Moments and couples

AS · 16 min

A force can do two things to an object: make it accelerate, and make it turn. Newton's laws handle the first; this note handles the second. It covers the centre of gravity, the moment of a force, couples and the torque of a couple. The two definitions here are examined word for word in Paper 2, and Paper 1 regularly sets questions where the only difficulty is finding the correct perpendicular distance.

Centre of gravity

Every particle of an object has weight. For a ruler, a ladder or a bridge, those tiny weights are spread along its whole length. Adding up the turning effect of thousands of small weights would be impractical, so we replace them all with a single force.

Definition

The centre of gravity of an object is the point at which the whole weight of the object may be considered to act.

For a uniform object (one whose mass is spread evenly) of regular shape, the centre of gravity is at the geometric centre: the midpoint of a uniform rod or metre rule, the centre of a uniform disc, the intersection of the diagonals of a uniform rectangular sheet. For a non-uniform object, such as a baseball bat or a crane jib with an engine at one end, the centre of gravity is shifted towards the heavier part and has to be found by experiment or by calculation.

The centre of gravity does not have to lie inside the material. The centre of gravity of a ring is at its centre, in the empty space.

Tip

The centre of mass is the point at which the whole mass may be considered to be concentrated. In a uniform gravitational field, which is the only situation at AS Level, the centre of mass and the centre of gravity are the same point. Questions use the two names interchangeably.

Finding the centre of gravity of a flat sheet (lamina)

Apparatus. An irregular sheet of card with three small holes punched near its edge; a pin clamped horizontally in a stand; a plumb line (a thread with a small mass on the end); a pencil and ruler.

Method.

Hang the card from the pin through one hole so that it swings freely, and let it come to rest.
Hang the plumb line from the same pin. Mark two points on the card along the thread and join them with a ruled line.
Repeat, hanging from the other two holes.
The centre of gravity is where the lines cross.

Why it works. A freely suspended object comes to rest with its centre of gravity directly below the point of suspension. In any other position the weight would have a moment about the pin and would turn the card. So each vertical line passes through the centre of gravity.

Sources of error and improvements. Friction at the pin stops the card hanging truly freely: use a smooth thin pin and tap the stand before marking. A thick pencil line or a thread that is not in contact with the card gives an uncertain line: keep the thread close to the card and use a sharp pencil. The third line is a check: if the three lines do not meet at a point, the small triangle they form shows the uncertainty.

The moment of a force

Try to open a door by pushing near the hinge and it barely moves. Push at the handle, the same distance from the hinge, but at a steep angle to the door, and it is still hard. Push at the handle, at right angles to the door, and it swings easily. The turning effect depends on the size of the force and on how far its line of action passes from the hinge.

Definition

The moment of a force about a point is the product of the force and the perpendicular distance of the line of action of the force from the point.

Key result

\text{moment} = F \times d

where $d$ is the perpendicular distance from the pivot to the line of action of the force. Unit: $\text{N m}$ . A moment is described as clockwise or anticlockwise about the chosen point.

The line of action of a force is the straight line through the point where the force acts, in the direction of the force, extended as far as needed in both directions. Moving a force along its own line of action does not change its moment.

The unit is the newton metre, $\text{N m}$ , written with a space. In base units it is $\text{kg m}^{2}\ \text{s}^{-2}$ . Do not write it as joules: the joule is the same combination of base units, but it is reserved for energy and work. A moment is not an energy.

Forces at an angle

If the force is not perpendicular to the line joining the pivot to the point where it acts, there are two equivalent ways to find the moment. Suppose the force $F$ acts at a distance $r$ from the pivot, at an angle $\theta$ to the line joining the pivot to that point.

A force F acts at distance r from the pivot O, at angle θ to the bar. The dashed line is the line of action extended backwards. The perpendicular distance from O to the line of action is d = r sin θ.

Method 1: perpendicular distance. The perpendicular distance from the pivot to the line of action is $d = r\sin\theta$ , so the moment is $F r\sin\theta$ .

Method 2: perpendicular component. Resolve $F$ into a component $F\sin\theta$ perpendicular to the bar and a component $F\cos\theta$ along it. The component along the bar passes straight through the pivot and has no moment. The perpendicular component has moment $(F\sin\theta)\,r$ .

Both methods give the same answer, $Fr\sin\theta$ . Use whichever makes the geometry clearer. When $\theta = 90^\circ$ the moment is a maximum, $Fr$ ; when $\theta = 0$ the force points straight through the pivot and has no turning effect.

