Forces CoreExtended adds F=ma, circular motion & more moments
This is one of the biggest sections in the whole syllabus, so it’s split into four clear parts: how forces stretch, squash and change the motion of objects (including friction and, for Extended, circular motion); how forces turn things around a pivot (moments); and where an object’s weight effectively acts from (centre of gravity) and what that means for stability.
1. Effects of Forces
A force can change the size and shape of an object (stretching, squashing, bending, twisting) and/or change its motion (starting, stopping, speeding up, slowing down, or changing direction).
Load-extension graphs for a spring
In the standard experiment, a spring is hung vertically from a clamp, and a ruler is fixed alongside it to measure its length. Masses (the “load”) are added one at a time, and the new length is recorded after each addition. The extension is the increase in length from the spring’s original, unloaded length. Plotting load (y-axis) against extension (x-axis) gives a load-extension graph.
For an elastic spring, the graph starts as a straight line through the origin — extension is directly proportional to load. This continues up to a point called the limit of proportionality (Extended); beyond this point, the graph curves, and extension is no longer proportional to load.
k = F / x
It is the gradient of the straight-line part of a load-extension graph, and tells you how “stiff” a spring is — a larger k means a stiffer spring that needs more force to stretch by the same amount.
k = F / x = 6 / 4 = 1.5 N/cm
- A spring extends by 5 cm under a load of 10 N, within the limit of proportionality. Calculate its spring constant.
- Explain, in terms of extension and load, what “the limit of proportionality” means on a load-extension graph.
- A different spring has a spring constant of 4 N/cm. Calculate the force needed to extend it by 7 cm.
Show answers
- k = F/x = 10/5 = 2 N/cm
- It is the point on the graph beyond which extension is no longer directly proportional to load — up to that point, doubling the load doubles the extension, but beyond it the graph starts to curve and this relationship breaks down.
- F = k × x = 4 × 7 = 28 N
2. Resultant Force and Motion
When more than one force acts on an object along the same straight line, they can be combined into a single resultant force. Forces in the same direction add together; forces in opposite directions subtract.
An object either remains at rest or continues moving in a straight line at constant speed unless a resultant force acts on it. A resultant force may change an object’s velocity — either its speed, its direction, or both.
Resultant force = 30 − 12 = 18 N to the right
Since there is a resultant force, the box’s velocity will change (it will accelerate in the direction of the resultant).
Resultant force = 8 − 8 = 0 N
Since there is no resultant force, the frame stays at rest (it does not accelerate).
3. Force, Mass and Acceleration Extended only
The connection between resultant force, mass and acceleration is given by:
The force and the acceleration produced are always in the same direction. Here, F should be taken as the resultant force acting on the object, not just any single force acting on it.
a = F / m = 20 / 4 = 5 m/s²
F = m × a = 1200 × 2.5 = 3000 N
Resultant force = 900 − 300 = 600 N
a = F / m = 600 / 500 = 1.2 m/s²
- A resultant force of 15 N acts on a 3 kg object. Calculate its acceleration.
- A 2 kg ball is accelerating at 6 m/s². Calculate the resultant force acting on it.
- A 800 kg car’s engine provides a driving force of 2400 N forward. Resistive forces (friction and air resistance) total 900 N backward. Calculate the car’s acceleration.
Show answers
- a = F/m = 15/3 = 5 m/s²
- F = ma = 2 × 6 = 12 N
- Resultant force = 2400 − 900 = 1500 N. a = F/m = 1500/800 = 1.875 m/s²
4. Friction
Solid friction is the force between two surfaces that may impede (resist) motion and produce heating. It acts to oppose relative motion between the two surfaces.
Friction (drag) also acts on objects moving through fluids: through a liquid, and through a gas (for example, air resistance on a moving car or falling object — recall this from terminal velocity in the Motion note). In both cases, drag increases as speed increases, and always acts to oppose the direction of motion.
5. Motion in a Circular Path Extended only
An object moving in a circle at constant speed is still accelerating, because its direction is constantly changing (remember: velocity is a vector, so a change in direction is a change in velocity, and any change in velocity is an acceleration). This requires a resultant force acting perpendicular to the motion, directed toward the centre of the circle.
| What changes | What stays the same | Effect on the motion |
|---|---|---|
| Force increases | mass and radius constant | speed increases |
| Force increases | mass and speed constant | radius decreases |
| Mass increases | speed and radius constant | an increased force is needed to maintain the circle |
Note: the equation F = mv²/r is not required at this level — you only need to describe these relationships qualitatively.
