Full chapter
Turning effects of forces
All 3 topics and the revision summary on one page.
01
Moments and perpendicular distance
The moment of a force describes its turning effect about a pivot. A force can have a turning effect even if other forces prevent the object from rotating.
A door turns about its hinges. A spanner turns about the centre of a nut. The pivot is the point or axis about which we consider the turning. For a given force, pushing farther from the pivot can produce a larger moment.
Use force in newtons and distance in metres. A force is different from a mass; review forces and weight or unit conversions if needed.
Moment = F x dF is the force in N. d is the perpendicular distance in m from the pivot to the force's line of action. The moment is measured in newton metres, N m.
A moment is also called a torque. State its turning sense: clockwise or anticlockwise in the view being used. N m means force multiplied by distance; N/m is not the unit of moment.
The same push at different distances
Imagine a 12 N push perpendicular to a door. At 0.75 m from the hinge, its moment is 12 x 0.75 = 9.0 N m. At 0.25 m, the same perpendicular push gives 12 x 0.25 = 3.0 N m.
The farther push has three times the turning effect because its perpendicular distance is three times as large. This explains why a door handle is usually placed far from the hinge and why a longer spanner can make a nut easier to turn.
The line of action determines the distance
The line of action is the straight line along which the force acts. Imagine extending the force arrow straight in both directions. Measure the shortest distance from the pivot to that line: it must meet the line at a right angle.
Measure to the force's line of action
Moment = 12 x 0.40 = 4.8 N m clockwise.
Dashed lines are construction guides. The blue curve shows turning sense, not an extra force.
A force through the pivot has zero moment
Perpendicular distance = 0. Moment about O = 0 N m.
Worked example
A tilted handle
Point A is 0.40 m to the right and 0.30 m above pivot O. The handle OA is 0.50 m long. A 12 N force acts vertically downwards at A.
- Identify the line of action: it is vertical through A.
- Select the perpendicular distance: the horizontal distance from O to that line is 0.40 m.
- Calculate: moment = 12 x 0.40 = 4.8 N m.
- Give the turning sense: a downward force to the right of O turns the handle clockwise in this view.
Using 12 x 0.50 would give 6.0 N m, but the handle's length is not the required perpendicular distance. The curved turning arrow indicates a sense of rotation; it is not another force.
A force can have zero moment
If the force at A is directed along the handle towards O, its line of action passes through the pivot. The perpendicular distance is zero, so its moment about O is zero. The force still exists; it simply has no turning effect about that point.
Worked example
Find a force from a required moment
A perpendicular lever arm is 0.20 m. What force gives a moment of 4.8 N m?
F = moment / d = 4.8 N m / 0.20 m = 24 N
The metre units cancel, leaving newtons. If the force is given and the lever arm is unknown, rearrange to d = moment / F.
Choose the distance before multiplying. It is not automatically the length of the handle or the distance to the point where the force is applied. Locate the line of action and its perpendicular from the pivot.
Optional check Viewed from the front, a 10 N force acts vertically downwards at a point to the right of a pivot. The handle is 0.50 m long, but the perpendicular distance from the pivot to the force's line of action is 0.30 m. What is the moment?
02
Balancing moments
For a body in equilibrium, the total clockwise moment equals the total anticlockwise moment about the same point.
Use force x perpendicular distance for each moment. When a load is given as a mass, use W = mg first: a total load-and-hanger mass of 200 g is 0.200 kg and weighs 2.0 N where g = 10 N/kg.
For a stationary body, the forces must balance as well as the moments. Moment balance concerns turning; force balance concerns changes in velocity. Zero total moment alone does not establish zero resultant force.
Include the rule's own weight
A uniform horizontal metre rule weighs 1.2 N. Its weight acts at its midpoint, the 50.0 cm mark. This point is its centre of gravity. A pivot supports the rule at 40.0 cm.
