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How do forces determine how objects move?

Scalars and vectors, resolving and combining forces, free-body diagrams, Newton's three laws of motion, weight, friction and the conditions for equilibrium and moments.

A focused answer to the Edexcel 9PH0 forces content, covering scalars and vectors, resolving and combining forces, free-body diagrams, Newton's three laws, weight, friction, and the conditions for equilibrium and moments.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
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What this dot point is asking

Edexcel wants you to distinguish scalars from vectors, resolve and combine forces, draw free-body diagrams, state and apply Newton's three laws, work with weight and friction, and use the conditions for equilibrium including the principle of moments.

The answer

Scalars, vectors and resolving

To add vectors, either draw them tip-to-tail and measure the resultant, or resolve each into components, add the components separately, and recombine using Pythagoras and trigonometry. Resolving is the workhorse of mechanics: choose convenient axes (often along and perpendicular to an incline) so that one component lines up with the motion.

Newton's three laws

The second law is the key calculating tool: find the resultant force, then a=Fma = \frac{F}{m}. The third law explains rocket propulsion, walking, and why a book on a table is not the same Newton pair as the table pushing the book up (that is the normal contact pair with the book pushing down on the table).

Weight, friction and free-body diagrams

Weight is the gravitational force W=mgW = mg, acting from the centre of mass. Friction opposes relative sliding and depends on the surfaces and the normal contact force, not on the contact area. A free-body diagram shows one chosen body with every force acting on it (weight, normal contact force, tension, friction, applied force) drawn as arrows from the body; it is the essential first step before resolving and applying F=maF = ma.

Equilibrium and moments

For a body in equilibrium under three forces, the three force vectors form a closed triangle. Choosing to take moments about the point where an unknown force acts eliminates that force from the equation, simplifying the algebra.

Examples in context

A climber's rope tension and the normal force from the rock are resolved to check equilibrium on a slope. A crane's jib is analysed with moments to keep the load balanced about the pivot. Seat belts and crumple zones apply Newton's second law in the momentum form to reduce the force during a crash. A rocket rises by Newton's third law, expelling exhaust gas backwards so the gas pushes the rocket forwards. A ladder against a wall is a classic three-force equilibrium-and-moments problem.

Try this

Q1. State Newton's second law in its momentum form. [1 mark]

  • Cue. The resultant force equals the rate of change of momentum, F=ΔpΔtF = \frac{\Delta p}{\Delta t}.

Q2. A force of 2020 N acts at 6060 degrees to the horizontal. Find its horizontal component. [2 marks]

  • Cue. Fx=20cos60=10F_x = 20\cos 60 = 10 N.

Q3. A 0.500.50 m spanner needs a moment of 1515 N m to loosen a bolt. Find the perpendicular force required at the end. [2 marks]

  • Cue. F=momentd=150.50=30F = \frac{\text{moment}}{d} = \frac{15}{0.50} = 30 N.

Exam-style practice questions

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

Edexcel 20174 marksA box of mass 1212 kg is pulled along a rough horizontal floor by a rope at 3030 degrees above the horizontal with a tension of 4040 N. The box accelerates at 0.500.50 m per second squared. Calculate the frictional force on the box.
Show worked answer →

Resolve the tension horizontally: Tx=40cos30=40×0.866=34.6T_x = 40\cos 30 = 40 \times 0.866 = 34.6 N.

Apply Newton's second law horizontally: TxFfriction=ma=12×0.50=6.0T_x - F_{\text{friction}} = ma = 12 \times 0.50 = 6.0 N.

So Ffriction=34.66.0=28.629F_{\text{friction}} = 34.6 - 6.0 = 28.6 \approx 29 N.

Markers reward resolving the tension, applying F=maF = ma to the horizontal direction, and the value about 2929 N.

Edexcel 20204 marksA uniform beam of weight 8080 N and length 4.04.0 m rests on a pivot 1.01.0 m from one end. A 120120 N weight hangs from that nearer end. Calculate the force needed at the far end to balance the beam.
Show worked answer →

Take moments about the pivot. The beam's weight acts at its centre, 1.01.0 m on the far side of the pivot; the 120120 N hangs 1.01.0 m on the near side; the unknown force FF acts 3.03.0 m from the pivot on the far side.

Clockwise (far side) moments: 80×1.0+F×3.080 \times 1.0 + F \times 3.0. Anticlockwise (near side): 120×1.0120 \times 1.0.

For equilibrium: 120×1.0=80×1.0+3.0F120 \times 1.0 = 80 \times 1.0 + 3.0F, so 3.0F=403.0F = 40, giving F=13F = 13 N.

Markers reward taking moments about the pivot, including the beam's weight at its centre, and the value about 1313 N.

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