Physics

1.16 - Moments - The Turning Effect

9 min

What a moment is

Calculating a moment

Clockwise and anticlockwise moments

Balancing a seesaw

note 1 of 5

A moment is the turning effect of a force

When a force acts on an object that can turn about a fixed point (the pivot), it produces a turning effect called the moment of the force.
Everyday examples include pushing a door open, loosening a nut with a spanner, and using a wheelbarrow.

Formula for the moment of a force

$M = F d$

Where:

$M$ = Moment of the force, in newton metres (Nm)
$F$ = Force, in newtons (N)
$d$ = Perpendicular distance from the pivot to the line of action of the force, in metres (m)

A spanner gripping a nut labelled pivot, with a downward arrow labelled force at the end of the handle and a double arrow labelled perpendicular distance between the force and the pivot.

What changes the size of a moment

Larger force - Increases the moment.
Greater distance from the pivot - Increases the moment.
Angle of the force - Applying the force at right angles (90°) gives the largest moment. Any other angle reduces the perpendicular distance to the pivot.

Moments have a direction

A moment turns an object either clockwise or anticlockwise, depending on the direction of the force and which side of the pivot it acts on.

note 2 of 5

Worked example - Moment of a force on a nut

A spanner is used to undo a tight nut. A force of 50 N is applied to the spanner at a perpendicular distance of 0.3 m from the nut, which acts as the pivot. What is the moment of the force on the nut?

A spanner gripping a hexagonal nut, with a blue arrow pointing down from the end of the handle labelled 50 N and the distance to the nut marked 0.3 m.

Step 1 - Write down the values

Force, $F = 50$ N
Perpendicular distance, $d = 0.3$ m

Step 2 - Formula

$M = F d$

Step 3 - Substitution and calculation

$M = 50 \times 0.3 = 15 Nm$

The moment of the force on the nut is 15 Nm.

note 3 of 5

The principle of moments

An object that can turn about a pivot is balanced (in rotational equilibrium) when the turning effects on each side cancel out.
Principle of moments - When an object is balanced:

Total clockwise moment = total anticlockwise moment

If the clockwise and anticlockwise moments are not equal, there's a resultant moment and the object turns.

A beam balanced on a triangular pivot, with a 450 N weight 1 m to the left of the pivot and a 300 N weight 1.5 m to the right.

note 4 of 5

Worked example - Tension in a supporting rope

A uniform wooden beam 10 m long and weighing 1,600 N rests horizontally on a support 2 m from one end. A rope is attached vertically to the other end and holds the beam so that it balances. What is the tension in the rope?

The support acts as the pivot. The weight acts downwards at the centre of mass (the middle of the beam); the rope tension acts upwards at the far end.

A wooden beam on a pivot 2 m from its left end, with a 1,600 N arrow down at its centre and a rope at the right end pulling up with an unknown tension, the spans marked 2 m, 8 m and 10 m.

Step 1 - Find the perpendicular distances from the pivot

Centre of mass is at the middle of the 10 m beam - 5 m from each end. The pivot is 2 m from one end, so the weight acts 5 - 2 = 3 m from the pivot.
The rope is at the far end, 10 - 2 = 8 m from the pivot.

Step 2 - Apply the principle of moments

$W \times d_{W} = T \times d_{T}$

$1,600 \times 3 = T \times 8$

Step 3 - Rearrange and solve for T

$T = \frac{1,600 \times 3}{8} = 600 N$

The tension in the rope is 600 N.

note 5 of 5

Worked example - Reaction forces on a beam

A light beam 8.0 m long rests on two supports, A and B, one at each end. A load of 1,200 N hangs from the beam 5.0 m from support A (3.0 m from support B). Find the upward force from support A ( $R_{A}$ ) and from support B ( $R_{B}$ ). (The beam is light, so its own weight is ignored.)

A horizontal bar labelled light beam resting on two supports from below, the left labelled support B and the right support A, with a 1,200 N arrow pointing down between them, 3 m from B and 5 m from A.

Step 1 - Take support B as the pivot and apply the principle of moments

Taking moments about B removes $R_{B}$ because its distance from B is zero:

$R_{A} \times 8.0 = 1,200 \times 3.0$

Step 2 - Solve for R_A

$R_{A} = \frac{1,200 \times 3.0}{8.0} = 450 N$

Step 3 - Use the balance of vertical forces to find R_B

The upward forces must balance the downward load:

$R_{A} + R_{B} = 1,200$

$R_{B} = 1,200 - 450 = 750 N$

Support A pushes up with 450 N and support B with 750 N.

key terms

TermDefinition

MomentThe turning effect of a force, equal to the force multiplied by the perpendicular distance from the pivot to the line of the force.

PivotThe fixed point that an object turns about.

Principle of momentsWhen an object is balanced, the total clockwise moment about a pivot equals the total anticlockwise moment.

end of notes

Key points to remember

Take a quiz View flashcards

1.17 - Moments - SeesawsNext up