4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
The figure shows a wrench on a bolt at O, seen from above, and a force exerted on the wrench in the plane of the page. What is the magnitude of the torque exerted by the force about O? Use sin 37° = 3/5 and cos 37° = 4/5.
Answer and reasoning
A8.00 N·m A student who multiplies the whole force by its distance from O picks this: (0.400 m)(20.0 N). The force is not perpendicular to the wrench, and its component along the wrench produces no torque.
B4.80 N·mCorrect Only the component of the force perpendicular to the wrench produces torque: F sin 37° = (20.0 N)(3/5) = 12.0 N, exerted 0.400 m from O. τ = (0.400 m)(12.0 N) = 4.80 N·m.
C6.40 N·m A student who uses cos 37° picks this: (0.400 m)(20.0 N)(4/5). F cos 37° is the component along the wrench, whose line of action passes through O; the perpendicular component is F sin 37°.
D6.00 N·m A student who uses the full length of the wrench, 0.500 m, picks this. The force is exerted 0.400 m from O, so that is the distance to use.
Working Only the component of F perpendicular to the wrench (and so to r⃗) produces torque: F sin 37° = (20.0 N)(3/5) = 12.0 N, exerted 0.400 m from O. τ = (0.400 m)(12.0 N) = 4.80 N·m. (Whole force: 8.00 N·m. cos 37°: 6.40 N·m. Full length 0.500 m: 6.00 N·m.)
The figure shows, seen from above, an L-shaped bracket that can rotate about a fixed axle at O, drawn on a grid of squares 0.10 m on a side, and the horizontal force exerted at its end P. What is the magnitude of the torque exerted by the force about O?
Answer and reasoning
A6.0 N·m A student who uses the straight-line distance from O to P, √(0.40² + 0.30²) m = 0.50 m, picks this. That distance is r, not the lever arm, because the force is not perpendicular to OP.
B4.8 N·m A student who uses 0.40 m, the distance from O measured along the direction of the force, picks this. That distance does not affect the torque; only the perpendicular distance to the line of action, 0.30 m, counts.
C3.6 N·mCorrect The lever arm is the perpendicular distance from O to the force's line of action. The force is horizontal, so its line of action is the horizontal line through P, 3 squares (0.30 m) from O: τ = (12 N)(0.30 m) = 3.6 N·m.
D8.4 N·m A student who measures along the bracket, 0.40 m + 0.30 m = 0.70 m, picks this. The lever arm is a straight, perpendicular distance from O to the line of action, not a distance along the object.
Working The force is horizontal, so its line of action is the horizontal line through P, 3 squares = 0.30 m from O: lever arm 0.30 m. τ = (12 N)(0.30 m) = 3.6 N·m. (Straight-line distance OP = √(0.40² + 0.30²) m = 0.50 m: 6.0 N·m. Distance along the force's direction, 0.40 m: 4.8 N·m. Distance along the bracket, 0.70 m: 8.4 N·m.)
A student is drawing a force diagram to analyze the torques exerted on a rigid beam. Which statement about the diagram is correct?
Answer and reasoning
AIt shows the forces that the beam exerts on its supports as well as the forces exerted on it. A student who includes both members of each third-law pair picks this. The beam's push on a support is exerted on the support; only forces exerted on the beam belong in its diagram.
BIt shows each torque exerted on the beam as a curved arrow, drawn in place of the force. A student who thinks torques are separate interactions picks this. The diagram shows the forces; the torque of each force is worked out from where and in what direction the force is exerted.
CIt shows every force exerted on the beam, with each drawn from the beam's center of mass. A student who draws every force from the center of mass, as in a free-body diagram for an object, picks this. For torques the point where each force is exerted matters, so contact forces are drawn where the contact is.
DIt shows only forces that other objects exert on the beam, as a free-body diagram does.Correct A force diagram is like a free-body diagram: it shows only forces exerted on the chosen system by other objects. Unlike a free-body diagram for an object, it also shows where on the beam each force is exerted, so that torques can be found.
The torque exerted by a force about an axis is τ⃗ = r⃗ × F⃗. Which describes the direction of τ⃗?
Answer and reasoning
AAt right angles to the plane that contains both r⃗ and F⃗Correct A cross product is perpendicular to both of its vectors, so τ⃗ is perpendicular to the plane that contains r⃗ and F⃗: along the axis about which the force tends to turn the system. The right-hand rule picks which of the two perpendicular directions.