Once the string snaps, there is no longer a force pulling the ball toward the centre, so the ball travels in a straight line (in the direction it was moving at that instant) — it does not fly directly away from the centre.
- A car goes around a bend of fixed radius. If the driver speeds up while keeping the same radius, explain what must happen to the force needed to keep the car on that curved path.
- Two identical balls are swung in horizontal circles on strings of the same length, but one ball has twice the mass of the other, and both move at the same speed. Which ball needs a greater force to keep it moving in its circle, and why?
Show answers
- The force needed must increase, since increasing speed while keeping the mass and radius constant requires a greater force directed toward the centre of the bend.
- The heavier ball needs a greater force, because for the same speed and radius, an increased mass requires an increased force to keep the object moving on that circular path.
6. Turning Effect of Forces: Moments
The moment of a force is a measure of its turning effect. Everyday examples include using a spanner to turn a nut, pushing open a door, or using a see-saw.
Note carefully: it must be the perpendicular distance from the pivot to the line of action of the force — if a force is applied at an angle, you need the shortest (perpendicular) distance, not just any distance along the object.
Moment = force × perpendicular distance = 25 × 0.4 = 10 N m
The Principle of Moments
When an object is balanced (in equilibrium), the sum of the clockwise moments about a pivot equals the sum of the anticlockwise moments about that same pivot. For a simple beam balancing with one force on each side, this becomes straightforward to apply.
Clockwise moment = anticlockwise moment
300 × 1.5 = W × 1
450 = W × 1 → W = 450 N
Total anticlockwise moment (left) = (20 × 0.6) + (10 × 1.2) = 12 + 12 = 24 N m
Clockwise moment (right) = W × 0.8
Setting them equal: W × 0.8 = 24 → W = 30 N
An object is in equilibrium when there is no resultant force and no resultant moment acting on it — this means it will neither accelerate in any direction, nor start rotating.
- A force of 40 N acts at a perpendicular distance of 0.25 m from a pivot. Calculate the moment.
- A see-saw balances with a 250 N child sitting 2 m from the pivot. A second child sits 2.5 m from the pivot on the other side. Find their weight.
- A beam pivots at its centre. On the left: 15 N at 0.5 m and 8 N at 1.0 m from the pivot. On the right: a single force F at 0.4 m from the pivot. Find F for the beam to balance.
Show answers
- Moment = 40 × 0.25 = 10 N m
- 250 × 2 = W × 2.5 → 500 = 2.5W → W = 200 N
- Left total moment = (15 × 0.5) + (8 × 1.0) = 7.5 + 8 = 15.5 N m. F × 0.4 = 15.5 → F = 38.75 N
7. Centre of Gravity
The centre of gravity of an object is the single point through which its entire weight can be considered to act. For a symmetrical object made of uniform material (like a ruler or a sphere), this is at its geometric centre.
2. Hang a plumb line (a thread with a weight on the end) from the same pin, and let it settle.
3. Mark the line of the thread on the lamina, or trace along it.
4. Repeat from a different hole, elsewhere on the edge of the lamina.
5. The centre of gravity is located at the point where the two marked lines cross.
A third hole is often used to check the result — all three lines should cross at the same point.
Centre of gravity and stability
The position of an object’s centre of gravity affects how stable it is. In general, an object with a lower centre of gravity and a wider base is more stable, because it can be tilted further before its centre of gravity moves outside its base and causes it to topple over.
The racing car has a lower centre of gravity and a wider base, so it can lean much further before its centre of gravity passes beyond its base — this is why it is far less likely to tip over when cornering quickly than the taller, narrower bus.
- Describe the experimental method for finding the centre of gravity of an irregularly shaped piece of card.
- Explain why a table lamp with a heavy, wide base is less likely to topple over than one with a light, narrow base.
- A delivery truck is loaded with heavy boxes stacked high above the truck bed rather than spread out at floor level. Explain why this makes the truck more likely to tip over on a sharp bend.
Show answers
- Hang the lamina freely from a pin through a hole near its edge so it can swing freely, hang a plumb line from the same pin, and mark where the line falls across the lamina. Repeat from a different hole. The centre of gravity is where the two marked lines intersect.
- A heavy, wide base gives the lamp a lower centre of gravity and a wider base of support, meaning it has to be tilted much further before its centre of gravity moves outside its base and it topples — making it more stable.
- Stacking heavy boxes high raises the position of the truck’s overall centre of gravity. A higher centre of gravity means the truck doesn’t need to tilt as far before its centre of gravity moves beyond its base (its wheelbase), making it more likely to topple, especially when cornering.
- Forgetting that F = ma requires the resultant force, not just one of several forces acting on an object.
