Couples, Torque and the Conditions for Equilibrium – Moments and Equilibrium of Bodies (Senior 4 Physics) | Ganzaa

Moments and Equilibrium of Bodies

Lessons

Senior 4 · Physics

Moments and Equilibrium of Bodies

4 lessons

Lesson 2 of 4

Couples, Torque and the Conditions for Equilibrium

25 min

Why the bridge does not move

A bridge across the Nyabarongo carries lorries, a builder's plank rests on two trestles under a load of bricks, a painter climbs a ladder leaning on a wall. None of these moves: every force on them is balanced, and so is every turning effect. Engineers design such structures by writing down exactly two conditions and solving them. This lesson gives you those conditions, and first introduces a special pair of forces, the couple, that turns a body without moving it along.

Couples and torque

When you turn a tap, a key, a screwdriver or a steering wheel, you apply two equal forces in opposite directions on opposite sides of the axis (Figure 2.1). Such a pair is a couple. Its resultant force is zero, so it does not accelerate the body as a whole, but it does make it turn.

Figure 2.1: A couple: two equal, opposite, parallel forces whose lines of action are a distance d apart.

The moment of a couple is the same about every point. Take moments about a point P on the bar between the forces, a distance $x$ from one of them and so $(d - x)$ from the other (Figure 2.1, right). Both forces turn the bar the same way (anticlockwise) about P, so

M = F x + F (d - x) = F d

whatever the value of $x$ . (If P lies outside the pair, the two moments have opposite senses, and their difference is again $F d$ .) The moment (or torque) of a couple is

\tau = F\,d

where $d$ is the perpendicular distance between the lines of action. "Torque" is the usual name for a turning effect produced by a couple, for example by an engine on a wheel or by a motor on its shaft.

Definition

A couple is a pair of forces that are equal in size, opposite in direction and parallel, with lines of action that do not coincide. Its torque (moment of the couple) is $\tau = Fd$ , the size of one force times the perpendicular distance between them; it is the same about every point. Example: two hands each pushing with 15 N on opposite sides of a steering wheel of diameter 0.40 m apply a torque of $15 \times 0.40 = 6.0\,\mathrm{N\,m}$ .

The two conditions for equilibrium

A rigid body is in equilibrium when it has no linear acceleration and no angular acceleration. That needs two conditions:

First condition (no resultant force): $\sum F_x = 0$ and $\sum F_y = 0$ . Upward forces balance downward forces; leftward forces balance rightward ones.
Second condition (no resultant torque): $\sum M = 0$ about any point. Total clockwise moment = total anticlockwise moment.

A couple shows why both are needed: it satisfies the first condition but not the second. For forces in one plane these give three independent equations, so you can find up to three unknown forces. A good choice of the point for moments, one where an unknown force acts, removes that unknown from the moment equation.

Three forces. If a body is in equilibrium under three non-parallel forces, their lines of action must pass through one point (otherwise any two of them would have a moment about the meeting point of the other pair), and the three forces drawn tip to tail form a closed triangle.

A strategy for every statics problem

Draw a free-body diagram of the one body you study, showing every force where it acts (weight at the centre of gravity).
Resolve forces and write $\sum F_x = 0$ and $\sum F_y = 0$ .
Choose a pivot point that eliminates an unknown and write $\sum M = 0$ .
Solve, and check that the answers are sensible (a support cannot pull; friction cannot exceed $\mu N$ ).

Worked example 1: a sign on two ropes

A sign of weight 50 N hangs from two ropes making 30° and 60° with the horizontal. Find the tension in each rope.

Horizontal: $T_1\cos 30^\circ = T_2\cos 60^\circ$ , so $T_2 = T_1\frac{\cos 30^\circ}{\cos 60^\circ} = 1.732\,T_1$ .

Vertical: $T_1\sin 30^\circ + T_2\sin 60^\circ = 50 \Rightarrow 0.500\,T_1 + 1.732 \times 0.866\,T_1 = 2.00\,T_1 = 50$ .