BAlong F⃗, in the direction in which the force pushes or pulls A student who gives a torque the direction of the force that produces it picks this. τ⃗ = r⃗ × F⃗ is perpendicular to F⃗, so a nonzero torque is never along it.
CIn the plane of r⃗ and F⃗, pointing between the two vectors A student who treats the cross product like a vector sum picks this. A sum of two vectors lies in their plane; a cross product is perpendicular to that plane.
DAlong the path of the point where the force is exerted A student who takes the direction in which the point of application moves as the torque's direction picks this. That point moves tangent to its circle, in the plane of rotation; τ⃗ is perpendicular to that plane.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
5.3.A.1 Perpendicular component of a force Fix
Perpendicular component of a force
The component of a force perpendicular to r⃗, F sin θ. Only this component produces a torque about the axis; the component along r⃗ produces none, so τ = rF sin θ.
Students often think The sine and cosine of an angle in the figure are interchangeable in a torque calculation, so the component of the force along r⃗ (or the wrong trigonometric factor) can be used in place of the perpendicular component. In fact No. θ is the angle between r⃗ and F⃗, and only the component F sin θ, perpendicular to r⃗, produces torque. If the angle shown is measured from another line, the right factor may be its cosine; using the wrong one gives the component along r⃗, which produces no torque.
5.3.A.2 Line of action Fix
Line of action
The line through the point where a force is exerted, drawn along the direction of the force.
Lever arm
The perpendicular distance from the axis of rotation to a force's line of action, equal to r sin θ, so τ = F × (lever arm). A force whose line of action passes through the axis has a zero lever arm and exerts no torque about it. SI unit: m.
Students often think The torque of a force is always the force multiplied by the distance from the axis to the point where it is exerted, whatever the force's direction. In fact Only when the force is perpendicular to r⃗. In general τ = rF sin θ = F × (lever arm), where the lever arm is the perpendicular distance from the axis to the force's line of action; it is shorter than r unless θ = 90°, and zero if the line of action passes through the axis.
Students often think Every force on an object acts as if exerted at its far end, so the distance used for torque is the distance from the axis to the far end of the object (its whole length when the axis is at the other end). In fact No. The distance is measured from the axis to the point where the force is exerted, and then the lever arm is the perpendicular distance to the force's line of action. The gravitational force on a uniform rod is exerted at its center, and a push may be applied partway along an object.
5.3.B.1 Force diagram Fix
Force diagram
A diagram that, like a free-body diagram, shows the forces exerted on a rigid system with their relative magnitudes and directions, and that also shows where each force is exerted relative to the axis of rotation, so that lever arms and torques can be found.
Describing torques from a force diagram
For each force in the diagram, the torque about the chosen axis is found from the force's magnitude, its direction and its point of application; for rotation in a plane, its sense is described as clockwise or counterclockwise about the axis.
Students often think A force's torque depends only on the force itself, its size and direction, not on where on the object it is exerted. In fact Yes. A force's torque depends on its line of action relative to the axis. The same force exerted at a different point (not on the same line of action) has a different lever arm and a different torque.
Students often think A force diagram should include the forces the system exerts on other objects, such as its push on its supports, as well as the forces exerted on it. In fact No. Like a free-body diagram, it includes only forces exerted on the system by other objects. The forces the system exerts on its supports are exerted on the supports.
5.3.B.2 Torque, τ Fix
Torque, τ
A measure of how effectively a force tends to rotate a rigid system about a chosen axis or pivot: τ⃗ = r⃗ × F⃗, with magnitude τ = rF sin θ, where θ is the angle between r⃗ and F⃗. SI unit: N·m.
Position vector of the point of application, r⃗
The vector from the axis of rotation (pivot) to the point where a force is exerted. Its length r equals the lever arm only when the force is perpendicular to r⃗. SI unit: m.
Cross product, A⃗ × B⃗
A vector of magnitude AB sin θ, where θ is the angle between A⃗ and B⃗. Reversing the order reverses the direction: B⃗ × A⃗ = −(A⃗ × B⃗).
Direction of the torque vector
Perpendicular to both r⃗ and F⃗, so normal to the plane that contains them: along the axis about which the force tends to rotate the system.
Right-hand rule for τ⃗ = r⃗ × F⃗
Point the fingers of the right hand along r⃗ and curl them toward F⃗ through the smaller angle between them; the thumb then points along τ⃗. For r⃗ and F⃗ in the plane of the page, a counterclockwise tendency gives a torque out of the page.
