Forces, Movement & Changing Shape
Forces & Motion
Balanced forces
- The forces on an object are balanced when they add up to a zero resultant
- A balanced object either stays still or carries on at a constant in a straight line, because its motion does not change at all
- Example: a book at rest on a desk feels a downward weight and an equal upward reaction force from the desk. The two cancel out
Unbalanced forces and Newton's second law
- The forces are unbalanced when they do not add to zero, so there is a non-zero resultant
- A non-zero resultant force makes the object accelerate: it can speed up, slow down, or change direction
- The size of the acceleration is given by Newton's second law of motion:
F = m × a
- where:
- F = resultant force (N)
- m = mass (kg)
- a = acceleration (m/s²)
- Rearranges to a = F / m and m = F / a
Example A — a delivery lorry of mass 1200 kg accelerates uniformly from rest to 18 m/s in 6.0 s. Calculate (i) the acceleration and (ii) the resultant force driving the lorry forward.
- (i) a = (v − u) / t = (18 − 0) / 6.0 = 3.0 m/s²
- (ii) F = m × a = 1200 × 3.0 = 3600 N
Example B — a cyclist plus bicycle have a combined mass of 85 kg. They brake from 12 m/s to a halt in 4.0 s. Calculate the size of the braking force.
- a = (v − u) / t = (0 − 12) / 4.0 = −3.0 m/s²
- F = m × a = 85 × (−3.0) = −255 N
- The negative sign tells you the resultant force points opposite to the motion, i.e. it is the brakes pushing backwards against the cyclist's forward travel
Common exam question
Calculating acceleration or force with F = ma
Question: State the formula linking force, mass and acceleration, then calculate the acceleration from a resultant force and a mass, or the force from a mass and an acceleration (1–4 marks).
Asked in 7 of the 24 papers. The one-mark formula part takes F = m × a in symbols or words and in any rearrangement. The calculation marks are often one for substituting into the formula, one for rearranging and one for evaluating, so write the substituted line (1500 = 600 × a) before the division. A slip in an earlier part is carried forward through a correct method, and where the recall mark sits inside the calculation rather than in a part of its own, a valid substitution implies it. The mass must be in kilograms, so convert a mass given in grams first: a power-of-ten error costs one mark. The sign of a deceleration is ignored, and when the question asks for a set number of significant figures that rounding is an independent mark.
Worked example
F = ma with a unit-conversion step
A resultant force of 6 N acts on a mass of 200 g. Calculate the acceleration.
Solution:
- Convert the mass to kilograms first: 200 g = 200 ÷ 1000 = 0.20 kg
- Rearrange F = m × a to make a the subject: a = F ÷ m
- Substitute and evaluate: a = 6 ÷ 0.20 = 30 m/s²
Convert grams to kilograms before substituting: leaving the mass as 200 g gives 0.03 m/s², a power-of-ten error that costs a mark. Write the substituted formula rather than F = m × a on its own, because the substitution is where the first calculation mark sits.