A downward 2.0 N load acts at 10.0 cm, and an unknown downward load F acts at 70.0 cm. The load forces include their hangers.
Forces on a uniform, balanced metre rule
The loads include their hangers. The rule's own weight acts at its 50 cm mark.
The rule's weight has an arm of 50 - 40 = 10 cm = 0.10 m. The support passes through the pivot, so its moment about the pivot is zero.
2.0 x 0.30 = 1.2 x 0.10 + F x 0.30, so F = 1.6 N.
Worked example
Find the load that balances the rule
- Anticlockwise: the 2.0 N force is left of the pivot, so its downward pull gives 2.0 x 0.30 = 0.60 N m.
- Clockwise from the rule: its 1.2 N weight is 0.10 m right of the pivot, giving 1.2 x 0.10 = 0.12 N m.
- Clockwise from F: this force acts 0.30 m right of the pivot, giving F x 0.30.
- Equate the total moments: 0.60 = 0.12 + 0.30F.
- Solve: 0.48 = 0.30F, so F = 1.6 N downwards.
The upward force from the pivot has zero moment about the pivot because its line of action passes through that point. It still acts on the rule.
For vertical force balance, the support is 2.0 + 1.2 + 1.6 = 4.8 N upwards. This equals the total downward force.
Omitting the rule's own weight would give F = 2.0 N, which is too large. Its weight has a clockwise moment because the pivot is not directly below the midpoint.
What if the 70.0 cm load is reduced?
The clockwise moment becomes smaller while the anticlockwise moment stays at 0.60 N m. The rule initially turns anticlockwise, with its left end moving downwards. You can predict the direction without calculating a new balance position.
Investigate balancing forces and distances
First balance an unloaded rule to locate its own centre of gravity. Place the pivot there, so the rule's weight has zero moment about the pivot. Keep a 2.0 N load 0.20 m to one side: its moment is 0.40 N m.
| Opposing load / N | Distance from pivot / m |
|---|---|
| 1.0 | 0.40 |
| 2.0 | 0.20 |
| 2.5 | 0.16 |
The products are 1.0 x 0.40, 2.0 x 0.20 and 2.5 x 0.16. Each gives 0.40 N m. A larger balancing force needs a shorter lever arm when the required moment is fixed. These are idealised model values; real readings vary.
In a practical investigation, record the loads and positions with quantity/unit headings, then calculate the pivot distances and moments separately. Keep the fixed-side load and distance unchanged while changing the opposing load and finding its balancing position.
- Use a narrow pivot with low friction, and identify its actual position on the rule.
- Keep the rule horizontal so vertical loads have horizontal perpendicular lever arms. Read the force positions at eye level.
- Include hanger weights and allow loads to hang freely, clear of the bench.
- Secure the support and let oscillations settle before recording a balance position.
Pivot friction may let the rule appear balanced over a range of positions. Approach balance from both directions to judge that range and reduce the pivot friction where possible. Repeating a reading alone does not remove this source of bias.
Optional check A uniform horizontal metre rule weighs 2.0 N and balances on a pivot at 40 cm. A downward 3.0 N load acts at 20 cm and a downward load F at 80 cm. What is F? All load forces include their hangers.
03
Centre of gravity
A body's weight can be represented as a single force acting at its centre of gravity, G.
Gravity acts throughout the body's matter. The centre of gravity lets us replace those distributed effects with one resultant weight arrow. Draw that arrow from G, vertically downwards near Earth.
A weight produces a moment about a pivot unless its line of action passes through the pivot. Use the body's total weight in newtons in the moment calculation.
The centre of gravity need not be the middle of the outline
For a uniform, symmetrical object in a uniform gravitational field, G is at its geometric centre. A uniform metre rule has G at the 50 cm mark. If more mass is concentrated towards one end, G shifts towards that end.
Some shapes have G outside their material: for a uniform ring, it is at the centre of the empty opening. G represents where the combined weight acts; it does not have to be a point containing material.