Students often think Torque is the same thing as force: forces of equal magnitude exert equal torques, and the larger force always exerts the larger torque. In fact No. Torque depends on the lever arm as well as on the force: equal forces with different lever arms exert different torques, and a smaller force with a longer lever arm can exert the larger torque.
Students often think The torque of a force is found by multiplying matching components of r⃗ and F⃗ and adding them (the dot product). In fact No. Multiplying matching components and adding gives the dot product, r⃗ · F⃗ = xFx + yFy, a scalar. The torque is the cross product; for vectors in the xy-plane, τz = xFy − yFx.
11 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 11
A student pushes on a door at its handle, a distance d from the hinges, with a force of magnitude F perpendicular to the door, exerting a torque τ about the hinges. She then pushes at a point d/2 from the hinges with a force of magnitude 3F directed at 37° to the door's surface. In terms of τ, what is the magnitude of the torque that the second push exerts about the hinges? Use sin 37° = 3/5 and cos 37° = 4/5.
Answer and reasoning
A0.90τCorrect The first push gives τ = dF. For the second, τ′ = r(3F) sin θ with r = d/2 and θ = 37°, the angle between the force and the door: τ′ = (d/2)(3F)(3/5) = 0.90dF = 0.90τ.
B1.50τ A student who ignores the direction of the second push picks this: (d/2)(3F) = 1.50τ. Only the component of the force perpendicular to the door, 3F sin 37°, produces torque.
C1.20τ A student who uses cos 37° picks this: (d/2)(3F)(4/5) = 1.20τ. 3F cos 37° is the component along the door, directed toward the hinges, which produces no torque.
D1.80τ A student who accounts for the size and direction of the second push but not for where it is exerted picks this: 3F sin 37° = 1.80F, compared with F. The second push is exerted half as far from the hinges, which halves the torque.
Working τ = dF. Second push: the angle between r⃗ (along the door) and the force is 37°, so τ′ = (d/2)(3F) sin 37° = (3/2)(3/5)dF = 0.90dF = 0.90τ. (Angle ignored: 1.50τ. Cosine used: 1.20τ. Point of application ignored: 3 × 3/5 = 1.80τ.)
The force diagram shows a rod that can rotate about a fixed axle at O, seen from above, and three forces, A, B and C, exerted on it perpendicular to the rod. The labels and arrow lengths show the forces' relative magnitudes, and the labeled tick marks show distances from O. Which ranking of the magnitudes of the torques about O is correct?
Answer and reasoning
AτC > τB > τA A student who ranks the torques by the sizes of the forces picks this. C is the largest force, but it is exerted closest to O; torque depends on the product of force and lever arm.
BτA > τB > τC A student who ranks the torques by distance from O alone picks this. A is exerted farthest out but is the smallest force, and its torque, 4Fd, equals B's.
CτA = τB > τCCorrect Each force is perpendicular to the rod, so its torque is its magnitude times its distance from O: A, F × 4d = 4Fd; B, 2F × 2d = 4Fd; C, 3F × d = 3Fd. So τA = τB > τC.
DτA > τB = τC A student who adds each force's multiple of F to its multiple of d picks this: A, 1 + 4 = 5; B, 2 + 2 = 4; C, 3 + 1 = 4. Torque is a product, not a sum: A and B both give 4Fd, and C gives 3Fd.
Working Each force is perpendicular to the rod, so τ = F × (distance from O). A: F × 4d = 4Fd. B: 2F × 2d = 4Fd. C: 3F × d = 3Fd. τA = τB > τC. (By force alone: C > B > A. By distance alone: A > B > C. Adding the multiples of F and d: A 5, B 4, C 4.)
A horizontal uniform beam is attached to a wall by a hinge at its left end and held up by a cable attached to its right end. To find the torques exerted on the beam about the hinge, a student draws the diagram shown. Which statement explains why the diagram is not suitable for this purpose?
Answer and reasoning
AIt does not show where on the beam each force is exerted, so no lever arm can be found.Correct All three forces are drawn from one dot, as in a free-body diagram for an object. A torque depends on where a force is exerted relative to the axis, so a force diagram for torques draws each force from its point of application: the hinge force at the hinge, the gravitational force at the beam's center and the cable's force at the right end.
BIt does not split each force into components parallel and perpendicular to the beam. A student who thinks a force diagram must show components picks this. A force diagram shows each force as a single arrow; components can be found afterward, and leaving them out is not what prevents the torques from being found.
