P21: Moments and Centre of Mass
Covering the moment of a force M=Fd, the principle of moments, levers and gears, centre of mass, stability, and force multipliers.
Moments
Moment = force x perpendicular distance from the pivot
M = Fd
M = moment (Nm), F = force (N), d = perpendicular distance from the pivot to the line of action of the force (m)
A moment is the turning effect of a force. It depends on both the size of the force and the perpendicular distance from the pivot (the point about which the object can rotate).
Worked Example
A force of 40 N is applied at a distance of 0.3 m from a pivot. Calculate the moment.
M = Fd = 40 x 0.3 = 12 Nm
Worked Example
A spanner is 0.25 m long. A mechanic applies a force of 80 N at the end of the spanner. Calculate the moment.
M = Fd = 80 x 0.25 = 20 Nm
To increase the moment, you can either increase the force or increase the distance from the pivot. This is why longer spanners make it easier to undo tight nuts - the same force produces a larger moment because the distance from the pivot is greater.
The distance in the moment equation must be the perpendicular distance from the pivot to the line of action of the force. If the force is applied at an angle, use the component of force perpendicular to the lever, or use the perpendicular distance.
The Principle of Moments
The principle of moments states that for an object in equilibrium (not rotating), the sum of the clockwise moments equals the sum of the anticlockwise moments about any pivot.
Total clockwise moments = Total anticlockwise moments
Sum of clockwise moments = Sum of anticlockwise moments
Worked Example
A uniform metre ruler is balanced on a pivot at the 40 cm mark. A weight of 6 N hangs from the 10 cm mark. Calculate the weight needed at the 90 cm mark to balance the ruler.
Anticlockwise moment = 6 x (40 - 10) = 6 x 30 = 180 Ncm
Clockwise moment = W x (90 - 40) = W x 50
W x 50 = 180
W = 180 / 50 = 3.6 N
Worked Example
A see-saw has a pivot at its centre. A child of weight 300 N sits 2 m from the pivot on the left. An adult of weight 750 N sits on the right. How far from the pivot must the adult sit for the see-saw to balance?
Anticlockwise moment = 300 x 2 = 600 Nm
Clockwise moment = 750 x d
750d = 600
d = 0.8 m
For an object to be in equilibrium, two conditions must be met: (1) the resultant force is zero, and (2) the resultant moment about any point is zero (principle of moments).
Levers
A lever is a simple machine that uses a pivot to multiply a force. The force applied at a greater distance from the pivot produces a larger moment, allowing a smaller effort force to overcome a larger load force.
Key terms for levers:
- Effort - the force applied to the lever
- Load - the force to be overcome
- Pivot (fulcrum) - the point about which the lever rotates
A lever acts as a force multiplier when the effort is applied further from the pivot than the load. A small effort can produce a large load force because the effort has a larger perpendicular distance from the pivot.
| Levers in everyday life | Effort distance from pivot | Load distance from pivot |
|---|---|---|
| Scissors | Long handles (large) | Blades near pivot (small) |
| Wheelbarrow | Handles (large) | Load near wheel (small) |
| Crowbar | Long end (large) | Short end near pivot (small) |
| Door handle | Far from hinge (large) | Near hinge (small) |
Worked Example
A wheelbarrow has a load of 600 N at a distance of 0.2 m from the wheel (pivot). The handles are 1.2 m from the wheel. Calculate the effort needed to lift the load.
Clockwise moment (load) = 600 x 0.2 = 120 Nm
Anticlockwise moment (effort) = E x 1.2
E x 1.2 = 120
E = 100 N
The 600 N load is lifted with only 100 N of effort - the lever multiplies the force by 6.
When describing how a lever works as a force multiplier, use the moment equation: since M = Fd, if the effort is applied at a larger distance from the pivot, a smaller force produces the same moment to balance the load.