T_1 = 25\,\mathrm{N}, \qquad T_2 = 43\,\mathrm{N}

The steeper rope carries more of the load.

Applications: beams, see-saws and cranes

Figure 2.2: Free-body diagram of a loaded beam on two supports.

Worked example 2: reactions at the supports

For the beam in Figure 2.2, find the forces $R_{A}$ and $R_{B}$ exerted by the supports.

Take moments about A (this removes $R_{A}$ ):

R_B \times 4.0 = 200 \times 2.0 + 600 \times 1.0 = 1000 \;\Rightarrow\; R_B = 250\,\mathrm{N}

Vertical forces: $R_{A} + R_{B} = 200 + 600$ , so $R_A = 800 - 250 = 550\,\mathrm{N}$ .

Check by taking moments about B: $R_A \times 4.0 = 200 \times 2.0 + 600 \times 3.0 = 2200$ , giving $R_A = 550\,\mathrm{N}$ . The support nearer the load takes more of it.

Worked example 3: a tower crane's counterweight

A tower crane lifts a load of $2.0 \times 10^{4}\,\mathrm{N}$ at 30 m from its tower. Its counterweight sits 10 m from the tower on the other side. Ignoring the weight of the jib, what counterweight balances the load?

W_c \times 10 = 2.0 \times 10^{4} \times 30 \;\Rightarrow\; W_c = 6.0 \times 10^{4}\,\mathrm{N}

In practice the counterweight is set to balance a typical load, and the crane's rated load falls as the trolley moves out along the jib, because the load's moment grows with distance.

The ladder problem

Figure 2.3: Forces on a ladder leaning on a smooth wall. A smooth surface can only push at right angles to itself.

Worked example 4: will the ladder slip?

The ladder in Figure 2.3 is 5.0 m long, weighs 200 N and leans at 60° to the ground against a smooth wall. (a) Find the forces at the wall and at the foot. (b) Find the least coefficient of friction that stops it slipping. (c) A 600 N painter climbs to 3.0 m along the ladder. Find the new least coefficient of friction.

(a) Moments about the foot (removes $N$ and $F$ ): the wall force acts at height $5.0\sin 60^\circ = 4.33\,\mathrm{m}$ , the weight at horizontal distance $2.5\cos 60^\circ = 1.25\,\mathrm{m}$ .

S \times 4.33 = 200 \times 1.25 \;\Rightarrow\; S = 57.7\,\mathrm{N}; \qquad F = S = 57.7\,\mathrm{N}, \qquad N = W = 200\,\mathrm{N}

(b) Friction can be at most $\mu N$ , so $\mu \ge \frac{F}{N} = \frac{57.7}{200} = 0.29$ .

S \times 4.33 = 200 \times 1.25 + 600 \times 1.5 = 1150 \;\Rightarrow\; S = 266\,\mathrm{N}, \qquad N = 800\,\mathrm{N}, \qquad \mu \ge \frac{266}{800} = 0.33

The higher the painter climbs, the larger the friction needed: a ladder that is safe at the bottom can slip when someone reaches the top. That is why a helper should stand on the foot of the ladder, and why a steeper ladder is safer.

Warning

Common mistakes. (1) Checking only that forces balance: a body can have zero resultant force and still turn (a couple). (2) Forgetting the weight of the beam or ladder itself. (3) Drawing a force at a smooth wall at an angle: a smooth surface pushes only at right angles to itself. (4) Using the length along the ladder instead of the perpendicular distance when taking moments. Safety: on a real ladder, keep it steep (about 4 m up for every 1 m out), have someone foot it, and never stand on the top rungs.

Tip

Take moments about the point with the most unknown forces through it. For a ladder this is the foot: both $N$ and $F$ vanish from the moment equation, leaving only $S$ .

Static equilibrium: the ladder problem (Step by Step Science). Follow how the free-body diagram, the force equations and one moment equation give the friction needed at the foot of a ladder.

