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

Free-body diagrams, equilibrium & resolving forces

IB Physics · Unit 1

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Contents

  1. 1Forces, free-body diagrams & equilibrium
  2. 2Resolving a force into components
  3. 3Exam-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.
Equilibrium: That balance is called equilibrium. Here you'll draw the forces on an object and check they add up to zero.
A force needs two bodies: Every force is an interaction between two bodies: the table pushes on the book, the Earth pulls on the book.

Take one of the two bodies away and that force is gone.
ForceWhat it isWhich way it points
Weight (Fg)the pull of gravityalways straight down
Normal (FN or N)a surface pushing back: the part of the contact force at right angles to itperpendicular to the surface
Tension (FT or T)a pull along a rope or stringalong the rope, away from the object
Friction / draga surface or fluid resisting motionopposes the motion

Contact forces

  • The two bodies touch
  • Normal force FN: the part of the contact force at right angles to the surface
  • Friction Ff, tension FT, a stretched spring's elastic force
  • Drag in a fluid and buoyancy (upthrust)

Field forces

  • Act across a gap, no touching needed
  • Weight Fg = mg: the gravitational pull of the Earth
  • Electric force Fe between charges
  • Magnetic force Fm on magnets and moving charges

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.

On a slope the normal force tilts with the surface. Draw the forces one at a time:

A box resting on a slope with its weight drawn straight down, the normal force at right angles to the slope and friction up the slope
Three forces, three labels: that is a complete free-body diagram.
When the forces do not balance: Add the arrows in each direction. A 30 N pull to the right and 12 N of friction to the left give a resultant force of 18 N to the right.

A resultant that is not zero means the object is not in equilibrium: it speeds up, slows down or changes direction.

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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.

The exact question below: only the 40 N across-part drags the sledge along; the 30 N up-part just lifts it.

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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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On a slope, tilt your axes: split the weight along the slope and at right angles to it.

The weight of a box on a slope split into mg sin theta down the slope and mg cos theta into the slope
Tilt your axes to match the slope: one along it, one at right angles to it.
The slope pair: Along the slope: mg sin θ. Into the slope: mg cos θ (θ is the angle of the slope).

Divide by m and the part of g along the slope is g sin θ: on an 8.0° ramp, 9.8 × sin 8.0° = 1.4 m s⁻².

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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 in the right directions — the two classics below.

Paper 2

  • An object in equilibrium held by ropes.
  • Resolve the tensions, set each direction to zero.

Paper 1A's two favourites — pick (or draw) the diagram whose arrows balance.

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The classic trap: The sag is how far a rope dips in the middle. A rope pulled nearly straight (a tiny sag) still has to balance the weight with only a tiny vertical part — so the tension becomes huge.
The exam's favourite scene: A bird lands on a washing line and the line barely dips. Watch what that does to the tension:

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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.

Back to that bird. In the exam you don't draw the posts and the line — you redraw the scene as a free-body diagram: the bird becomes a simple box, with one arrow for each force acting on it:

The two halves of the line pull up-and-out along the rope (tension T each side); gravity pulls down (weight 6.0 N). The T arrows are drawn long because the tension ends up much bigger than the weight.

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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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A cork floats at rest, partly submerged, on the surface of still water.

the two forces acting on the cork and the relationship between their magnitudes.
[2 marks]

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