- Using a distance that isn’t perpendicular to the force’s line of action when calculating a moment.
- Forgetting that “equilibrium” requires both zero resultant force AND zero resultant moment — one without the other is not enough.
- Assuming circular motion at constant speed means there’s no acceleration — direction is still changing, so there is still an acceleration (and a resultant force) toward the centre.
- Thinking friction/drag only opposes solids moving on solids — it also acts on objects moving through liquids and gases.
- Believing an object automatically topples over as soon as it is tilted at all — it only topples once its centre of gravity moves beyond the edge of its base.
8. Quick-Fire Challenge Round
- An object moving in a circle at constant speed has zero acceleration.
- The moment of a force depends on the perpendicular distance from the pivot to the line of action of the force.
- F = ma uses the resultant force acting on an object.
- A lower centre of gravity and a wider base make an object more stable.
Show answers
- False — even at constant speed, the direction is continually changing, so the velocity (a vector) is changing, meaning there is an acceleration directed toward the centre of the circle.
- True — this is exactly why the distance used in a moment calculation must be perpendicular to the force.
- True — F in F = ma refers to the resultant (net) force, not any single individual force.
- True — both factors mean the object can be tilted further before its centre of gravity passes beyond its base and it topples.
- “A 10 N driving force and a 4 N resistive force act on an object, so to find its acceleration, use a = 10/m.”
- “A force of 15 N acts 0.3 m along a beam from the pivot, but at an angle, so the moment is 15 × 0.3 = 4.5 N m.”
- “A ball on a string moving in a circle at constant speed isn’t accelerating because its speed never changes.”
Show answers
- The resultant force should be used, not just the driving force. Resultant = 10 − 4 = 6 N, so a = 6/m.
- Since the force acts at an angle, 0.3 m is not necessarily the perpendicular distance from the pivot to the force’s line of action — the perpendicular distance must be used, which may be shorter than 0.3 m depending on the angle.
- Acceleration depends on velocity, which is a vector including direction, not just speed. Even though speed is constant, direction is continually changing, so velocity is changing and the ball is accelerating (toward the centre of the circle).
9. Mixed Practice — Bringing It All Together
- A spring extends by 6 cm under a 12 N load, within the limit of proportionality. Find its spring constant.
- A resultant force of 45 N acts on a 9 kg object. Find its acceleration.
- A force of 18 N acts at a perpendicular distance of 0.5 m from a pivot. Calculate the moment.
- A beam pivots at its centre. On the left: a 12 N force at 0.6 m from the pivot. On the right: a 9 N force at distance d from the pivot. Find d for the beam to balance.
- Explain why a bus with a low centre of gravity and wide wheelbase is less likely to tip over than one with a high centre of gravity and narrow wheelbase.
Show answers
- k = F/x = 12/6 = 2 N/cm
- a = F/m = 45/9 = 5 m/s²
- Moment = 18 × 0.5 = 9 N m
- 12 × 0.6 = 9 × d → 7.2 = 9d → d = 0.8 m
- A low centre of gravity and wide wheelbase mean the bus can be tilted much further before its centre of gravity passes beyond its base of support and causes it to topple, making it more stable than a bus with a higher centre of gravity and narrower base.
Practice at Home
A large topic, so a large set — take it steadily across a few sittings. Parts C and E are Extended-only content.
- Describe the method for obtaining a load-extension graph for a spring.
- A spring extends 8 cm under a load of 16 N. Find its spring constant.
- State two ways a force can change an object.
- A box is pulled with 50 N to the right and friction acts with 20 N to the left. Find the resultant force.
- Explain why a resultant force is needed to change an object’s velocity.
- Give one example each of solid friction, liquid drag, and gas (air) drag.
- A resultant force of 60 N acts on a 12 kg object. Find its acceleration.
- A car’s engine provides 3500 N forward while resistive forces total 1400 N backward, for a 1500 kg car. Find its acceleration.
- Explain why an object moving in a circle at constant speed still requires a resultant force acting on it.
- A force of 22 N acts at a perpendicular distance of 0.35 m from a pivot. Calculate the moment.
- A see-saw balances with a 400 N adult sitting 1 m from the pivot and a child sitting 2 m from the pivot on the other side. Find the child’s weight.
- State the two conditions needed for an object to be in equilibrium.
- A beam pivots at its centre. On the left: 10 N at 0.4 m and 6 N at 0.6 m from the pivot. On the right: a single force F at 0.5 m. Find F for the beam to balance.
- Explain, using the idea of centre of gravity, why a loaded shelf is more likely to tip forward if heavy books are placed at the very top rather than at the bottom.