Turning effects of forces

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.

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.

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.

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.

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.

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.

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.

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.

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.

1. Make a small hole near the edge of the lamina. Suspend it from a pin through that hole, allowing it to rotate freely.
2. Hang a plumb line from the same suspension point. Keep the thread free and wait until both the lamina and line are still.
3. 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.
4. Suspend the same lamina from a different hole. Let it settle into its new orientation, then mark the new plumb-line position.
5. Find the intersection of the two traces on the card. This estimates G. A third suspension from another hole provides a check.

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.

04 / Pure

Stability and toppling

To decide whether a tilted object tends to return or topple, compare the vertical line of its weight with the edge it could turn about.

The weight acts downwards through the centre of gravity. Its moment depends on which side of the pivot its line of action lies.

Imagine gently tilting a box clockwise about its bottom-right edge, then releasing it. Assume it does not slide. The edge is the possible toppling pivot; gravity still acts vertically downwards while the box tilts.

• If the weight's line of action lies to the left of that edge, weight produces an anticlockwise, restoring moment. The box tends to return towards its original position.
• If the line passes through the edge, weight has zero moment about the edge at that instant. This is the boundary between the two cases.
• If the line lies to the right of the edge, weight produces a clockwise moment and promotes further toppling.

A lower centre of gravity allows more tilt

The illustrated bodies are each 0.60 m wide and 1.0 m high. When upright, G is at mid-width and either 0.20 m or 0.80 m above the base. Their different mass distributions place G at different heights.

At the same tilt, the lower G has not moved far enough sideways for its weight line to pass the edge. The higher G has. For otherwise comparable objects, lowering G allows a greater tilt before the weight starts to promote toppling.

A wider base can increase stability

Keeping the height of G unchanged, a wider base in the direction of the tilt places the toppling edge farther from the upright weight line. More tilt is then needed before that line passes beyond the edge.

For example, placing a heavy load low in a vehicle can lower its overall G. Increasing the support width can also help resist toppling. In each case, explain what happens to the position of G or the edge, rather than only saying the object is "more stable".

Greater weight alone does not determine the toppling boundary. With the same geometry and G position, a larger weight changes the size of its moment, but not which side of the edge its line acts on. The line-of-action argument here assumes the object does not slide.