Torque: Study Guide (Physics C: Mechanics 5.3)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 5.3 called Torque. This guide is the one for the calculus-based course. It uses the same three ways of finding the size of a torque, and it adds the vector form, τ = r × F, with a direction found by the right-hand rule. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

What makes a rigid system turn

Topics 5.1 and 5.2 described rotation. Now you ask what changes it. A rigid system is one whose shape does not change, so every part turns through the same angle about the axis. Push on it and it may move along, turn, or both. The turning effect of a force about an axis is its torque, τ.

Three things set the torque of one force:

• How big the force is. Double F and τ doubles.
• Where it acts. The position vector r runs from the axis to the point where the force is exerted. Push far from the axis and the torque is larger.
• Which way it points. Push straight towards or away from the axis and nothing turns, however hard you push.

So torque is always torque about a chosen axis. Before any calculation, write the axis down, for example “axis: the hinge line at O”.

Only the perpendicular component turns

Split the force into two components, relative to r:

• F∥ = F cos θ, along r. It pulls straight away from (or pushes towards) the axis. It exerts no torque.
• F⊥ = F sin θ, perpendicular to r. This is the only part that turns the object.

Here θ is the angle between r and F when the two vectors are drawn tail to tail. That gives the size of the torque:

τ = rF⊥ = rF sin θ

There is a second way to read the same product. Extend the force in both directions to get its line of action. The lever arm r⊥ is the perpendicular distance from the axis to the line of action. Geometry gives r⊥ = r sin θ, so

τ = r⊥F

All three forms give the same number. Pick the one that suits the information you have. If you can see the line of action and the axis, the lever arm is often quickest. If you know r and θ, use rF sin θ.

Two special cases follow at once:

• θ = 90°: the force is perpendicular to r and τ = rF, the largest torque that force can give at that point.
• θ = 0° or 180°, or r = 0: the line of action passes through the axis and τ = 0. A hinge or axle force always exerts zero torque about its own hinge or axle.

Force diagrams for rigid systems

For a point object, a free-body diagram shows forces at a single dot. For a rigid system that can turn, where a force acts matters, so you draw a force diagram instead. It is like a free-body diagram with three extra rules:

1. Draw the object’s outline, roughly to scale, and mark the axis.
2. Start each force arrow at the point where that force is exerted, with its length showing the relative size and its direction correct.
3. Label distances from the axis to each point of application, or to each line of action.

Weight acts at the centre of mass when g is uniform (Topic 2.1), so a uniform rod’s weight arrow starts at its midpoint. Contact forces start where the contact is. A rope’s pull starts where the rope is tied. Figure 2 below is a force diagram.

Torque as a vector: the cross product

Torque has a direction as well as a size. It is defined by the cross product of r and F:

τ = r × F

For any two vectors A and B with angle θ between them (tail to tail), the cross product A × B is a vector with:

• Size |A × B| = AB sin θ. That is why τ = rF sin θ.
• Direction perpendicular to both A and B, so normal to the plane they define.
• Sense given by the right-hand rule: point the fingers of your right hand along A, curl them towards B through the smaller angle, and your thumb points along A × B.

Order matters: B × A = −(A × B). Always write r first, then F.

For forces in the plane of the page, r × F points straight out of the page or straight into it. Use the right-hand rule a second way to link that to the turning: curl your fingers the way the object would turn, and your thumb gives the torque direction. So:

• out of the page (+z) ↔ tends to turn the object counterclockwise;
• into the page (−z) ↔ tends to turn it clockwise.

The usual sign convention is counterclockwise positive, matching +z out of the page. State your convention whenever you add torques.

Working with components. With unit vectors, î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ, and any vector crossed with itself is zero. For r = xî + yĵ and F = F_xî + F_yĵ in the xy-plane, this gives

τ = (xF_y − yF_x) k̂

Each term is a “distance times perpendicular component”. The term xF_y is the turning effect of F_y with lever arm x. The term yF_x is the turning effect of F_x with lever arm y, and its minus sign shows that a positive F_x above the axis turns things clockwise.

Net torque about an axis

Several forces usually act at once. The net torque about an axis is the sum of the individual torques about that same axis, with signs. For forces in one plane, add the z-components:

τ_net = τ₁ + τ₂ + τ₃ + … (counterclockwise +, clockwise −)

Torques about different axes must never be added together. Topics 5.5 and 5.6 use the net torque to decide whether rotation stays steady or changes.

Worked example 1: a guy rope on a tent pole

Question. A tent pole 2.5 m tall is hinged at its base, which is the axis. A guy rope tied to the top pulls with a tension of 140 N. The angle between the pole and the rope is 25°. Find the size of the torque the rope exerts about the base three ways, and the largest torque a 140 N pull at the top could give.

