Moments

Section: Motion, Forces & Energy | Syllabus: Cambridge IGCSE Physics (0625)

Moment of a force is a measure of its turning effect about a pivot (or axis)
everyday examples of turning effects include:
- pushing down on one end of a see-saw
- turning a spanner to tighten a nut
- pushing a door open near the hinge vs. near the handle
moment is calculated using:
- M = F × d
Where:
- M = moment (N·m)
- F = force (N)
- d = perpendicular distance from the pivot to the line of action of the force (m)
only the perpendicular distance from the pivot counts — not the slanted or total length of the object

The moment of a force depends on both the size of the force and its perpendicular distance from the pivot

Muumbe applies a force of 10 N to a spanner, at a perpendicular distance of 0.3 m from the pivot (the nut). Calculate the moment produced.
Muumbe's spanner: force of 10 N at 0.3 m from the pivot
Step 1: write down the formula
- M = F × d
Step 2: substitute the values
- M = 10 × 0.3
Step 3: calculate the answer
- M = 3 N·m

Equilibrium and the Principle of Moments

an object is in equilibrium when there is no resultant force and no resultant moment acting on it
- it does not accelerate and does not rotate
the principle of moments states that, for an object in equilibrium:
- sum of clockwise moments = sum of anticlockwise moments
this applies to situations with one force each side of the pivot, such as balancing a beam

A beam balanced with one force on each side of the pivot

Museli balances a uniform beam on a pivot at its midpoint. A weight of 20 N is hung 0.4 m from the pivot on the left side. Calculate the distance from the pivot on the right side at which a 16 N weight must be placed for the beam to be in equilibrium.
Museli's beam: 20 N at 0.4 m balanced against 16 N at distance d
Step 1: apply the principle of moments
- clockwise moment = anticlockwise moment
Step 2: substitute known values
- 16 × d = 20 × 0.4
Step 3: solve for d
- d = (20 × 0.4/16) = 0.5 m

Extended Only

the principle of moments also applies when there is more than one force acting on each side of the pivot
to check for equilibrium, calculate the total clockwise moment and the total anticlockwise moment separately, then compare them:
- total clockwise moment = total anticlockwise moment

Equilibrium with more than one force on each side of the pivot

Muchindu balances a beam on a central pivot. On the left side, a 12 N weight acts 0.5 m from the pivot and a 6 N weight acts 0.2 m from the pivot. On the right side, a 10 N weight acts 0.6 m from the pivot. Determine whether the beam is in equilibrium.
Muchindu's beam: two weights on the left, one on the right: checking for equilibrium
Step 1: calculate the total anticlockwise moment (left side)
- (12 × 0.5) + (6 × 0.2) = 6 + 1.2 = 7.2 N·m
Step 2: calculate the total clockwise moment (right side)
- 10 × 0.6 = 6 N·m
Step 3: compare the two totals
- 7.2 N·m ≠ 6 N·m, so the beam is not in equilibrium — it will turn anticlockwise
- it will turn anticlockwise

Extended Only

an experiment to show there is no resultant moment on an object in equilibrium:
- set up a metre rule balanced on a pivot at its midpoint, so it is initially balanced with no loads
- hang known weights at measured distances from the pivot on each side, using string over the rule
- adjust the positions of the weights until the rule is exactly horizontal (balanced) again
- record each force and its distance from the pivot
- calculate the sum of the clockwise moments and the sum of the anticlockwise moments
- repeat with different combinations of weights and distances, including more than one weight on a side
in each case, the sum of clockwise moments is found to equal the sum of anticlockwise moments, confirming there is no resultant moment when the rule is balanced (in equilibrium)

Testing the principle of moments using a balanced metre rule

centre of gravity is the point at which the entire weight of an object appears to act
for a regular, uniform object, the centre of gravity is at its geometric centre
for an irregularly shaped object, the centre of gravity must be found experimentally

procedure to find the centre of gravity of an irregularly shaped plane lamina:
- make a small hole near the edge of the lamina and hang it freely from a pin so it can swing
- hang a plumb line (a string with a small weight tied to the end) from the same pin, in front of the lamina
- once the lamina and plumb line are still, mark the position of the string on the lamina, then draw a straight line along it
- hang the lamina from a different hole and repeat the process to draw a second line
- the point where the two lines cross is the centre of gravity
a third hole can be used to check the result — all three lines should cross at the same point

Finding the centre of gravity of an irregular lamina using a plumb line

an object is more stable when it has a low centre of gravity and a wide base
an object topples over if the line of action of its weight (a vertical line down from its centre of gravity) falls outside its base
raising an object's centre of gravity, or narrowing its base, makes it less stable and more likely to topple

A lower centre of gravity and a wider base both increase stability

when describing the plumb-line experiment, always mention that the lamina must be free to swing/rotate about the pin, and that the process must be repeated from a second (or third) hole — a single line only gives one possible line through the centre of gravity, not the exact point

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