A horizontal rule can balance on a narrow support directly below its G. The weight's line of action then passes through the support, so weight has no moment about it. The upward support also balances the downward weight.
Locate G by suspending a lamina
A plane lamina is a thin, flat sheet, such as a rigid card cut into an irregular outline. Its centre of gravity can be found without assuming the outline is symmetrical.
When it hangs freely and settles, G lies vertically below the suspension point. Otherwise the weight would have a moment about that point and turn the lamina. A plumb line, made from a freely hanging thread and small weight, shows the vertical direction.
- Make a small hole near the edge of the lamina. Suspend it from a pin through that hole, allowing it to rotate freely.
- Hang a plumb line from the same suspension point. Keep the thread free and wait until both the lamina and line are still.
- Mark the plumb-line position on the lamina using a thin line or two well-separated marks that can be joined. G lies somewhere on this traced line.
- Suspend the same lamina from a different hole. Let it settle into its new orientation, then mark the new plumb-line position.
- Find the intersection of the two traces on the card. This estimates G. A third suspension from another hole provides a check.
Two suspensions of the same lamina
Let the lamina and plumb line hang freely and settle. G is shown to explain the result of the measurement.
1. Suspend from hole A
2. Suspend from hole B
3. Compare the marked lines on the lamina
The thin solid vertical line is the plumb line; red W is the lamina's weight. Use a third suspension as a check on the inferred position.
Changing the suspension hole changes how the lamina hangs, not where G lies relative to the card. One suspension gives a line of possible positions. A second, differently directed trace locates their common point.
Make the intersection useful
- Allow free rotation: do not clamp the lamina against the pin. Friction could hold it before G is vertically below the suspension point.
- Use thin marks and a still plumb line: thick or moving marks leave a broad range of possible lines.
- View perpendicular to the card: reduce parallax between the thread and the surface being marked.
- Choose useful suspension positions: traces with distinctly different directions locate an intersection more clearly than nearly parallel traces.
Real traces may meet within a small region rather than at one exact point. Use that region to judge the uncertainty in G, and check the result with another suspension. Do not choose the geometric centre simply because the lines are imperfect.
Draw one weight at G. G is not an extra force. When replacing distributed weight with its resultant, do not also keep a second full-weight arrow elsewhere on the same body.
Optional check A freely suspended irregular lamina has settled, and its vertical plumb-line position has been marked on the card. Why repeat the suspension from a different hole?
Revision summary
- Moment of a force
- Moment = F x d. F in N; d is the perpendicular distance in m from the pivot to the force's line of action. Give N m and the clockwise or anticlockwise turning sense.
- Principle of moments
- For equilibrium, total clockwise moment = total anticlockwise moment about the same point. A stationary body also needs zero resultant force.
Before calculating a moment
- Name the pivot and extend the force's line of action.
- Choose the perpendicular distance to that line, not automatically the handle length.
- Convert distances to metres for a result in N m. Convert mass to force using W = mg when needed.
- A force whose line of action passes through the pivot has zero moment about it; the force may still be non-zero.
Build a balance equation
Include every force with a turning effect, including the object's own weight. Measure all lever arms from one common pivot and classify their turning senses. For a uniform metre rule, its weight acts at the 50 cm mark; a support elsewhere does not automatically remove that moment.
Centre of gravity and suspension
Represent the body's combined weight by one downward force at G. A uniform symmetrical object's G is at its geometric centre, but an irregular object needs evidence.
- Suspend the lamina freely and let it settle.
- Mark the vertical plumb-line position from that same suspension point.
- Repeat from another hole. The intersection of the two traces estimates G.
- Use a third suspension to check; thin, well-separated trace directions give a more useful intersection.
For a moments investigation, include hanger weights, keep forces vertical and measure from the actual pivot. For suspension, avoid clamping, parallax and a moving plumb line. Explain a specific limitation and its effect.
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