_2) is zero.

Question 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:

  1. at the pivot point, with a very large magnitude.
  2. at the midpoint between the pivot and the right end.
  3. at the far right end of the plank. (correct answer)
  4. at any point, as the location does not affect the force needed.
Explanation: For rotational equilibrium, the sum of the torques must be zero. The block creates a counter-clockwise torque of (L/2)mg(L/2)mg. The applied force F creates a clockwise torque of rFrF, where r is the distance from the pivot. To balance, rF=(L/2)mgrF = (L/2)mg, so F=(L/2r)mgF = (L/2r)mg. To minimize the required force FF, the distance rr must be maximized. The maximum possible value for rr is L/2L/2, which occurs when the force is applied at the far right end.

Question 6

A 1.00 m door pivots about its hinge; two forces act at the handle 1.00 m from hinge: F1=10 NF_1=10\,\mathrm{N} perpendicular, F2=10 NF_2=10\,\mathrm{N} at 30∘30^\circ to the door. Based on the setup, which force produces the maximum torque?

  1. F1F_1 produces greater torque (correct answer)
  2. F2F_2 produces greater torque
  3. They produce equal torque
  4. Neither produces torque about hinge
Explanation: 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°.

Question 7

A non-uniform rod of length LL and mass MM is pivoted at one end. Its center of mass is located at a distance of L/3L/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?

  1. MgLMgL
  2. MgL/2MgL/2
  3. MgL/3MgL/3 (correct answer)
  4. 2MgL/32MgL/3
Explanation: 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 MgMg. The torque is τ=rFsin⁡θ\tau = rF\sin\theta. Here, the distance from the pivot to the center of mass is r=L/3r = L/3. Since the rod is horizontal, the force of gravity is perpendicular to the rod, so θ=90∘\theta = 90^\circ and sin⁡θ=1\sin\theta = 1. Therefore, the torque is τ=(L/3)(Mg)(1)=MgL/3\tau = (L/3)(Mg)(1) = MgL/3.

Question 8

A force of magnitude FF is applied at an angle of 60∘60^\circ to the free end of a uniform rod of length LL. The rod is pivoted at its other end. What is the lever arm for this force with respect to the pivot?

  1. Lsin⁡(60∘)L \sin(60^\circ) (correct answer)
  2. Lcos⁡(60∘)L \cos(60^\circ)
  3. Fsin⁡(60∘)F \sin(60^\circ)
  4. Fcos⁡(60∘)F \cos(60^\circ)
Explanation: 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⊥=rsin⁡θr_\perp = r \sin\theta, where rr is the distance from the pivot to the point of force application and θ\theta is the angle between the position vector and the force vector. Here, r=Lr=L and θ=60∘\theta = 60^\circ, so the lever arm is Lsin⁡(60∘)L \sin(60^\circ).

Question 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?

  1. 15 N·m clockwise (correct answer)
  2. 25 N·m clockwise
  3. 15 N·m counter-clockwise
  4. 25 N·m counter-clockwise
Explanation: 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: τ1=−(10 N)(2 m)=−20 N\cdotpm\tau_1 = -(10 \text{ N})(2 \text{ m}) = -20 \text{ N·m}. The 5 N force is applied at a distance of 1 m from the pivot, creating a counter-clockwise torque: τ2=+(5 N)(1 m)=+5 N\cdotpm\tau_2 = +(5 \text{ N})(1 \text{ m}) = +5 \text{ N·m}. The net torque is the sum: τnet=τ1+τ2=−20 N\cdotpm+5 N\cdotpm=−15 N\cdotpm\tau_{net} = \tau_1 + \tau_2 = -20 \text{ N·m} + 5 \text{ N·m} = -15 \text{ N·m}. The negative sign indicates a net clockwise torque of magnitude 15 N·m.

