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Moments and turning effects

Key idea: A force's moment is its size times the perpendicular distance from the pivot to its line of action. A larger force or distance gives a greater turning effect.

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Try opening a door by pushing right next to its hinges, and it barely moves; push the handle with the same force and it swings open. A force can turn an object as well as push it along, and how well it turns depends on where it acts as much as on how big it is. That turning effect, the moment of the force, explains doors, spanners, seesaws and wheelbarrows, and you can calculate it with one multiplication.

The turning effect of a force

An object that can turn rotates about a fixed point called the pivot. A door turns about its hinges. A seesaw turns about its central support.

The moment of a force about a pivot is the turning effect of that force. It is also called the torque. Every moment has a direction:

  • clockwise: the way the hands of a clock turn;
  • anticlockwise: the opposite way.

Name the pivot and the side you view from before you name a direction. A door that turns clockwise seen from above turns anticlockwise seen from below. In a diagram, use the view as drawn.

Two things decide the size of a moment:

  1. the size of the force: a larger force gives a larger turning effect;
  2. how far from the pivot the force acts: the same force gives a larger turning effect when it acts farther from the pivot.

A force whose line passes through the pivot has no turning effect. Push on the edge of an open door straight towards the hinges and it does not swing.

The same push on a door, near the hinge and at the handleA door seen from above. The hinge, which is the pivot, is at the left end. The door is 0.80 m wide. Two equal 10 N pushes act at right angles to the door: one 0.20 m from the hinge and one at the handle, 0.80 m from the hinge. The push near the hinge gives a moment of 10 × 0.20 = 2.0 N m. The push at the handle gives 10 × 0.80 = 8.0 N m.A door seen from abovehinge(pivot)handle10 N10 N0.20 m0.80 mmoments: 10 × 0.20 = 2.0 N m and 10 × 0.80 = 8.0 N m
Scroll across the figure to read all labels.
The same force turns the door more when it acts farther from the hinge.

Check your understanding 1 A pull through the pivot

A 30 N pull acts along a line through a lever’s pivot. One student doubles the pull without changing its line; another moves the line so that it misses the pivot. Which change can give a turning effect, and why?

Show answer

Moving the line away from the pivot can give a turning effect. Doubling the pull along the original line still gives no moment because that line passes through the pivot. Increasing force alone cannot create a turning effect when its line still goes through the pivot.

Turning effects in everyday objects

Each object below uses distance from the pivot to make a job easier. To explain any of them, name the pivot, say where the force acts, and say how that distance changes the turning effect.

Pivots and forces in everyday objects
ObjectPivotWhere you apply the forceWhy it helps
DoorThe hingesThe handle, on the edge farthest from the hinges

The handle is far from the pivot, so a small push gives a large moment. Pushing near the hinges needs a much larger push.

SpannerThe centre of the nutThe end of the handle

A longer spanner puts your hand farther from the nut, so the same push gives a larger moment on a tight nut.

SeesawThe central supportThe weight of each person sitting on it

A person’s weight has a larger turning effect when they sit farther from the centre, so a lighter child can balance a heavier one by sitting farther out.

WheelbarrowThe wheel’s axleAn upward lift at the ends of the handles

The load sits near the wheel and your hands are far from it, so a small lift at the handles turns the barrow about the wheel and raises a heavy load.

Bottle openerWhere the tip of the opener presses on top of the capAn upward lift at the end of the handle

Your hand is far from the pivot and the lip that grips the edge of the cap is close to it, so a small lift makes the lip pull hard on the cap.

Check your understanding 2 Pivots, distances and turning directions

You push an open door at right angles with the same force at three places. Where does it turn most easily?
Why is a tight nut easier to undo with a longer spanner?
A seesaw is drawn side on with its pivot in the middle. A child pushes down on the right-hand end. Which way does the push turn the seesaw, as drawn?
Show answer

The door turns most easily at the edge farthest from the hinges: the same perpendicular force has its greatest distance from the pivot. A longer spanner increases that perpendicular distance; it does not increase your push or move the pivot from the nut. As drawn side on, pushing down on the right of the seesaw gives a clockwise moment.

