Torque: Study Guide (Physics 1 5.3)

This guide is for the algebra-based Physics 1 course. Topics 5.1 and 5.2 described rotation: angles, angular velocity and angular acceleration. Now you start asking what causes a change in rotation. The answer is torque. No calculus is needed.

Rigid systems and the axis of rotation

In this unit we model objects such as doors, spanners, see-saws and wheels as rigid systems. A rigid system keeps its shape: the distance between any two points on it does not change. That lets you describe the whole object with one angle, one angular velocity and one angular acceleration.

Every torque calculation starts with an axis of rotation: the line the object turns about (or could turn about). For a door, the axis runs through the hinges. For a spanner on a bolt, it runs along the bolt. When nothing is fixed, you may choose an axis. Always say which axis you are using, just as you state “+x to the right” in kinematics. A torque has no meaning until the axis is named.

What makes an object turn?

Try opening a heavy door. Three things change how easily it swings:

• How hard you push. A bigger force turns it more readily.
• Where you push. A push of 20 N at right angles at the handle, 0.75 m from the hinges, is far more effective than the same push 0.25 m from the hinges.
• Which way you push. Push straight towards the hinges, along the door, and nothing turns at all, however hard you push.

Torque combines all three. For the door, the push at the handle gives a torque of 0.75 m × 20 N = 15 N·m. The same push near the hinge gives 0.25 m × 20 N = 5.0 N·m: three times the distance, three times the torque. The push along the door gives 0 N·m.

The torque equation

Draw a vector r from the axis to the point of application of the force F, the exact point where the force acts. Let θ be the angle between r and F (put the two vectors tail to tail to measure it). The size of the torque is:

τ = rF sin θ

The unit is the newton-metre, N·m. Torque is not energy, so do not write joules.

The sin θ factor is the key. Here is how it behaves:

Angle θ between r and F	sin θ	Torque
0° (F points straight away from the axis)	0	zero
30° or 150°	0.50	half the maximum
60° or 120°	0.87	87% of the maximum
90° (F at right angles to r)	1	maximum, rF
180° (F points straight at the axis)	0	zero

Two other ways to read the same equation

You can group the factors in two ways. Both give the same number.

1. Perpendicular component. Split F into a part along r and a part at right angles to r. The part along r only pulls or pushes on the axis; it cannot turn anything. The perpendicular part, F⊥ = F sin θ, does all the turning: τ = rF⊥
2. Lever arm. Extend the force arrow into a long straight line in both directions. This is the line of action of the force. The lever arm r⊥ is the perpendicular distance from the axis to that line. Because r⊥ = r sin θ: τ = r⊥F

The lever arm is not the distance from the axis to the point of application (that is r). It is the shortest distance from the axis to the line of action. The two are equal only when the force is at 90° to r.

Clockwise or counterclockwise?

When several forces act, some tend to turn the object one way and some the other. Label each torque clockwise or counterclockwise as seen in your diagram. This bookkeeping lets you add torques in Topics 5.5 and 5.6. You do not need the vector direction of torque in this course; you only work with sizes and this turning sense.

Force diagrams for rigid systems

In Unit 2 you drew free-body diagrams, with every force starting from one dot. That was fine for an object that only moves along. For rotation, where a force acts matters, so you draw a force diagram instead.

A force diagram is like a free-body diagram in most ways:

• Show every force exerted on the system, and only those.
• Label each force with what exerts it, for example “weight of rod (Earth on rod)”.
• Draw the arrows with their correct directions and roughly correct relative lengths.

The difference: each arrow starts at the point where the force acts. Draw the object (often as a simple bar), mark the axis clearly, and mark the distances from the axis. Gravity on a rigid object acts at its centre of mass (Topic 2.1); for a uniform rod that is the middle.

Once the diagram is drawn, go through the forces one by one and ask: what is r, what is θ, and so what is τ about this axis?

Predicting changes: functional dependence

Because τ = rF sin θ is a product, factor-of-change questions are quick:

• Double F (same r and θ): torque doubles.
• Double r (same F and θ): torque doubles.
• Double r and change θ from 90° to 30°: 2 × sin 30° = 1, so the torque is unchanged.
• Halve the lever arm and double the force: torque unchanged.

Write the ratio τ_new / τ_old as a product of factors, and cancel anything that stays the same.

Worked example 1: a spanner at an angle

Question. In Figure 1, the axis is the centre of the bolt. A mechanic pulls on a spanner with a force F = 80 N at a point 0.30 m from the axis. The angle between r and F is 60°. Find the size of the torque on the bolt three ways, and the largest torque this force could give.

1. Equation: τ = rF sin θ = 0.30 m × 80 N × sin 60° = 0.30 × 80 × 0.866 = 20.8 N·m ≈ 21 N·m.
2. Perpendicular component: F⊥ = 80 N × sin 60° = 69.3 N. τ = rF⊥ = 0.30 m × 69.3 N = 21 N·m.
3. Lever arm: r⊥ = r sin θ = 0.30 m × 0.866 = 0.260 m. τ = r⊥F = 0.260 m × 80 N = 21 N·m.
4. Largest torque: at θ = 90°, sin θ = 1, so τ_max = 0.30 m × 80 N = 24 N·m.