Question 10

A force F⃗=(3i^−5j^)\vec{F} = (3\hat{i} - 5\hat{j}) N is applied at a point with position vector r⃗=(2i^+4j^)\vec{r} = (2\hat{i} + 4\hat{j}) m relative to an axis of rotation. What is the torque produced by this force?

  1. (−22k^) N\cdotpm(-22\hat{k}) \text{ N·m} (correct answer)
  2. (22k^) N\cdotpm(22\hat{k}) \text{ N·m}
  3. (−14k^) N\cdotpm(-14\hat{k}) \text{ N·m}
  4. (2k^) N\cdotpm(2\hat{k}) \text{ N·m}
Explanation: Torque is calculated using the cross product: τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}. For vectors in the xy-plane, the z-component is τz=rxFy−ryFx\tau_z = r_x F_y - r_y F_x. Substituting the given values: τz=(2 m)(−5 N)−(4 m)(3 N)=−10 N\cdotpm−12 N\cdotpm=−22 N\cdotpm\tau_z = (2 \text{ m})(-5 \text{ N}) - (4 \text{ m})(3 \text{ N}) = -10 \text{ N·m} - 12 \text{ N·m} = -22 \text{ N·m}. The torque vector is τ⃗=(−22k^) N\cdotpm\vec{\tau} = (-22\hat{k}) \text{ N·m}.

Question 11

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:

  1. 0
  2. FL/2FL/2
  3. FLFL (correct answer)
  4. 2FL2FL
Explanation: Let the pivot be the center of the rod. Each force is applied at a distance of L/2L/2 from the pivot. The first force creates a torque of magnitude (L/2)F(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(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(L/2)F + (L/2)F = FL.

Question 12

A force vector is given by F⃗=5i^\vec{F} = 5\hat{i} N. It is applied at a point with a position vector r⃗=(3i^+4j^)\vec{r} = (3\hat{i} + 4\hat{j}) m relative to the origin. What is the magnitude of the torque about the origin?

  1. 0 N·m
  2. 15 N·m
  3. 20 N·m (correct answer)
  4. 25 N·m
Explanation: The torque is calculated by the cross product τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}. τ⃗=(3i^+4j^)×(5i^)=(3i^×5i^)+(4j^×5i^)\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 i^×i^=0\hat{i} \times \hat{i} = 0 and j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k}, the expression becomes τ⃗=0+20(−k^)=−20k^ N\cdotpm\vec{\tau} = 0 + 20(-\hat{k}) = -20\hat{k} \text{ N·m}. The magnitude of this torque is 20 N·m.

Question 13

A wrench is used to tighten a bolt. The wrench handle makes an angle of 30°30° with the horizontal, and a force of 5050 N is applied vertically downward at the end of the 0.30.3 m handle. If the same torque is to be produced by applying a force horizontally at the end of the handle, what force magnitude is required?

  1. 2525 N
  2. 25325\sqrt{3} N (correct answer)
  3. 50350\sqrt{3} N
  4. 503\frac{50}{\sqrt{3}} N
Explanation: Initially, the perpendicular component is F1cos⁡(30°)=50⋅(3/2)F_1 \cos(30°) = 50 \cdot (\sqrt{3}/2), giving torque τ1=50⋅(3/2)⋅0.3\tau_1 = 50 \cdot (\sqrt{3}/2) \cdot 0.3. For horizontal force, the perpendicular component is F2sin⁡(30°)=F2⋅(1/2)F_2 \sin(30°) = F_2 \cdot (1/2). Setting torques equal: 50⋅(3/2)=F2⋅(1/2)50 \cdot (\sqrt{3}/2) = F_2 \cdot (1/2), so F2=503F_2 = 50\sqrt{3} N.

Question 14

A uniform rod of mass MM and length LL is hinged at one end and hangs vertically in equilibrium. A horizontal force FF is applied at a distance dd from the hinge to hold the rod at an angle θ\theta from the vertical. If the force is moved to distance 2d2d from the hinge, what force magnitude is needed to maintain the same angle?

