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How do forces combine, balance and turn objects, and what keeps a body in equilibrium?

Forces in action: types of force and free-body diagrams, density and pressure, moments and couples, the principle of moments, centre of mass and the conditions for equilibrium, and terminal velocity.

A focused answer to the OCR H556 forces and motion content on forces in action, covering types of force and free-body diagrams, density and pressure, moments, couples and the principle of moments, centre of mass, the conditions for equilibrium, and terminal velocity.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

OCR wants you to identify the forces acting on a body with a free-body diagram, use density and pressure, calculate moments and the turning effect of a couple, apply the principle of moments, locate the centre of mass, state the conditions for equilibrium, and explain terminal velocity.

The answer

Forces and free-body diagrams

Density and pressure

Moments and couples

Equilibrium and centre of mass

Choosing to take moments about the point where an unknown force acts removes that force from the equation, which is the standard trick for beam and ladder problems.

Examples in context

Cranes and bridges are designed by balancing moments about supports. Hydraulic brakes and jacks multiply force using p=FAp = \frac{F}{A} across pistons of different areas. A parachutist reaches terminal velocity when drag rises to equal weight, then a lower terminal velocity once the canopy opens and increases the drag. Stability questions use the centre of mass: a vehicle tips when the line of action of its weight falls outside the wheelbase.

Try this

Q1. State the two conditions for a body to be in equilibrium. [2 marks]

  • Cue. The resultant force is zero and the resultant moment about any point is zero.

Q2. A force of 25 N25\ \text{N} acts at the end of a 0.40 m0.40\ \text{m} spanner, perpendicular to it. Find the moment about the bolt. [2 marks]

  • Cue. moment=Fd=25×0.40=10 N m\text{moment} = Fd = 25 \times 0.40 = 10\ \text{N m}.

Q3. Explain why a skydiver reaches a constant velocity during free fall. [2 marks]

  • Cue. Drag increases with speed until it balances weight; the resultant force is then zero, so the velocity is constant.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20184 marksA uniform plank of weight 60 N60\ \text{N} and length 3.0 m3.0\ \text{m} rests horizontally on two supports, one at each end. A child of weight 240 N240\ \text{N} stands 1.0 m1.0\ \text{m} from the left support. Calculate the upward force from the right-hand support.
Show worked answer →

Take moments about the left support to eliminate its unknown force. The plank's weight acts at its centre (1.5 m1.5\ \text{m} from the left), the child is 1.0 m1.0\ \text{m} from the left, and the right support is 3.0 m3.0\ \text{m} from the left.

Clockwise moments: 60×1.5+240×1.0=90+240=330 N m60 \times 1.5 + 240 \times 1.0 = 90 + 240 = 330\ \text{N m}.

Anticlockwise moment from the right support RR: R×3.0R \times 3.0.

For equilibrium R×3.0=330R \times 3.0 = 330, so R=110 NR = 110\ \text{N}. Markers reward taking moments about a support, including the plank weight at its centre, and the value 110 N110\ \text{N}.

OCR 20213 marksA hydraulic press has a small piston of area 2.0×10−4 m22.0 \times 10^{-4}\ \text{m}^2 and a large piston of area 5.0×10−2 m25.0 \times 10^{-2}\ \text{m}^2. A force of 40 N40\ \text{N} is applied to the small piston. Calculate the force produced at the large piston.
Show worked answer →

Pressure is transmitted equally through the fluid, so p=F1A1=402.0×10−4=2.0×105 Pap = \frac{F_1}{A_1} = \frac{40}{2.0 \times 10^{-4}} = 2.0 \times 10^{5}\ \text{Pa}.

Force at the large piston: F2=pA2=2.0×105×5.0×10−2=1.0×104 NF_2 = pA_2 = 2.0 \times 10^{5} \times 5.0 \times 10^{-2} = 1.0 \times 10^{4}\ \text{N}.

Markers reward calculating the pressure on the small piston, applying the same pressure to the large area, and the value 1.0×104 N1.0 \times 10^{4}\ \text{N}.

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