Watch out

The most common error in moments questions is using the distance along the object instead of the perpendicular distance to the line of action. If the force is not at right angles to the object, a $\sin$ or $\cos$ must appear somewhere. Sketch the line of action and drop a perpendicular from the pivot to it before calculating.

Tightening a nut

A mechanic pulls on the end of a spanner of length $0.25\ \text{m}$ with a force of $40\ \text{N}$ at right angles to the spanner. Calculate the moment of the force about the nut.

Solution

The force is perpendicular to the spanner, so the perpendicular distance is the full length.

\text{moment} = Fd = 40 \times 0.25 = 10\ \text{N m}

A force at an angle

A door is $0.80\ \text{m}$ wide. A rope tied to the handle, at the edge of the door, pulls with a force of $60\ \text{N}$ at $30^\circ$ to the plane of the door. Calculate the moment of the force about the hinge.

Solution

Perpendicular distance from the hinge to the line of action:

d = r\sin\theta = 0.80 \times \sin 30^\circ = 0.40\ \text{m}

\text{moment} = Fd = 60 \times 0.40 = 24\ \text{N m}

Check with components: perpendicular component $= 60\sin 30^\circ = 30\ \text{N}$ , and $30 \times 0.80 = 24\ \text{N m}$ . The same.

Several forces: resultant moment

When several forces act, take moments about the same point for each one, decide whether each is clockwise or anticlockwise, and subtract. The resultant moment is the difference between the total anticlockwise and total clockwise moments, and the object starts to turn in the direction of the larger total.

The weight of the object itself is one of the forces. Put it at the centre of gravity, and do not forget it: it is the force most often left out.

An unbalanced see-saw

A uniform plank of length $3.0\ \text{m}$ and weight $120\ \text{N}$ is pivoted at a point $1.0\ \text{m}$ from its left-hand end. A child of weight $200\ \text{N}$ sits at the left-hand end.

(a) Calculate the resultant moment about the pivot and state its direction.

(b) A second child, of weight $100\ \text{N}$ , sits on the right-hand side. How far from the pivot must she sit for the plank to balance?

Solution

(a) The plank is uniform, so its weight acts at its midpoint, $1.5\ \text{m}$ from the left end. That is $0.5\ \text{m}$ to the right of the pivot.

Anticlockwise moment (child on the left): $200 \times 1.0 = 200\ \text{N m}$ .

Clockwise moment (plank's weight on the right): $120 \times 0.5 = 60\ \text{N m}$ .

\text{resultant moment} = 200 - 60 = 140\ \text{N m} \text{ anticlockwise}

(b) For balance the second child must provide an extra $140\ \text{N m}$ clockwise:

100 \times x = 140 \quad\Rightarrow\quad x = 1.4\ \text{m}

to the right of the pivot. This is possible, since the plank extends $2.0\ \text{m}$ to the right of the pivot. (Balancing moments like this is the principle of moments, developed in Equilibrium of forces.)

A rod held at an angle

A uniform rod of length $1.2\ \text{m}$ and weight $30\ \text{N}$ is hinged at its lower end and held at $40^\circ$ above the horizontal. Calculate the moment of its weight about the hinge.

Solution

The weight acts vertically downwards at the midpoint, $0.60\ \text{m}$ along the rod. The perpendicular distance from the hinge to this vertical line of action is the horizontal distance from the hinge to the midpoint:

d = 0.60\cos 40^\circ = 0.460\ \text{m}

\text{moment} = 30 \times 0.460 = 13.8\ \text{N m}

clockwise if the rod points up to the right. Notice that the angle is between the rod and the horizontal, so the perpendicular distance to a vertical force uses $\cos 40^\circ$ , not $\sin 40^\circ$ . Always work out which side of the triangle is perpendicular to the force rather than reaching for a formula.

Couples

Turn a steering wheel with both hands and you push up with one hand and down with the other. The two forces are equal in size and opposite in direction, so the resultant force on the wheel is zero: it does not accelerate sideways. But both forces turn it the same way, so it rotates. This pair of forces is a couple.

Definition

A couple is a pair of forces that acts to produce rotation only.

The two forces of a couple are equal in magnitude, opposite in direction and parallel but not along the same line (their lines of action are separated).

Each part of that description matters:

Equal and opposite, so the resultant force is zero and there is no linear acceleration (Newton's second law).
Not along the same line, so their moments do not cancel: both turn the object the same way.

A single force can never produce rotation only. A single force off the centre of mass both turns the object and accelerates it.

Definition

The torque of a couple is the product of one of the forces and the perpendicular distance between the forces.

