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NotesPhysicsTopic 1.2Free-body diagrams, equilibrium & resolving forces
Back to Physics Topics
1.2.12 min read

Free-body diagrams, equilibrium & resolving forces

IB Physics • Unit 1

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Contents

  • Forces, free-body diagrams & equilibrium
  • Resolving a force into components
  • Exam-style question
The big idea: A book rests on a table and doesn't move. Why? Gravity pulls it down, the table pushes up just as hard — the two forces cancel.

That balance is called equilibrium. Here you'll draw the forces on an object and check they add up to zero.
ForceWhat it isWhich way it points
Weight (Fg)the pull of gravityalways straight down
Normal (N)a surface pushing backperpendicular to the surface
Tension (T)a pull along a rope or stringalong the rope, away from the object
Friction / draga surface or fluid resisting motionopposes the motion

Free-body diagram of a box on a rough surface: one arrow per force acting ON the box — weight down, the table's normal push up, the applied pull along, and friction opposing it.

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Spot it: Draw only the forces on the object — not the forces it pushes back on.

Equilibrium = the forces balance, so the net force is zero. That can mean staying still or moving at steady speed.

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A slanted force is hard to add up. The trick is to resolve it — split it into a horizontal part and a vertical part that, together, do the same job. Resolve just means 'break into perpendicular pieces'.

Splitting a slanted force into a sideways part F cos θ and an up/down part F sin θ.

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Horizontal component. Given in the data booklet. θ is measured from the horizontal.
Vertical component. Given in the data booklet. θ is measured from the horizontal.
size (magnitude) of the force (N)
angle the force makes with the horizontal (°)
horizontal component of the force (N)
vertical component of the force (N)
Which is cos, which is sin?: Measure the angle θ from the horizontal.

cos goes with the side next to the angle (the horizontal one); sin goes with the side across from it (the vertical one).

If the angle is given from the vertical instead, swap them.
IB-style questionCalculate[2 marks]

A child pulls a sledge with a 50 N force on a rope at 37° above the horizontal. Find the horizontal and vertical components of the pull. (cos 37° = 0.80, sin 37° = 0.60)

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How this is tested — forces in equilibrium are everywhere in Theme A. It comes up two ways:

Paper 1A

  • Draw or pick the free-body diagram.
  • The right arrows, the right directions — a floating cork, a hanging sign.

Paper 2

  • An object in equilibrium held by ropes.
  • Resolve the tensions, set each direction to zero.
The classic trap: A rope pulled nearly straight still has to balance the weight with only a tiny vertical part — so the tension becomes huge. A small sag means a very big pull.
The equilibrium recipe: Equilibrium means the forces balance, so the net force is zero — and that must be true in each direction on its own.

So left pull = right pull and up pull = down pull.

Resolve every slanted force first, then balance each direction.

This is the free-body diagram the exam wants: the bird as a dot, the two rope tensions pulling up-and-out at a shallow angle, and its weight down. The ropes are drawn long because T turns out much bigger than the 6.0 N weight (the sag angle is exaggerated here so the arrows are readable).

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IB-style questionDetermine[3 marks]

A small bird of weight 6.0 N lands at the exact middle of a washing line. The line sags so each half makes 5.0° with the horizontal. Find the tension T in the line. (sin 5.0° = 0.087)

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what is meant by an object being in translational equilibrium. [1 mark]

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