Gears
Gears are toothed wheels that transmit rotational forces. When two gears are meshed together, the driven gear rotates in the opposite direction to the driver gear.
Key principles of gears:
- A larger gear driving a smaller gear increases the speed of rotation but decreases the moment
- A smaller gear driving a larger gear decreases the speed of rotation but increases the moment
- The ratio of teeth determines the gear ratio
- Gears act as force multipliers when a small gear drives a large gear
Gear ratio = number of teeth on driven gear / number of teeth on driver gear
When a small gear drives a large gear: the large gear rotates more slowly but with a greater moment. This is because the force at the teeth is the same for both gears, but the larger gear has a larger radius (greater distance from pivot), so M = Fd gives a larger moment.
Worked Example
A gear with 10 teeth drives a gear with 30 teeth. The driver gear has a moment of 5 Nm applied. Calculate the moment on the driven gear.
Gear ratio = 30/10 = 3
The driven gear has 3 times the moment (but rotates at 1/3 the speed).
Moment on driven gear = 5 x 3 = 15 Nm
Gears are often examined alongside moments. The key link is that the same force acts on both gear teeth at the point of contact, but the different radii of the gears mean different moments are produced.
Centre of Mass
The centre of mass of an object is the single point at which the entire mass of the object can be considered to be concentrated. For a uniform regular object, the centre of mass is at its geometric centre.
Finding the centre of mass:
- For a regular uniform shape (rectangle, circle, sphere), it is at the centre of the shape
- For an irregular shape, it can be found by suspending the object from different points and drawing plumb lines - the intersection of the plumb lines gives the centre of mass
- For a triangle, the centre of mass is at the intersection of the medians (1/3 from the base)
Worked Example - Irregular Shape
To find the centre of mass of an irregular flat object:
- Suspend the object from one point and let it hang freely
- Use a plumb line to draw a vertical line on the object from the suspension point
- Suspend the object from a different point and draw another plumb line
- The centre of mass is where the two lines cross
For a uniform regular solid, the centre of mass is at the centre of the object. For example, a uniform sphere has its centre of mass at its geometric centre. A uniform cuboid has its centre of mass at the intersection of its diagonals.
Stability
An object is stable if a line of action of its weight (drawn downwards from the centre of mass) falls within its base. If the line of action of the weight falls outside the base, the object will topple.
Factors affecting stability:
- Width of the base - a wider base makes an object more stable
- Height of the centre of mass - a lower centre of mass makes an object more stable
To make an object more stable: lower the centre of mass (e.g. add weight at the bottom) and/or widen the base (e.g. spread the support points further apart).
| Feature | Stable object | Unstable object |
|---|---|---|
| Base width | Wide base | Narrow base |
| Centre of mass height | Low centre of mass | High centre of mass |
| Examples | Racing car, Bunsen burner, pyramid | Tall lamp post, tall thin glass, pencil standing on end |
Worked Example
Explain why a double-decker bus is less stable than a sports car.
A double-decker bus has a higher centre of mass (passengers on the upper deck) and a relatively narrow base compared to its height. A sports car has a low centre of mass (heavy engine low down) and a wider wheelbase relative to its height. This means the bus will topple at a smaller tilt angle than the sports car.
An object will topple when the line of action of its weight moves outside the base area. This happens when the object is tilted far enough that the centre of mass is no longer above the base. The greater the tilt needed, the more stable the object.
When explaining stability, always mention both the position of the centre of mass and the width of the base. An object with a low centre of mass and a wide base is most stable because it requires the greatest tilt angle for the weight line to fall outside the base.
Simple Pendulum
A simple pendulum consists of a mass (bob) suspended from a fixed point by a string. The pendulum oscillates back and forth. The time for one complete oscillation is called the period.
Period = 1 / frequency
T = 1 / f
T = period (s), f = frequency (Hz)
The period of a pendulum depends only on the length of the string and the gravitational field strength. It does NOT depend on the mass of the bob or the amplitude of the swing (for small angles).
