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How do forces determine the motion of an object, and how do you apply Newton's laws to solve problems?

Force as a vector, the resultant of forces, Newton's three laws of motion, weight and the relationship between mass and weight, connected particles, and resolving forces in two dimensions.

A focused answer to the AQA A-Level Mathematics forces content, covering force as a vector, resultant forces, Newton's three laws, weight, connected particles, and resolving forces in two dimensions.

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  1. What this dot point is asking
  2. Force as a vector and the resultant
  3. Newton's laws
  4. Resolving forces in two dimensions
  5. Connected particles

What this dot point is asking

AQA wants you to treat force as a vector, find resultant forces, apply Newton's three laws, relate mass and weight, solve problems involving connected particles, and resolve forces into components in two dimensions. This is the heart of mechanics: almost every problem reduces to drawing a force diagram and applying F=maF = ma in chosen directions.

Force as a vector and the resultant

Forces add as vectors. If several forces act on a body, the resultant is their vector sum, found by adding components in two perpendicular directions. A body is in equilibrium precisely when the resultant force is zero, which means the components in each direction separately sum to zero. Drawing a clear force diagram, with weight, normal reaction, tension, friction and any applied force, is the essential first step and is often worth marks in its own right.

Newton's laws

Newton's first law is the special case of the second with zero resultant force, giving equilibrium (rest or constant velocity). Newton's third law says that if body A exerts a force on body B, then B exerts an equal and opposite force on A. The key point for exams is that these two forces act on different bodies, so they never cancel when you analyse a single body.

Resolving forces in two dimensions

In two dimensions you resolve each force into perpendicular components, usually horizontal and vertical, or along and perpendicular to a slope, then apply F=maF = ma in each direction independently. A force FF at angle θ\theta to a chosen axis has component FcosθF\cos\theta along that axis and FsinθF\sin\theta perpendicular to it. On an inclined plane it is almost always easier to resolve along and perpendicular to the slope than horizontally and vertically.

Connected particles

The light string assumption means its mass is negligible (so tension is uniform), and inextensible means the connected particles share one acceleration. For a system on a table with a hanging mass, the hanging mass drives the motion; for two hanging masses, the heavier one descends.

Exam-style practice questions

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

AQA 20187 marksPaper 2, Section B. Two particles AA of mass 55 kg and BB of mass 33 kg are connected by a light inextensible string passing over a smooth pulley at the edge of a smooth horizontal table. AA rests on the table and BB hangs vertically. The system is released from rest. Take g=9.8g = 9.8 metres per second squared. (a) Find the acceleration of the system. (b) Find the tension in the string.
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Let the acceleration be aa and the tension TT. For the hanging particle BB (taking downward as positive): 3gT=3a3g - T = 3a. For AA on the smooth table (horizontal): T=5aT = 5a. Adding the two equations eliminates TT: 3g=8a3g = 8a, so a=3×9.88=3.675a = \frac{3 \times 9.8}{8} = 3.675 metres per second squared. Then T=5a=5×3.675=18.375T = 5a = 5 \times 3.675 = 18.375 newtons. Markers reward applying F=maF = ma separately to each particle with the shared acceleration and common tension, and eliminating TT to solve.

AQA 20215 marksPaper 2, Section A. A block of mass 88 kg rests on a smooth plane inclined at 2525 degrees to the horizontal. It is held in equilibrium by a force PP acting up the line of greatest slope. Take g=9.8g = 9.8 metres per second squared. (a) By resolving along the slope, find PP. (b) Find the normal reaction between the block and the plane.
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Resolve along the slope. The component of weight down the slope is 8gsin258g\sin 25^\circ. For equilibrium, P=8gsin25=8×9.8×0.422633.1P = 8g\sin 25^\circ = 8 \times 9.8 \times 0.4226 \approx 33.1 newtons. Resolve perpendicular to the slope: the normal reaction balances the perpendicular weight component, R=8gcos25=8×9.8×0.906371.1R = 8g\cos 25^\circ = 8 \times 9.8 \times 0.9063 \approx 71.1 newtons. Markers reward correct resolution into components parallel and perpendicular to the slope, with sin\sin used for the along-slope component and cos\cos for the normal reaction.

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