CIt does not include a curved arrow to show the torque that each force exerts on the beam. A student who thinks torques are drawn as separate items picks this. Torques are not separate interactions; they are worked out from the forces once each force is drawn at its point of application.
DIt draws the arrows with different lengths, but only the directions of forces matter. A student who thinks arrow lengths carry no information picks this. A force diagram shows the forces' relative magnitudes, so drawing arrows of different lengths is correct.
A force F⃗ = (6.00î + 1.00ĵ) N is exerted on a rigid object at a point whose position relative to the object's axis of rotation is r⃗ = (0.300î + 0.400ĵ) m. The axis is perpendicular to the xy-plane. What is the magnitude of the torque exerted by the force about the axis?
Answer and reasoning
A2.20 N·m A student who multiplies matching components and adds, xFx + yFy = (0.300)(6.00) + (0.400)(1.00), picks this. That is the dot product r⃗ · F⃗, a scalar; torque is the cross product r⃗ × F⃗.
B3.04 N·m A student who multiplies the magnitudes, |r⃗||F⃗| = (0.500 m)(6.08 N), picks this, as if the force were perpendicular to r⃗. Only the part of the force perpendicular to r⃗ produces torque, so the angle between the vectors matters.
C2.70 N·m A student who adds the magnitudes of the two components' torques, 0.300 N·m + 2.40 N·m, picks this. One is counterclockwise and the other clockwise, so they partly cancel: 2.40 − 0.300 = 2.10 N·m.
D2.10 N·mCorrect For vectors in the xy-plane, τz = xFy − yFx = (0.300 m)(1.00 N) − (0.400 m)(6.00 N) = 0.300 N·m − 2.40 N·m = −2.10 N·m. The y-component tends to turn the object counterclockwise and the x-component clockwise, so their torques partly cancel; the magnitude is 2.10 N·m.
A uniform rod of mass M and length L is pivoted at its lower end and held at rest at 37° above the horizontal. What is the magnitude of the torque exerted on the rod by the gravitational force about the pivot? Take g as the acceleration due to gravity, and use sin 37° = 3/5 and cos 37° = 4/5.
Answer and reasoning
A0.30 MgL A student who uses sin 37° picks this: Mg(L/2)(3/5). With the rod at 37° above the horizontal, the perpendicular distance to the vertical line of action is (L/2)cos 37°, so the factor is 4/5.
B0.40 MgLCorrect The gravitational force Mg is exerted at the rod's center, L/2 from the pivot. Its lever arm is the horizontal distance from the pivot to its vertical line of action, (L/2)cos 37°, so τ = Mg(L/2)(4/5) = 0.40 MgL.
C0.80 MgL A student who places the gravitational force at the far end of the rod picks this: Mg × L cos 37°. For a uniform rod, the gravitational force can be treated as exerted at the center of mass, L/2 from the pivot.
D0.50 MgL A student who multiplies Mg by the distance to the center, L/2, without accounting for the direction of the force picks this. The force is not perpendicular to the rod, so the lever arm is (L/2)cos 37°, not L/2.
Working The gravitational force Mg is exerted at the center of mass, r = L/2 from the pivot. The angle between r⃗ (along the rod) and the downward force is 90° + 37° = 127°, and sin 127° = cos 37°, so τ = (L/2)Mg cos 37° = (1/2)(4/5)MgL = 0.40 MgL. Equivalently, the lever arm is the horizontal distance (L/2)cos 37°. (sin 37°: 0.30 MgL. Force at the far end: L cos 37° → 0.80 MgL. Angle ignored: 0.50 MgL.)
A rope pulls on a lever with tension of magnitude FT at a distance r from the pivot, perpendicular to the lever, exerting a torque τ about the pivot. The rope is moved to a point 2r from the pivot and now pulls at 30° to the lever with the same tension. In terms of τ, what is the magnitude of the torque that the rope now exerts about the pivot?
Answer and reasoning
A2τ A student who doubles the torque for the doubled distance but ignores the new angle picks this. At 30° only FT sin 30° = FT/2 is perpendicular to the lever, which cancels the effect of the doubled distance.
B√3 τ A student who uses cos 30° picks this: (2r)FT cos 30° = √3 rFT. FT cos 30° is the component along the lever, toward the pivot, which produces no torque.
Cτ/2 A student who accounts for the change in the rope's direction but not for where it is attached picks this: rFT sin 30° = τ/2. The rope now pulls twice as far from the pivot, which doubles the torque of the perpendicular component.