T squared is proportional to L
A longer pendulum has a longer period (swings more slowly)
Worked Example
A pendulum has a length of 1.0 m and a period of 2.0 s. Another pendulum has a length of 0.25 m. Predict its period.
Since T squared is proportional to L, reducing the length from 1.0 m to 0.25 m (factor of 4) means T squared decreases by a factor of 4, so T decreases by a factor of 2.
T = 2.0 / 2 = 1.0 s
Required Practical: Investigating Moments
The principle of moments can be verified using a balanced metre ruler and known weights.
Method:
- Balance a metre ruler on a pivot at its centre (the 50 cm mark for a uniform ruler).
- Hang known weights at different positions on one side of the pivot.
- Hang weights on the other side to balance the ruler.
- Record the forces and distances from the pivot.
- Calculate clockwise and anticlockwise moments to verify they are equal.
- Repeat with different combinations of weights and positions.
The pivot should be at the centre of mass for a uniform ruler. If the ruler is not perfectly balanced initially, note the offset and account for it. To reduce errors, use a sharp pivot (like a knife edge) to minimise friction at the pivot point.
Practice Questions
1. A force of 50 N acts at a perpendicular distance of 0.4 m from a pivot. Calculate the moment.
M = Fd = 50 x 0.4 = 20 Nm.
2. A uniform see-saw is balanced at its centre. A child of weight 400 N sits 1.5 m from the pivot. Where must a second child of weight 300 N sit on the other side to balance the see-saw?
Anticlockwise moment = 400 x 1.5 = 600 Nm. Clockwise moment = 300 x d. d = 600/300 = 2.0 m from the pivot.
3. Explain how a crowbar acts as a force multiplier.
The effort is applied at a large distance from the pivot, while the load is close to the pivot. Since moment = force x distance, the same moment is produced by a small effort at a large distance and a large load at a small distance. The effort force is multiplied in proportion to the ratio of distances.
4. A gear with 20 teeth drives a gear with 60 teeth. The driver gear has a moment of 8 Nm. Calculate the moment on the driven gear.
Gear ratio = 60/20 = 3. Moment on driven gear = 8 x 3 = 24 Nm.
5. Explain why a racing car has a low centre of mass and a wide wheelbase.
A low centre of mass and wide wheelbase make the car more stable. The weight acts from a lower point and the base is wider, so a greater tilt angle is needed before the line of action of the weight falls outside the base. This means the car is less likely to topple during fast cornering.
Maths Skills
Moment Calculations: M = Fd
The moment of a force depends on both the force and the perpendicular distance from the pivot. Rearrange to find unknowns: F = M/d or d = M/F. Remember the distance must be perpendicular to the line of action of the force. If a force is applied at an angle, use the perpendicular component of the force or the perpendicular distance.
Worked Example
A force of 25 N is applied at the end of a 0.6 m spanner at 30° to the spanner. Calculate the moment.
Perpendicular component of force = 25 cos 30° = 25 × 0.866 = 21.65 N
M = 21.65 × 0.6 = 12.99 Nm
Alternatively: perpendicular distance = 0.6 cos 30° = 0.520 m, M = 25 × 0.520 = 12.99 Nm (same result).
Principle of Moments Calculations
For equilibrium: sum of clockwise moments = sum of anticlockwise moments. Set up the equation with all forces and their perpendicular distances from the pivot. Solve for the unknown. If there are multiple forces on one side, add their moments together. Always state the direction (clockwise or anticlockwise).
Worked Example
A uniform beam of weight 40 N and length 2.0 m is balanced on a pivot 0.5 m from the left end. A 60 N weight hangs from the left end. Calculate the additional force needed at the right end to balance the beam.
