MathsMechanics › Forces and Newton's laws

Forces and Newton's laws

Draw the forces, add them up, divide by the mass: Newton's second law is three instructions long, and almost every mechanics mark hangs off the first one, the diagram.

Year 12-13EDEXCEL 9MA0 M8

Builds on Kinematics with constant acceleration.

IN THIS TOPIC

  • Draw free body diagrams with weight, normal reaction, tension, thrust and friction correctly placed.
  • Apply F = ma along a chosen direction, and the equilibrium condition when a = 0.
  • Combine F = ma with suvat when a constant force produces constant acceleration.

WHAT YOU PROBABLY THINK

A moving object always has a force pushing it along in the direction of motion.

The diagram is the physics

A free body diagram of a dragged block: weight down, normal reaction up, pull forward, friction backW = mgRTF
FIG. 1Four arrows on a dot: get this picture right and the algebra writes itself.

Every force on the object, and only forces on the object: weight mg down, normal reaction R perpendicular to the surface, tension pulling along a string, thrust pushing along a rod, friction opposing sliding. Forces the object exerts on other things belong on their diagrams, not this one.

Newton's laws at work

F = maIN THE FORMULAE BOOKLET

The first law says no resultant force means no change of velocity; rest and steady speed are the same condition, equilibrium. The second law quantifies the rest: resultant force equals mass times acceleration, applied along a chosen direction with consistent signs. The third law says forces come in equal and opposite pairs acting on different objects, which is why a diagram only ever shows one of each pair.

Resultant force 20 newtons on 5 kilograms gives acceleration 4: the arrows do the subtraction5 kg30 N10 Nresultant 20 N, so a = 20/5 = 4 m s⁻²
FIG. 2Thirty forward, ten back: the object only ever feels the twenty.

WORKED EXAMPLE

Resultant first, then divide

A 5 kg crate is dragged by a horizontal rope with tension 30 N against a constant resistance of 10 N. Find the acceleration, and the speed after 4 s from rest.

Along the motion: F = 30 − 10 = 20 N, so a = 20/5 = 4 m s⁻².

Constant force means constant acceleration, so suvat applies: v = 0 + 4 × 4 = 16 m s⁻¹.

Vertically, nothing moves: R = mg = 49 N, and the vertical equation quietly confirms the crate stays on the floor.

THE EXAM BIT

  • Draw the free body diagram first; examiners award it and everything downstream leans on it.
  • Write F = ma along a declared direction and keep every sign consistent with it.
  • Equilibrium means resultant zero in each direction: resolve twice and solve.
  • F = ma then suvat is the standard two-step; say when the force, and so the acceleration, is constant.

CHECK YOURSELF

A 1200 kg car accelerates at 2.5 m s⁻² against a total resistance of 600 N. Find the driving force.

Show a hint

F in F = ma is the resultant.

Show the answer

Resultant needed: ma = 1200 × 2.5 = 3000 N.

Driving force = 3000 + 600 = 3600 N.

Diagram first: every force on the object, nothing it exerts on anything else.

Resolve, sum with signs, apply F = ma; equilibrium is the a = 0 special case.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

CHECK YOUR PROGRESS

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  • Draw free body diagrams with weight, normal reaction, tension, thrust and friction correctly placed.
  • Apply F = ma along a chosen direction, and the equilibrium condition when a = 0.
  • Combine F = ma with suvat when a constant force produces constant acceleration.

No animated video for this topic yet; these notes stand alone.