✦

Practice

State the two conditions for a rigid body to be in equilibrium, and give an example of a body that satisfies only the first.
A tap handle is turned by a couple of two 12 N forces 6.0 cm apart. Find the torque.
A uniform beam 6.0 m long and weighing 300 N rests on two supports, each 1.0 m from an end. A 500 N load sits 2.0 m from the left end. Find the force on each support.
A 40 N lamp hangs from the midpoint of a wire whose two halves each make 10° with the horizontal. Find the tension in the wire.
A uniform plank of weight 150 N and length 4.0 m projects 1.5 m over the edge of a flat roof, resting on the roof edge and held down by a bolt at its inner end. What force must the bolt provide when a 600 N worker stands at the outer end?
A uniform ladder 4.0 m long and weighing 160 N leans against a smooth wall at 70° to the ground. Find the friction force at the foot.
Explain, using moments, why it is more dangerous to climb to the top of a ladder than to stand on its bottom rung, and suggest two ways to make it safer.

Show answersHide answers

$\sum F = 0$ (in every direction) and $\sum M = 0$ about any point. A steering wheel turned by a couple has zero resultant force but a resultant torque.
$\tau = 12 \times 0.060 = 0.72\,\mathrm{N\,m}$ .
Supports at 1.0 m and 5.0 m from the left end. Moments about the left support: $R_2 \times 4.0 = 300 \times 2.0 + 500 \times 1.0 = 1100$ , so $R_2 = 275\,\mathrm{N}$ and $R_1 = 800 - 275 = 525\,\mathrm{N}$ .
$2T\sin 10^\circ = 40$ , so $T = \frac{40}{2 \times 0.174} = 115\,\mathrm{N}$ : a nearly straight wire needs a large tension.
Moments about the roof edge: the worker gives $600 \times 1.5 = 900\,\mathrm{N\,m}$ one way; the plank's weight acts 0.5 m inside the edge ( $150 \times 0.50 = 75\,\mathrm{N\,m}$ ) and the bolt 2.5 m inside: $B \times 2.5 + 75 = 900$ , so $B = 330\,\mathrm{N}$ downwards.
Moments about the foot: $S \times 4.0\sin 70^\circ = 160 \times 2.0\cos 70^\circ$ , so $S = \frac{160 \times 0.684}{3.76} = 29\,\mathrm{N}$ ; friction $F = S = 29\,\mathrm{N}$ .
The climber's weight has a moment about the foot proportional to his horizontal distance from it, which grows as he climbs; the wall force, and so the friction needed at the foot, grows with it, until friction can no longer supply it and the foot slides out. Safer: set the ladder steeper (smaller horizontal distances), have a person stand on the foot or tie it, use rubber feet on a rough floor (larger $μ \mu$ ).

Quick check

A couple acts on a body that is free to move. What happens?

Key points

A couple is two equal, opposite, parallel forces; its torque $\tau = Fd$ is the same about every point, and its resultant force is zero.
Equilibrium needs $\sum F_x = \sum F_y = 0$ and $\sum M = 0$ about any point; three non-parallel forces in equilibrium are concurrent.
Solve statics problems with a free-body diagram, force equations and one moment equation about a well-chosen point.
On a ladder against a smooth wall, the friction needed at the foot rises as a person climbs.

Sources and credits

9.1: The First Condition for Equilibrium and 9.4: Applications of Statics, Including Problem-Solving Strategies, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Ladder leaning against the wall, Peabody Institute by Eli Pousson (Baltimore Heritage), Wikimedia Commons, CC0 1.0.
Photo: Construction site in Cabo San Lucas (near the port) with a tower crane.jpg by Alexis Doine, Wikimedia Commons, CC0 1.0.
Video: Physics, Torque (12 of 13) Static Equilibrium, Ladder Problem, Step by Step Science, embedded from YouTube.

The diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Why the bridge does not move

2Couples and torque

3The two conditions for equilibrium