Elastic Deformation, Moments and Centre of Gravity: Revision Notes

Condensed for the final weeks. For the full explanation, use the Elastic Deformation, Moments and Centre of Gravity study guide.

Hooke’s law

F = k x        F = force (N)
               k = spring constant (N/m) -- the STIFFNESS
               x = EXTENSION, not total length

Extension is proportional to load up to the limit of proportionality; beyond it, the graph curves. (Background note, not examined: beyond that is a further point called the elastic limit, beyond which the spring no longer returns to its original length — understanding it is not required at this level.)

On a load–extension graph: gradient = k. A steeper line means a stiffer spring.

Always subtract the original length to get extension — using total length is the standard lost mark.

Moments

moment = F x d       d = PERPENDICULAR distance from the pivot
                     units: N m

A moment is the turning effect of a force — everyday examples include a spanner turning a nut, a door opening about its hinges, a see-saw balancing, and a lever such as a crowbar prising up a heavy slab. It is zero if the force acts through the pivot, or along the line of the pivot.

Principle of moments

For an object in equilibrium:

sum of CLOCKWISE moments  =  sum of ANTICLOCKWISE moments

Plus, for full equilibrium, the resultant force must also be zero.

Worked example: a 2 m beam pivoted at its centre; 30 N at 0.8 m on the left. What force at 0.5 m on the right balances it?

anticlockwise = 30 x 0.8 = 24 N m
clockwise     = F x 0.5

F x 0.5 = 24   ->   F = 48 N

Worked example. A uniform see-saw is pivoted at its centre. A 300 N weight sits 1.2 m from the pivot on one side. How far from the pivot on the other side must a 400 N weight sit, to balance?

clockwise moment = anticlockwise moment
400 x d = 300 x 1.2
d = 360 / 400 = 0.9 m

Centre of gravity

The single point where the entire weight of an object appears to act.

Finding it for an irregular lamina: suspend from a point, hang a plumb line, draw the vertical. Repeat from a second point. The intersection is the centre of gravity.

For a symmetrical uniform object, it is at the geometric centre. The principle of moments only applies to an object that is genuinely in equilibrium and not turning — if the clockwise and anticlockwise totals are unequal, the object has a resultant moment and will rotate rather than balance.

Stability

An object is stable if its centre of gravity is low and its base is wide.

It topples when the vertical line through the centre of gravity falls outside the base.

Beyond the syllabus, not examinable: 5054 1.5.6 asks only for the qualitative effect of the position of the centre of gravity on stability; the terms below are background.

Equilibrium	Behaviour when tilted slightly
Stable	Returns to its original position
Unstable	Topples further away
Neutral	Stays in the new position (e.g. a ball on a flat surface)

Practical examples: racing cars (low, wide), Bunsen burners (heavy base), double-decker buses (low centre of gravity by design). Two objects of equal weight can have very different stability — stability depends on the position of the centre of gravity relative to the base, not on weight itself.

Worked example: combining Hooke’s law and moments

A uniform beam of weight 20 N and length 1.0 m rests horizontally, supported by a spring at one end and a pivot at the other. If the spring must supply a force such that the beam balances, and the beam’s weight acts at its centre (0.5 m from the pivot), find the force the spring must provide, given it acts at the far end (1.0 m from the pivot).

anticlockwise moment (weight) = 20 x 0.5 = 10 N m
clockwise moment (spring)     = F x 1.0

F x 1.0 = 10   ->   F = 10 N

If that same spring has a spring constant of 200 N/m, its extension under this 10 N load is x = F / k = 10 / 200 = 0.05 m. This kind of combined question – using the principle of moments to find a force, then Hooke’s law to find what that force does to a spring – is a common way exam papers link the two halves of this topic together in a single multi-part question.

Toppling: a step further than “low and wide”

A stability question often asks you to explain, not just state, why a specific object is more stable than another. The reasoning chain is: a wider base means the centre of gravity has to shift further sideways before its vertical line falls outside the base; a lower centre of gravity means, for the same amount of tilt, the vertical line through it moves a smaller horizontal distance before reaching the edge of the base. Both effects work in the same direction, which is why racing cars combine them deliberately rather than relying on just one.

Exam traps

• Use extension, not total length, in F = kx.
• Distance in a moment must be perpendicular to the force.
• Both conditions are needed for equilibrium: moments balance and forces balance.
• Convert cm to m before calculating moments.
• Say the object topples when the line of action of the weight falls outside the base.

Self-test

1. A spring of natural length 12 cm extends to 18 cm under 3 N. Find k.
2. State the principle of moments.
3. A 40 N force acts 0.25 m from a pivot. Find the moment.
4. How do you locate the centre of gravity of an irregular lamina?
5. Why is a racing car harder to topple than a bus?
6. A see-saw pivoted at its centre has a 300 N weight 1.2 m from the pivot on one side. Find the distance from the pivot for a balancing 400 N weight on the other side.

Answers: 1. Extension = 0.06 m; k = 3/0.06 = 50 N/m. 2. For an object in equilibrium, the sum of clockwise moments about a point equals the sum of anticlockwise moments about the same point. 3. 40 × 0.25 = 10 N m. 4. Suspend it freely from one point and mark the vertical with a plumb line; repeat from a second point; the centre of gravity is where the lines intersect. 5. It has a lower centre of gravity and a wider wheelbase, so the car must tilt much further before the line of action of its weight falls outside its base. 6. 400 × d = 300 × 1.2 → d = 360 ÷ 400 = 0.9 m.