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How do forces cause acceleration, and how do you analyse the forces acting on a body?

Newton's three laws of motion, weight and the relationship between mass and force, resolving forces, friction and the coefficient of friction, and connected particles.

A focused answer to the Edexcel A-Level Mathematics forces content, covering Newton's three laws of motion, weight, resolving forces, friction and the coefficient of friction, and connected particles such as pulleys.

Generated by Claude Opus 4.811 min answer

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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

Edexcel wants you to apply Newton's three laws of motion, use F=maF = ma, work with weight W=mgW = mg, resolve forces into components, handle friction using FμRF \le \mu R and the limiting case, and analyse connected particles such as masses joined by a string over a pulley.

The answer

Newton's laws

Newton's first law says a body stays at rest or moves at constant velocity unless a resultant force acts. The third law says that if body AA exerts a force on body BB, then BB exerts an equal and opposite force on AA.

Resolving forces

Forces are resolved into perpendicular components, often horizontal and vertical or along and perpendicular to a slope. The body is in equilibrium when the resultant in each direction is zero, and accelerates when there is a non-zero resultant.

Friction

Connected particles

For two masses joined by a light inextensible string over a smooth pulley, write F=maF = ma for each mass using the same acceleration magnitude and the same tension, then solve the simultaneous equations. The string being inextensible is what forces both masses to share the same acceleration magnitude aa, and the pulley being smooth is what forces the tension TT to be the same on both sides. If you let go of either assumption the two equations stop being coupled and the method collapses, so always state them.

A strategy for any force problem

The same routine handles almost every Edexcel forces question. First, draw a clear diagram and mark every force: weight mgmg downward, the normal reaction RR perpendicular to the surface, any applied force, tension along strings, and friction opposing the direction of (attempted) motion. Second, choose convenient perpendicular directions to resolve in. On a slope it is almost always cleanest to use along-the-slope and perpendicular-to-the-slope rather than horizontal and vertical. Third, resolve perpendicular to the motion to find the normal reaction RR, since that direction is usually in equilibrium. Fourth, write F=maF = ma along the direction of motion. Solving these gives the unknown.

Examples in context

Try this

Q1. A force of 1212 N acts on a 33 kg mass on a smooth surface. Find the acceleration. [2 marks]

  • Cue. a=Fm=123=4a = \dfrac{F}{m} = \dfrac{12}{3} = 4 m per second squared.

Q2. A 55 kg block on a rough horizontal surface has μ=0.4\mu = 0.4. Find the maximum friction. Take g=9.8g = 9.8. [3 marks]

  • Cue. R=mg=49R = mg = 49 N, so F=μR=0.4×49=19.6F = \mu R = 0.4 \times 49 = 19.6 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 20196 marksA block of mass 44 kg rests on a rough horizontal plane with coefficient of friction 0.30.3. A horizontal force PP is applied. Calculate the value of PP that gives the block an acceleration of 1.51.5 m s2^{-2}. Take g=9.8g = 9.8 m s2^{-2}.
Show worked answer →

Resolve vertically: the normal reaction R=mg=4×9.8=39.2R = mg = 4 \times 9.8 = 39.2 N (M1, A1).

The maximum (and here kinetic) friction is F=μR=0.3×39.2=11.76F = \mu R = 0.3 \times 39.2 = 11.76 N (M1).

Apply Newton's second law horizontally: PF=maP - F = ma (M1), so P11.76=4×1.5=6P - 11.76 = 4 \times 1.5 = 6 (A1).

Therefore P=6+11.76=17.76P = 6 + 11.76 = 17.76 N, approximately 17.817.8 N (A1).

Markers reward the vertical resolution, the friction value, the horizontal equation of motion, and the final force.

Edexcel 20227 marksTwo particles of mass 33 kg and 55 kg are connected by a light inextensible string over a smooth pulley. The system is released from rest. Find the acceleration of the system and the tension in the string. Take g=9.8g = 9.8 m s2^{-2}.
Show worked answer →

The heavier 55 kg mass descends and the 33 kg mass rises with common acceleration aa and common tension TT (B1).

For the 55 kg mass (downwards positive): 5gT=5a5g - T = 5a (M1).

For the 33 kg mass (upwards positive): T3g=3aT - 3g = 3a (M1).

Add the equations to eliminate TT (M1): 5g3g=8a5g - 3g = 8a, so 2g=8a2g = 8a and a=2×9.88=2.45a = \dfrac{2 \times 9.8}{8} = 2.45 m s2^{-2} (A1).

Substitute back: T=3g+3a=3(9.8)+3(2.45)=29.4+7.35=36.75T = 3g + 3a = 3(9.8) + 3(2.45) = 29.4 + 7.35 = 36.75 N (M1, A1).

Markers reward two equations of motion, eliminating TT, the acceleration, and the tension.

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