Key result

\text{torque of a couple} = F \times d

where $F$ is the size of one of the two forces and $d$ is the perpendicular distance between their lines of action. Unit: $\text{N m}$ .

A couple: two equal, opposite, parallel forces F whose lines of action are a perpendicular distance d apart. The resultant force is zero; the torque is F × d, clockwise here.

Why it is one force times the separation

Take moments about the centre C, midway between the forces. Each force is a perpendicular distance $d/2$ from C, and both turn the bar clockwise:

\text{total moment} = F\cdot\frac{d}{2} + F\cdot\frac{d}{2} = Fd

Now take moments about any other point, say a point a distance $x$ from the upward force, on the far side from the downward force. The upward force has moment $Fx$ anticlockwise and the downward force has moment $F(x + d)$ clockwise. The resultant is $F(x + d) - Fx = Fd$ clockwise. The answer does not depend on $x$ .

So the torque of a couple is the same about every point. That is why the definition does not mention a pivot, unlike the definition of the moment of a single force.

Watch out

Two errors appear every year:

Using $2Fd$ . If $d$ is the distance between the forces, the torque is $Fd$ . You get $2F \times (d/2)$ only when measuring from the centre, which is the same thing.
Using the distance between the points where the forces act when the forces are not perpendicular to the line joining them. The distance in the definition is the perpendicular distance between the lines of action.

Steering wheel

A driver turns a steering wheel of diameter $0.40\ \text{m}$ by applying forces of $25\ \text{N}$ , tangential to the rim, with each hand at opposite ends of a diameter. Calculate the torque applied.

Solution

The two tangential forces are parallel and opposite, and their lines of action are a diameter apart, so $d = 0.40\ \text{m}$ .

\text{torque} = Fd = 25 \times 0.40 = 10\ \text{N m}

A couple at an angle

Two forces, each of $15\ \text{N}$ , act at opposite ends of a bar of length $0.60\ \text{m}$ . They are parallel, opposite in direction and each at $50^\circ$ to the bar.

(a) Explain why the bar has no linear acceleration.

(b) Calculate the torque of the couple.

Solution

(a) The forces are equal in magnitude and opposite in direction, so the resultant force is zero. By Newton's second law there is no linear acceleration.

(b) The perpendicular distance between the lines of action is the component of the bar's length perpendicular to the forces:

d = 0.60\sin 50^\circ = 0.460\ \text{m}

\text{torque} = Fd = 15 \times 0.460 = 6.89\ \text{N m}

A common wrong answer is $15 \times 0.60 = 9.0\ \text{N m}$ , which uses the length of the bar instead of the perpendicular separation of the lines of action.

Moments compared with couples

	Moment of a force	Torque of a couple
Forces involved	One force	Two equal, opposite, parallel forces
Formula	force $\times$ perpendicular distance from the point to the line of action	one force $\times$ perpendicular distance between the lines of action
Depends on the chosen point?	Yes	No, the same about every point
Resultant force	Equal to the force	Zero
Effect on a free object	Turning and acceleration	Turning only
Unit	$\text{N m}$	$\text{N m}$

Exam tip

"Define the moment of a force" needs three things: the force, the perpendicular distance, and from the point (pivot) to the line of action of the force. "Force times distance" scores nothing.
"Define the torque of a couple": one of the forces times the perpendicular distance between the (lines of action of the) forces. Saying "the sum of the forces" or omitting "perpendicular" loses the mark.
"State what is meant by a couple": a pair of forces producing rotation only, or equivalently, two equal and opposite forces that are not in the same line. Include both "equal" and "not along the same line" to be safe.
In calculations, write the perpendicular distance on its own line with the trigonometry visible ( $d = 0.60\cos 40^\circ = 0.460\ \text{m}$ ). If the final answer is wrong, this line still earns credit.
State the direction (clockwise or anticlockwise) when asked for a resultant moment, and the unit $\text{N m}$ , never $\text{J}$ and never $\text{N m}^{-1}$ .

Summary

The centre of gravity is the point at which the whole weight of an object may be considered to act; for a uniform regular object it is the geometric centre.
Moment of a force $=$ force $\times$ perpendicular distance from the point to the line of action of the force. Unit $\text{N m}$ .
For a force at angle $\theta$ to the line from the pivot, moment $= Fr\sin\theta$ : use the perpendicular distance or the perpendicular component.
Include the object's own weight, acting at its centre of gravity, in every moments calculation.
A couple is a pair of forces that produces rotation only: equal, opposite, parallel, not in the same line, so the resultant force is zero.
Torque of a couple $=$ one force $\times$ perpendicular distance between the forces, and it is the same about any point.