1. τ = rF sin θ = 2.5 m × 140 N × sin 25° = 2.5 × 140 × 0.4226 = 148 N·m (147.9 N·m before rounding).
2. Perpendicular component. F⊥ = 140 sin 25° = 59.2 N. τ = rF⊥ = 2.5 × 59.2 = 148 N·m. The component along the pole, 140 cos 25° = 127 N, presses the pole into its base and turns nothing.
3. Lever arm. r⊥ = 2.5 sin 25° = 1.06 m. τ = r⊥F = 1.06 × 140 = 148 N·m.
4. Largest torque. A pull at right angles to the pole gives τ = rF = 2.5 × 140 = 350 N·m.

Interpretation. A guy rope at a small angle to the pole uses less than half of its pull for turning. Most of the tension squeezes the pole downwards.

Worked example 2: the cross product in components

Question. A flat bracket can turn about an axle along the z-axis, through the origin. Take +x to the right and +y up the page, so +z points out of the page. A force F = (−24î + 18ĵ) N acts at the point r = (1.2î + 0.50ĵ) m. Find τ = r × F, say which way it tends to turn the bracket, and find the lever arm.

1. Components. τ_z = xF_y − yF_x = (1.2)(18) − (0.50)(−24) = 21.6 + 12.0 = 33.6 N·m. So τ = +33.6k̂ N·m.
2. Direction. +k̂ is out of the page, so the force tends to turn the bracket counterclockwise. Check with the right-hand rule: fingers along r (right and a little up), curled towards F (up and to the left). Your thumb points out of the page.
3. Check with rF sin θ. |r| = √(1.2² + 0.50²) = 1.3 m and |F| = √(24² + 18²) = 30 N. Then sin θ = 33.6 ÷ (1.3 × 30) = 0.862. From the dot product, cos θ = (r · F)/(rF) = (−28.8 + 9.0) ÷ 39 = −0.508, so θ = 120.5°. Sin 120.5° = sin 59.5° = 0.862, which matches.
4. Lever arm. r⊥ = τ ÷ F = 33.6 ÷ 30 = 1.12 m, a little less than |r| = 1.3 m, as it must be.

Check the order. F × r would give −33.6k̂ N·m, the wrong sense. Always put r first.

Worked example 3: net torque about two different axes

Question. A uniform plank 3.0 m long lies on smooth ice. You look down on it from above (Figure 2). Take +x along the plank to the right, +y across it (up the page) and counterclockwise positive. Three people pull on ropes: A, 40 N in +y at the left end; B, 60 N in −y at 2.0 m from the left end; C, 50 N at the right end, with components 30 N in +x and 40 N in +y. Find the net torque (a) about axis 1 at the left end and (b) about axis 2 at the midpoint.

(a) Axis 1, the left end.

• A acts at the axis, so r = 0 and τ_A = 0.
• B: lever arm 2.0 m, pushing towards −y on the right of the axis, so clockwise. τ_B = −(2.0)(60) = −120 N·m.
• C: the 30 N component lies along the plank, through the axis, so it gives no torque. The 40 N component has lever arm 3.0 m and turns the plank counterclockwise. τ_C = +(3.0)(40) = +120 N·m. (Check: rF sin θ = 3.0 × 50 × 0.80 = 120 N·m.)
• Net torque about axis 1: 0.

(b) Axis 2, the midpoint. Use τ_z = xF_y − yF_x with x measured from the midpoint. Every force acts on the plank’s centre line, so y = 0 throughout.

• A: x = −1.5 m, F_y = +40 N. τ_A = (−1.5)(40) = −60 N·m (clockwise: an upward pull on the left half).
• B: x = +0.50 m, F_y = −60 N. τ_B = (0.50)(−60) = −30 N·m.
• C: x = +1.5 m, F_y = +40 N. τ_C = (1.5)(40) = +60 N·m.
• Net torque about axis 2: −30 N·m, that is, 30 N·m clockwise.

Interpretation. The net torque is zero about one axis and not about another. That is possible because the net force here is not zero: it is (30î + 20ĵ) N. The lesson for later topics: a zero net torque about one point does not by itself tell you the plank will not start to turn. You must state the axis, and Topic 5.5 shows when the choice stops mattering.

Common misconceptions

• “Torque = force × distance to the point.” Only if the force is perpendicular to r. In general τ = rF sin θ (Worked example 1).
• “A bigger force always gives a bigger torque.” A large force aimed at the axis gives zero torque (the 30 N component of C in Worked example 3).
• Measuring the lever arm to the point of application. The lever arm is the perpendicular distance to the line of action, not the distance r.
• Writing F × r. The order reverses the direction. The definition is r × F.
• “The torque vector points the way the object moves.” It points along the axis, perpendicular to the plane of r and F.
• Mixing axes. Every torque in one sum must be about the same axis.
• Forgetting signs. A clockwise torque is negative under the counterclockwise-positive convention. State the convention first.
• Writing N·m as J. Same base units, different quantity.

Where this leads

Topic 5.4, Rotational Inertia, describes how hard it is to change an object’s rotation. Topic 5.5 then uses net torque for rotational equilibrium, and Topic 5.6 links net torque, rotational inertia and angular acceleration. Try the practice questions now, then use the revision notes and the checklist. You can also go back to Topic 5.2 or the course roadmap.