  1. 2F2F
  2. F4\frac{F}{4}
  3. F2\frac{F}{2} (correct answer)
  4. FF
Explanation: 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 MgMg acts downward at the rod's center of mass (distance L/2L/2 from the hinge), creating a clockwise torque of Mg⋅L2⋅sin⁡θMg \cdot \frac{L}{2} \cdot \sin\theta. The horizontal force FF creates a counterclockwise torque. Initially, with force FF at distance dd: the torque is F⋅d⋅cos⁡θF \cdot d \cdot \cos\theta (since the force is perpendicular to the radial direction). For equilibrium: F⋅d⋅cos⁡θ=Mg⋅L2⋅sin⁡θF \cdot d \cdot \cos\theta = Mg \cdot \frac{L}{2} \cdot \sin\theta. When the force moves to distance 2d2d, let the new force be F′F'. The equilibrium condition becomes: F′⋅2d⋅cos⁡θ=Mg⋅L2⋅sin⁡θF' \cdot 2d \cdot \cos\theta = Mg \cdot \frac{L}{2} \cdot \sin\theta. Since the right side is identical in both equations: F⋅d=F′⋅2dF \cdot d = F' \cdot 2d, so F′=F2F' = \frac{F}{2}. Answer C is correct because doubling the moment arm halves the required force. Answer A (2F2F) incorrectly assumes force scales with distance. Answer B (F4\frac{F}{4}) might result from confusing area scaling with linear scaling. Answer D (FF) 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.

Question 15

A thin ring of radius RR and mass mm is free to rotate about a fixed axis through its center. A tangential force FF is applied to the ring for time tt, after which the ring rotates with angular velocity ω\omega. If instead the same force FF is applied at an angle θ\theta to the tangent for the same time tt, what will be the final angular velocity?

  1. ωsin⁡θ\omega \sin \theta (correct answer)
  2. ωcos⁡θ\omega \cos \theta
  3. ωtan⁡θ\omega \tan \theta
  4. ωcos⁡θ\frac{\omega}{\cos \theta}
Explanation: The torque depends on the component of force perpendicular to the radius. When tangential: τ1=FR\tau_1 = FR. When at angle θ\theta: τ2=FRsin⁡θ\tau_2 = FR\sin\theta. Since τ=Iα\tau = I\alpha and ω=αt\omega = \alpha t, the angular velocity is proportional to torque. Therefore ω2=ω1sin⁡θ=ωsin⁡θ\omega_2 = \omega_1 \sin\theta = \omega \sin\theta.

Question 16

A solid sphere of radius RR is subject to three forces applied at its surface. The forces have equal magnitudes FF but are applied in different ways: Force 1 is applied tangentially, Force 2 is applied radially inward, and Force 3 is applied at 45°45° to the radius. If the sphere can rotate about its center, which correctly ranks the magnitudes of the torques produced?

  1. τ1>τ2>τ3\tau_1 > \tau_2 > \tau_3
  2. τ1=τ3>τ2\tau_1 = \tau_3 > \tau_2
  3. τ3>τ1>τ2\tau_3 > \tau_1 > \tau_2
  4. τ1>τ3>τ2\tau_1 > \tau_3 > \tau_2 (correct answer)
Explanation: 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 τ=rFsin⁡θ\tau = rF\sin\theta, where θ\theta is the angle between the position vector and force vector. For Force 1 (tangential): The force is applied perpendicular to the radius, so θ=90°\theta = 90° and sin⁡θ=1\sin\theta = 1. This gives τ1=RF(1)=RF\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 θ=0°\theta = 0° and sin⁡θ=0\sin\theta = 0. This produces τ2=RF(0)=0\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 θ=45°\theta = 45°, so τ3=RFsin⁡(45°)=RF(22)≈0.71RF\tau_3 = RF\sin(45°) = RF(\frac{\sqrt{2}}{2}) \approx 0.71RF. Therefore: τ1>τ3>τ2\tau_1 > \tau_3 > \tau_2, confirming answer D. Answer A incorrectly suggests τ2>0\tau_2 > 0, but radial forces produce zero torque. Answer B incorrectly equates τ1\tau_1 and τ3\tau_3, ignoring that tangential forces produce maximum torque. Answer C incorrectly ranks τ3\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.