Calculate a moment

The size of a moment is the force multiplied by its perpendicular distance from the pivot:

moment of a force = force × perpendicular distance from the pivot
M = F × d
Quantities in the moment equation
SymbolQuantityUnit
Mmoment of the forcenewton metre, N m
Fforcenewton, N
d

perpendicular distance from the pivot to the force’s line of action

metre, m

Put the distance in metres before you multiply, so the moment comes out in N m. A moment is a turning effect, not an amount of energy, so do not write its unit as J.

The perpendicular distance

The line of action of a force is the straight line through its arrow, extended as far as you need in both directions. The distance d is the shortest distance from the pivot to that line, and it meets the line at 90°.

When the force acts at 90° to a straight object, as on the door, d is the distance along the object to where the force acts. When the force acts at any other angle, d is shorter than that distance.

Perpendicular distance for a force at an angleA spanner turns a nut, which is the pivot. A hand pushes on the spanner 0.30 m from the nut, but the 50 N push acts at 60° to the spanner, down and to the right. The force's line of action is drawn as a dashed line through the hand and extended back. A second line runs from the nut to the line of action and meets it at a right angle. That perpendicular distance is 0.26 m, shorter than the 0.30 m along the spanner.nut (pivot)line of action50 N60°d = 0.26 m0.30 m along the spanner
Perpendicular distance for a force at an angleA spanner turns a nut, which is the pivot. A hand pushes on the spanner 0.30 m from the nut, but the 50 N push acts at 60° to the spanner, down and to the right. The force's line of action is drawn as a dashed line through the hand and extended back. A second line runs from the nut to the line of action and meets it at a right angle. That perpendicular distance is 0.26 m, shorter than the 0.30 m along the spanner.nut (pivot)line of action50 N60°d = 0.26 m0.30 m along the spanner
Use the perpendicular distance from the pivot to the line of action: 50 N × 0.26 m = 13 N m. Multiplying by the 0.30 m along the spanner gives a moment that is too large.

To find d:

  1. Mark the pivot.
  2. Draw the force’s line of action, extending it if you need to.
  3. Draw a line from the pivot that meets the line of action at 90°. Its length is d.

Now make the decision yourself. The bar below is pivoted at O.

An inclined bar with a vertical forceA straight bar runs up to the right from pivot O to point P. OP is 0.25 m. P is 0.20 m to the right of O and 0.15 m above it. A 40 N force acts vertically downward at P. The force's vertical line of action is drawn dashed and extends down past the level of O.OP0.25 m40 N0.20 m0.15 m
Find the line of action first, then choose the distance.

Check your understanding 3 Choose the distance before calculating

Which distance is d for the 40 N force about O?
Unit: N m
Show answer

The force’s line of action is vertical through P. Its perpendicular distance from O is horizontal: 0.20 m. The 0.25 m runs along the bar and the 0.15 m is parallel to the force. The moment is 40 × 0.20 = 8.0 N m clockwise, as drawn.

Worked example 1

Finding a moment

Problem

A 24 N force acts at right angles to a spanner, 35 cm from the centre of the nut. Calculate the moment of the force about the nut.

Show full solution

  1. Find the perpendicular distance

    Method

    The force is at 90° to the spanner.

    Reason

    So the distance along the spanner is the perpendicular distance. Change it to metres.

    Working

    Answer

    d = 35 cm = 0.35 m

  2. Multiply

    Working

    Answer

    M = F × d = 24 N × 0.35 m = 8.4 N m

Worked example 2

Finding the force needed

Problem

A tight wheel nut needs a moment of 90 N m to loosen it. A wheel brace is pushed at right angles, 0.30 m from the nut. Find the smallest force that loosens the nut. Then find it for a brace pushed 0.45 m from the nut.

Show full solution

  1. Rearrange for the force

    Method

    You know M and d, and want F.

    Reason

    Divide both sides of M = F × d by d.

    Working

    Answer

    F = M/d

  2. Substitute for each brace

    Working

    Answer

    F = (90 N m)/(0.30 m) = 300 N and F = (90 N m)/(0.45 m) = 200 N

Check your understanding 4 Force on a spanner

A force at right angles to a spanner, 25 cm from the nut, gives a moment of 20 N m. Calculate the force.

Unit: m
Show the steps Optional working
Unit: m
Unit: N
Show answer

d = 0.25 m, so F = M ÷ d = 20 ÷ 0.25 = 80 N.