Interpretation. Pulling at 60° gives 87% of the best possible torque. The 40 N component along the spanner (80 N × cos 60°) only pulls on the bolt; it does no turning.

Check. All three methods agree, as they must: they are the same product grouped differently.

Worked example 2: identifying torques from a force diagram

Question. A uniform rod of mass 5.0 kg and length 2.0 m is hinged to a wall at its left end and held horizontal (Figure 2). A 3.0 kg lamp hangs from the rod 1.5 m from the hinge. A cable fixed to the right-hand end pulls on the rod with a tension of 93 N, at 30° above the rod. Taking the hinge as the axis, find the torque exerted by each force and state its turning sense.

1. List the forces on the rod (the system): the hinge force, the rod’s weight, the pull of the lamp’s string, and the cable tension. Draw them where they act (Figure 2).
2. Hinge force. It acts at the axis, so r = 0 and τ = 0. You do not need to know its size or direction.
3. Rod’s weight. W = mg = 5.0 kg × 9.8 m/s² = 49 N, acting at the centre, r = 1.0 m, at 90° to the rod. τ = 1.0 m × 49 N = 49 N·m, clockwise.
4. Lamp. The string pulls down with the lamp’s weight, 3.0 kg × 9.8 m/s² = 29.4 N, at r = 1.5 m, at 90°. τ = 1.5 m × 29.4 N = 44 N·m, clockwise.
5. Cable. r = 2.0 m and θ = 30°. τ = 2.0 m × 93 N × sin 30° = 93 N·m, counterclockwise. The lever-arm view gives the same: r⊥ = 2.0 m × sin 30° = 1.0 m, and 1.0 m × 93 N = 93 N·m.

Interpretation. The clockwise torques add to 49 + 44.1 = 93.1 N·m, about the same as the counterclockwise 93 N·m. That is no accident: the rod is not starting to rotate. Topic 5.5 turns this observation into the condition for rotational equilibrium.

Check the choice of axis. If you took the axis at the right-hand end instead, the cable would exert zero torque and the hinge force would not. The torques depend on the axis you choose, which is why you must always state it.

Worked example 3: deriving a torque for a raised barrier

Question. A car-park barrier is a uniform arm of mass M and length L, pivoted at one end. It is raised to an angle φ above the horizontal. (a) Derive an expression for the torque exerted by gravity on the arm about the pivot. (b) Evaluate it for M = 12 kg and L = 4.0 m at φ = 0 and φ = 60°. (c) What happens at φ = 90°?

(a) Gravity acts at the centre of mass, a distance L/2 along the arm from the pivot. The weight Mg points straight down. The line of action is vertical, so the lever arm is the horizontal distance from the pivot to the centre of mass:

r⊥ = (L/2) cos φ

so τ = ½MgL cos φ.

Check with τ = rF sin θ: the angle between r (along the arm, φ above horizontal) and the downward weight is 90° + φ, and sin(90° + φ) = cos φ. Same result.

(b) Mg = 12 kg × 9.8 m/s² = 117.6 N. At φ = 0: τ = ½ × 117.6 N × 4.0 m × 1 = 240 N·m (235.2 N·m). At φ = 60°: τ = 235.2 N·m × cos 60° = 120 N·m (117.6 N·m). The lever arm has shrunk from 2.0 m to 1.0 m.

(c) At φ = 90° the arm is vertical, cos 90° = 0, and the torque is zero: the weight’s line of action passes through the pivot.

Functional dependence. τ is proportional to L for a fixed mass. A barrier twice as long with the same mass needs twice the torque to hold horizontal. If the mass also doubled, the torque would quadruple.

Common misconceptions

• “Torque is just a bigger or smaller force.” Torque depends on force, distance and angle. A 60 N push straight towards a hinge gives no torque; a 30 N push at the edge can give plenty.
• “r is always the lever arm.” r is the distance to the point of application. The lever arm is the perpendicular distance to the line of action. They are equal only at 90° (Figure 1).
• “A large force at the axis must exert a large torque.” A force acting at the axis, or one whose line of action passes through it, exerts zero torque about that axis (Worked example 2, step 2).
• “Further out always means more torque.” Not if the angle changes too: doubling r while going from 90° to 30° leaves τ unchanged.
• Using the angle between F and the object’s surface without checking. θ must be the angle between r and F. If you are given a different angle, convert it first (Worked example 3).
• Drawing every force from the centre of the object. That is a free-body diagram. For torque, each force must start where it acts.
• Writing joules. Torque is in N·m. It is not work or energy.
• Forgetting to state the axis. The same force has different torques about different axes.

Where this leads

Torque is the “cause” in rotational dynamics. Topic 5.4, Rotational Inertia, describes how hard an object is to set turning; Topics 5.5 and 5.6 put torque and rotational inertia together in rotational versions of Newton’s first and second laws. 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.