Question 17

Two identical rods, each of length LL and mass mm, are connected at right angles to form an L-shape. The system rotates about an axis perpendicular to both rods and passing through their junction point. If a force FF is applied perpendicular to one rod at its free end, what torque does this force produce about the rotation axis?

  1. FLFL (correct answer)
  2. FL2\frac{FL}{2}
  3. FL2FL\sqrt{2}
  4. FL2\frac{FL}{\sqrt{2}}
Explanation: The torque is τ=rFsin⁡θ\tau = rF\sin\theta where rr is the distance from the axis to the point of force application, FF is the force magnitude, and θ\theta is the angle between r⃗\vec{r} and F⃗\vec{F}. Here, r=Lr = L (distance along the rod), θ=90°\theta = 90° (force perpendicular to rod), so τ=FLsin⁡(90°)=FL\tau = FL\sin(90°) = FL. The L-shape configuration doesn't affect this calculation.

Question 18

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?

  1. Force applied at a corner, perpendicular to the line connecting the center to that corner. (correct answer)
  2. Force applied at the midpoint of an edge, perpendicular to that edge.
  3. Force applied at a corner, directed parallel to an adjacent edge of the plate.
  4. Force applied at a corner, directed toward the center of the plate.
Explanation: The magnitude of the torque is τ=rFsin⁡θ\tau = rF\sin\theta. We want to maximize rsin⁡θr \sin\theta. In case A, the distance to a corner is r=(L/2)2+(L/2)2=L/2r = \sqrt{(L/2)^2 + (L/2)^2} = L/\sqrt{2}, and the force is perpendicular (sin⁡θ=1\sin\theta=1), so τA=(L/2)F≈0.707LF\tau_A = (L/\sqrt{2})F \approx 0.707LF. In case B, the distance is r=L/2r=L/2 and the force is perpendicular, so τB=(L/2)F=0.5LF\tau_B = (L/2)F = 0.5LF. In case C, the distance is r=L/2r = L/\sqrt{2} but the angle is not 90 degrees; the torque is LF/2=0.5LFLF/2 = 0.5LF. In case D, the force points toward the pivot, so θ=180∘\theta=180^\circ and the torque is zero. Comparing the magnitudes, case A produces the greatest torque.

Question 19

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:

  1. as close to the bolt as possible and parallel to the handle.
  2. as far from the bolt as possible and perpendicular to the handle. (correct answer)
  3. as far from the bolt as possible and parallel to the handle.
  4. at the midpoint of the handle and at a 45-degree angle to it.
Explanation: Torque is given by the expression τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}, and its magnitude is τ=rFsin⁡θ\tau = rF\sin\theta. To maximize the torque for a given force magnitude FF, the distance from the pivot (the bolt), rr, must be maximized, and the angle θ\theta between the position vector r⃗\vec{r} and the force vector F⃗\vec{F} must be 90∘90^\circ (so sin⁡θ=1\sin\theta = 1). Therefore, the force should be applied as far from the bolt as possible and perpendicular to the handle.

Question 20

A force F⃗=F0k^\vec{F} = F_0 \hat{k} is applied at a position r⃗=r0j^\vec{r} = r_0 \hat{j} from the origin. The resulting torque vector is in the direction of:

  1. +i^+\hat{i} (correct answer)
  2. −i^-\hat{i}
  3. +j^+\hat{j}
  4. −k^-\hat{k}
Explanation: Torque is calculated as the cross product τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}. Substituting the given vectors: τ⃗=(r0j^)×(F0k^)=r0F0(j^×k^)\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, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}. Therefore, the torque vector is in the positive x-direction (+i^+\hat{i}).