Worked example 3

Finding the distance needed

Problem

A heavy fire door needs a moment of 6.0 N m about its hinges to start opening. A child can push at right angles to the door with a force of 15 N. How far from the hinges must the child push?

Show full solution

  1. Rearrange for the distance

    Method

    You know M and F, and want d.

    Reason

    Divide both sides of M = F × d by F.

    Working

    Answer

    d = M/F

  2. Substitute and answer

    Working

    Answer

    d = (6.0 N m)/(15 N) = 0.40 m. The child must push at least 0.40 m from the hinges.

Check your understanding 5 How far out?

A 12 N force must give a moment of 3.0 N m. What perpendicular distance from the pivot is needed?

Unit: m
How is this distance measured?
Show answer

d = M ÷ F = 3.0 ÷ 12 = 0.25 m, measured from the pivot to the line of the force at 90°.

Guided practice 4

Will the nut turn?

About 6 min

Problem

A nut needs a moment of 18 N m to loosen it. A student pushes on a spanner with a force of 60 N. The student’s hand is 0.30 m from the nut, but the push is at an angle, so its line of action passes 0.20 m from the nut.

Try this before viewing the solution

Unit: N m
Does the nut loosen?
Unit: N

Hints

Hint 1

which distance

Two distances are given. Only one is measured from the nut to the line of action at 90°.

Hint 2

finding the force

Rearrange M = F × d to F = M ÷ d and keep d = 0.20 m.

Show solution step by step

  1. Choose d

    Method

    The line of action passes 0.20 m from the nut.

    Reason

    d is the perpendicular distance to the line of action, not the distance to the hand.

    Working

    Answer

    d = 0.20 m

  2. Compare with what is needed

    Method

    Calculate the moment and compare it with 18 N m.

    Reason

    The nut loosens only if the moment reaches 18 N m.

    Working

    Answer

    M = 60 × 0.20 = 12 N m, which is less than 18 N m, so the nut does not loosen.

  3. Find the force needed

    Working

    Answer

    F = 18/0.20 = 90 N. Pushing at 90° to the spanner would also work: then d = 0.30 m and 60 N gives 60 × 0.30 = 18 N m.

Check your understanding 6 Find the error

A 50 N force is applied 0.30 m from a pivot, but the perpendicular distance from the pivot to its line of action is 0.18 m. A learner writes M = 50 × 0.30 = 15 N m. Choose the correct distance and calculate the moment.

Distance to use
Unit: N m
Show answer

The learner used the distance to the point where the force is applied. The perpendicular distance to the line of action is 0.18 m, so M = 50 × 0.18 = 9.0 N m. In a sketch, the 0.18 m segment goes from the pivot to the extended line of action and meets that line at 90°. The 0.30 m segment goes to the application point and must not replace it. To name the turning direction, you also need the force direction.

Check your understanding 7 Opening a hinged lid

A lid needs a moment of 4.0 N m about its hinge to start opening. A child can push with at most 25 N. Their hand is 0.30 m from the hinge, but the push’s line of action passes 0.12 m from it. Find the largest moment this push can give and decide whether the lid opens. Find the force needed along the same line.

Unit: N m
Can this push open the lid?
Unit: N
Show answer

The distance to the hand is not the perpendicular distance. Use d = 0.12 m:

M = Fd = 25 × 0.12 = 3.0 N m.

This is less than 4.0 N m, so the lid stays closed. Along the same line, the smallest force needed is

F = M/d = 4.0/0.12 = 33.3 N ≈ 33 N.

That exceeds the child’s 25 N maximum. A push with a larger perpendicular distance could work; a longer distance to the hand alone is not enough.

Common mistakes

  • Using the distance to where the force is applied when the force is not at 90° to the object. Use the perpendicular distance to the line of action.
  • Thinking the larger force always has the larger moment. 20 N at 0.10 m gives 2.0 N m, but 5.0 N at 0.60 m gives 3.0 N m.
  • Multiplying by a distance in centimetres and writing N m. 24 N × 35 cm is 840 N cm, which is 8.4 N m.
  • Naming a direction, clockwise or anticlockwise, without saying which pivot and which view.
  • Giving a force whose line passes through the pivot a moment. Its perpendicular distance is zero, so its moment is zero.

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