DτCorrect Only the component of the tension perpendicular to the lever produces torque. At 2r and 30°, τ′ = (2r)FT sin 30° = (2r)(FT/2) = rFT, the same torque as before.
Working τ = rFT. New torque: only the component of the tension perpendicular to the lever, FT sin 30° = FT/2, produces torque: τ′ = (2r)(FT/2) = rFT = τ. (Angle ignored: (2r)FT = 2τ. Cosine used: (2r)FT cos 30° = √3 τ. New point of application ignored: FT sin 30° at r gives τ/2.)
A uniform ladder of length L leans against a smooth vertical wall, making an angle of 53° with the floor. The wall exerts a horizontal normal force of magnitude FN on the top of the ladder. What is the magnitude of the torque exerted by this force about the bottom of the ladder? Use sin 53° = 4/5 and cos 53° = 3/5.
Answer and reasoning
A0.60 FN L A student who uses the horizontal distance from the bottom to the top of the ladder, L cos 53°, picks this. The force is horizontal, so the perpendicular distance to its line of action is the vertical height, L sin 53°.
B1.00 FN L A student who multiplies FN by the distance L to the top of the ladder picks this, as if the force were perpendicular to the ladder. It is horizontal, at 53° to the ladder, so only FN sin 53° is perpendicular to it.
C0.80 FN LCorrect The lever arm is the perpendicular distance from the bottom of the ladder to the horizontal line of action through the top: the height of the top, L sin 53° = 0.80L. So τ = 0.80 FN L.
D0.40 FN L A student who draws the wall's force at the ladder's center of mass picks this, using half the height. The normal force is exerted where the ladder touches the wall, at the top.
Working The force is horizontal and is exerted at the top of the ladder, so its line of action is a horizontal line at the height of the top, L sin 53° = (4/5)L above the floor. Lever arm about the bottom: 0.80L, so τ = 0.80 FN L. (Horizontal distance L cos 53°: 0.60 FN L. Distance L, angle ignored: 1.00 FN L. Force placed at the center: (L/2) sin 53° → 0.40 FN L.)
The figure shows a lever in the plane of the page with its axle at O. The position vector r⃗ from O to point P and the force F⃗ exerted at P both lie in the plane of the page. What is the direction of the torque τ⃗ = r⃗ × F⃗ about O?
Answer and reasoning
AInto the page, perpendicular to r⃗ and F⃗ A student who starts the right-hand rule from F⃗ and curls toward r⃗ picks this, which gives F⃗ × r⃗ = −(r⃗ × F⃗). The order matters: for torque the fingers start along r⃗.
BOut of the page, perpendicular to r⃗ and F⃗Correct Point the fingers of the right hand along r⃗ (to the right) and curl them toward F⃗ (up the page): the thumb points out of the page. This matches the counterclockwise turning that the force tends to give the lever.
CUp the page, in the same direction as F⃗ A student who gives the torque the direction of the force picks this. τ⃗ = r⃗ × F⃗ is perpendicular to F⃗, so it cannot point along it.
DWithin the plane of the page, between r⃗ and F⃗ A student who treats the cross product like a vector sum picks this. A cross product is perpendicular to the plane of its two vectors, so here it points out of or into the page.
The figure shows, seen from above, a flat plate that can rotate about a fixed axle at O, drawn on a square grid, and three forces, F₁, F₂ and F₃, of equal magnitude exerted on it in the plane of the page. Which ranking of the magnitudes of the torques τ₁, τ₂ and τ₃ about O is correct?
Answer and reasoning
Aτ₁ > τ₃ > τ₂ A student who uses the distance from O to each point of application picks this: 5 squares for F₁, about 4.1 for F₃ and about 2.2 for F₂. F₃ is exerted far from O, but its line of action passes only 1 square from O.
Bτ₃ > τ₁ > τ₂ A student who measures each distance from O along the direction of the force picks this: 4 squares for F₃, 3 for F₁ and 1 for F₂. The lever arm is perpendicular to the force's direction, not along it.
Cτ₁ > τ₂ > τ₃Correct The forces are equal, so the torques rank as the lever arms, the perpendicular distances from O to the lines of action: F₁'s vertical line is 4 squares from O, F₂'s vertical line is 2 squares from O and F₃'s horizontal line is 1 square from O.