Anticlockwise moment (60 N weight) = 60 × 0.5 = 30 Nm
Anticlockwise moment (beam weight at centre) = 40 × 0.5 = 20 Nm (centre of beam is 0.5 m right of pivot)
Note: beam weight acts clockwise here. Clockwise moment (beam) = 40 × 0.5 = 20 Nm
Clockwise moment needed from force F: F × 1.5
Anticlockwise = clockwise: 30 = 20 + 1.5F → 1.5F = 10 → F = 6.67 N
Gear Ratio Calculations
Gear ratio = Ndriven / Ndriver (number of teeth). If the gear ratio is 3:1, the driven gear has 3 times the moment but rotates at 1/3 the speed. The force at the gear teeth is the same for both gears, but the different radii produce different moments.
Common Misconceptions
Levers and Work
A longer lever always makes work easier. A longer lever increases the moment for a given effort force, making it easier to overcome a load. However, the work done is the same: a smaller force applied over a greater distance at the lever end gives the same work as a larger force over a smaller distance at the load. Levers trade force for distance — they do not reduce the total work needed.
Centre of Mass Position
The centre of mass is always in the middle of an object. The centre of mass depends on how the mass is distributed. For a uniform regular shape, it is at the geometric centre. For irregular or non-uniform objects, the centre of mass can be anywhere — it shifts towards the heavier end. A hammer, for example, has its centre of mass much closer to the heavy metal head than the handle.
6-Mark Extended Question
Explain how levers and gears can be used as force multipliers. Give an everyday example of each. [6 marks]
A lever acts as a force multiplier when the effort is applied further from the pivot than the load (1). Since the moment is the same on both sides at equilibrium (M = Fd), a small effort at a large distance from the pivot can balance a large load at a small distance (1). For example, a wheelbarrow has the load close to the wheel (pivot) and the effort applied at the handles far from the wheel, so a 600 N load can be lifted with only 100 N of effort (1). Gears act as force multipliers when a small driver gear meshes with a larger driven gear (1). The contact force at the gear teeth is the same for both gears, but the larger gear has a greater radius, producing a larger moment (M = Fd, larger d means larger M) (1). For example, a car in low gear uses a small engine gear driving a large wheel gear, multiplying the engine's turning force to help the car climb a hill or accelerate from rest (1).
AO3: Analyse and Evaluate
A see-saw has a pivot at its centre. The see-saw is 4.0 m long. The following weights are placed on it:
- Child A: 300 N, sits 1.5 m left of the pivot
- Child B: 400 N, sits 1.0 m right of the pivot
- Child C: 250 N, sits at the far right end (2.0 m from pivot)
(a) Calculate the total clockwise and anticlockwise moments. Does the see-saw balance? (b) If the see-saw does not balance, where should child A move to so that it does balance? (c) The see-saw beam itself has a weight of 200 N. Assuming it is uniform, explain why this does not affect the balance when the pivot is at the centre.
Evaluation
(a) Anticlockwise moment (child A) = 300 × 1.5 = 450 Nm. Clockwise moment (child B) = 400 × 1.0 = 400 Nm. Clockwise moment (child C) = 250 × 2.0 = 500 Nm. Total clockwise = 400 + 500 = 900 Nm. The see-saw does not balance (900 Nm clockwise vs 450 Nm anticlockwise).
(b) For balance: 300 × d = 900 → d = 900/300 = 3.0 m. Child A would need to sit 3.0 m to the left of the pivot, but the see-saw is only 2.0 m on that side, so this is impossible. Alternatively, child C could move: 300 × 1.5 + 250 × d = 400 × 1.0 is not solvable for the right side. A practical solution would require repositioning multiple children.
(c) A uniform beam has its centre of mass at the geometric centre, which is where the pivot is located. The weight of the beam acts through the pivot, so its perpendicular distance from the pivot is zero, meaning the moment due to the beam's weight is zero (M = Fd = 200 × 0 = 0 Nm). Therefore, the beam's own weight does not create a turning effect and does not affect the balance.
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