Practice

Question

Define the moment of a force and state its SI unit in base units.
A nut needs a moment of $30\ \text{N m}$ to loosen it. A spanner of length $0.25\ \text{m}$ is used. (a) Calculate the minimum force needed. (b) Calculate the force needed if it is applied at the end of the spanner at $60^\circ$ to the spanner.
State what is meant by a couple, and explain why a couple does not cause the object to accelerate.
A tap handle is $6.0\ \text{cm}$ across. Forces of $15\ \text{N}$ are applied at each end, perpendicular to the handle and in opposite directions. Calculate the torque of the couple.
A uniform metre rule of weight $1.2\ \text{N}$ is pivoted at the $40.0\ \text{cm}$ mark. Find the mass that must be hung at the $5.0\ \text{cm}$ mark for the rule to balance.
A uniform rod of length $2.0\ \text{m}$ and weight $40\ \text{N}$ is hinged at one end and held at $30^\circ$ above the horizontal by a force applied at the other end, perpendicular to the rod. (a) Calculate the moment of the weight about the hinge. (b) Calculate the size of the applied force.
Show that the torque of a couple has the same value about any point on the line joining the two forces.
A cyclist pushes vertically down on a pedal with a force of $300\ \text{N}$ . The crank is $0.17\ \text{m}$ long. Calculate the moment about the axle when the crank is (a) horizontal, (b) at $40^\circ$ below the horizontal, (c) vertical.
A student says: "A force applied at the rim of a wheel always produces the same moment, because the radius is fixed." Explain why the student is wrong, and state the direction in which the force should act to give the largest moment.

Answers

The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. Unit $\text{N m} = \text{kg m}^{2}\ \text{s}^{-2}$ .
(a) The minimum force acts perpendicular to the spanner at its end: $F = 30/0.25 = 120\ \text{N}$ . (b) Only the perpendicular component turns the nut: $F\sin 60^\circ = 120$ , so $F = 120/0.866 = 139\ \text{N}$ .
A couple is a pair of forces that produces rotation only: two equal, opposite, parallel forces whose lines of action do not coincide. The forces are equal and opposite, so the resultant force is zero; by Newton's second law the acceleration of the centre of mass is zero.
Torque $= Fd = 15 \times 0.060 = 0.90\ \text{N m}$ .
The rule's weight acts at the $50.0\ \text{cm}$ mark, $0.100\ \text{m}$ from the pivot: clockwise moment $= 1.2 \times 0.100 = 0.12\ \text{N m}$ . The hanging mass is $0.35\ \text{m}$ from the pivot on the other side: $W \times 0.35 = 0.12$ , so $W = 0.343\ \text{N}$ and $m = 0.343/9.81 = 0.035\ \text{kg}$ ( $35\ \text{g}$ ).
(a) The weight acts at the midpoint, $1.0\ \text{m}$ along the rod. Horizontal (perpendicular) distance $= 1.0\cos 30^\circ = 0.866\ \text{m}$ ; moment $= 40 \times 0.866 = 34.6\ \text{N m}$ . (b) The applied force is perpendicular to the rod at $2.0\ \text{m}$ : $F \times 2.0 = 34.6$ , so $F = 17.3\ \text{N}$ .
Let the forces $F$ be a perpendicular distance $d$ apart and choose a point P a distance $x$ from one force, on the line joining them, outside the pair. One force has moment $Fx$ in one sense; the other has moment $F(x + d)$ in the opposite sense. Resultant $= F(x + d) - Fx = Fd$ . If P lies between the forces, at $x$ from one, both moments are in the same sense: $Fx + F(d - x) = Fd$ . Either way the torque is $Fd$ , independent of $x$ .
The perpendicular distance from the axle to the vertical line of action is the horizontal distance to the pedal. (a) $300 \times 0.17 = 51\ \text{N m}$ . (b) $300 \times 0.17\cos 40^\circ = 39.1\ \text{N m}$ . (c) The line of action passes through the axle: moment $= 0$ . (This is why pedalling is hardest at the top and bottom of the stroke.)
The moment depends on the perpendicular distance from the axle to the line of action, not just the radius. A force applied along a radius (towards or away from the axle) has zero moment. The largest moment, $Fr$ , is produced when the force acts tangentially to the rim, perpendicular to the radius.

Centre of gravity

The moment of a force

Forces at an angle

Several forces: resultant moment

Couples

Why it is one force times the separation

Moments compared with couples

Summary

Practice

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