Dτ₁ = τ₂ = τ₃ A student who thinks equal forces exert equal torques picks this. Torque also depends on the lever arm, which differs for the three forces.
Working Lever arm = perpendicular distance from O to each line of action, in grid squares: F₁ is vertical through the point 4 squares left of O, so 4; F₂ is vertical through the point 2 squares right of O, so 2; F₃ is horizontal through the point 1 square below O, so 1. Equal forces, so τ₁ > τ₂ > τ₃. (Distances from O to the points of application: 5, about 2.2 and about 4.1 squares. Distances along the forces' directions: 3, 1 and 4 squares.)
A uniform rod of mass M and length L is held horizontal and can rotate about an axle through a point L/4 from its left end. What is the magnitude of the torque exerted on the rod by the gravitational force about the axle? Take g as the acceleration due to gravity.
Answer and reasoning
A0.50 MgL A student who measures the distance to the center from the rod's left end instead of from the axle picks this: Mg(L/2). Distances for torque are measured from the axle, which is L/4 from that end.
B0.25 MgLCorrect For torque, the gravitational force on the uniform rod is exerted at its center, L/2 from the left end and so L/2 − L/4 = L/4 from the axle. The force is perpendicular to the rod: τ = Mg(L/4) = 0.25 MgL.
C0.75 MgL A student who places the gravitational force at the far end of the rod, 3L/4 from the axle, picks this. For a uniform rod it is exerted at the center, L/4 from the axle.
D0.31 MgL A student who splits the rod at the axle and adds the magnitudes of the two parts' torques picks this: (3M/4)g(3L/8) + (M/4)g(L/8) ≈ 0.31 MgL. The two torques are in opposite senses, so they subtract: 0.28 MgL − 0.03 MgL = 0.25 MgL.
Working The gravitational force Mg is exerted at the center of mass, L/2 from the left end, which is L/2 − L/4 = L/4 from the axle. The force is perpendicular to the rod: τ = Mg(L/4) = 0.25 MgL. (Distance measured from the left end, L/2: 0.50 MgL. Force at the far end, 3L/4 from the axle: 0.75 MgL. Rod split at the axle and the two torque magnitudes added: (3/4)Mg(3L/8) + (1/4)Mg(L/8) = (10/32)MgL ≈ 0.31 MgL.)
A thin rod of length L and total mass M is held horizontal and can rotate about a horizontal axle through its end B, perpendicular to the rod. Its linear mass density is λ = λ₀x/L, where x is the distance from its other end, A, and λ₀ is a positive constant. What is the magnitude of the torque exerted on the rod by the gravitational force about the axle? Take g as the acceleration due to gravity.
Answer and reasoning
A0.33 MgLCorrect Each element dm = λ dx is a distance L − x from B, so τ = ∫₀ᴸ (L − x)gλ₀(x/L) dx = gλ₀L²/6. With M = λ₀L/2 this is (1/3)MgL ≈ 0.33 MgL: the mass is concentrated toward B, so the center of mass is only L/3 from the axle.
B0.67 MgL A student who measures each element's distance from end A, where x = 0, instead of from the axle picks this: ∫₀ᴸ x gλ₀(x/L) dx = (2/3)MgL. The axle is at B, so the lever arm of the element at x is L − x.
C0.50 MgL A student who takes the gravitational force to be exerted at the rod's midpoint picks this: Mg(L/2). That holds for a uniform rod; here the density increases toward B, so the center of mass is L/3 from B.
D1.00 MgL A student who uses the whole length of the rod as the distance picks this: MgL, as though the whole weight acted at end A. Each element contributes according to its own distance from B, and most of the mass is near the axle.
Working M = ∫₀ᴸ λ₀x/L dx = λ₀L/2. The element dm = λ dx at x is a distance L − x from the axle at B, and the gravitational force on it, g dm, is perpendicular to the horizontal rod, so dτ = (L − x)gλ₀(x/L) dx. τ = (gλ₀/L)∫₀ᴸ (Lx − x²) dx = gλ₀L²/6 = (1/3)MgL ≈ 0.33 MgL. Equivalently, xcm = (∫x λ dx)/M = 2L/3 from A, which is L/3 from B, and τ = Mg(L/3). Distractors: distances measured from A, ∫₀ᴸ x g λ dx = (2/3)MgL ≈ 0.67 MgL; weight at the midpoint, Mg(L/2) = 0.50 MgL; weight at the far end A, 1.00 MgL. Key and distractors checked with sympy.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
