Forces and Newton's Laws

Overview

Forces questions convert a real-world situation into a model: a particle, a selection of external forces and a direction in which to apply Newton’s laws. In WJEC GCE Mathematics, 2.2.8 Forces and Newton’s laws sits in the 35-mark Mechanics section of AS Unit 2. It is a reliable source of marks because the method is highly visible: correct diagrams, a stated sign convention and a logically formed equation receive credit even before the final calculation.

The topic connects directly to 2.2.6 quantities and units, 2.2.7 kinematics and 2.2.9 vectors. You will meet lifts, towing, smooth horizontal planes and particles connected by light strings over smooth pulleys. The usual command words are state, show, calculate, find and explain. Candidates should think like modellers, not formula hunters: first identify the body or system, then identify every force acting on it, and only then write the resultant-force equation. The central rule is simple: a non-zero resultant force causes acceleration; zero resultant force means equilibrium or constant velocity.

Specification boundary: at 2.2.8, questions are restricted to two perpendicular directions or simple 2-D vector cases. The WJEC teaching guidance says that resolving forces and inclined-plane friction are developed later at 2.4.9. Do not add unnecessary trigonometry to an AS question.

Key Concepts

1. Force diagrams are your mark plan

A free-body diagram shows one chosen object as a particle and every external force acting on that object. It is not a sketch of the whole scene. A block on a smooth table has weight (W) down and normal reaction (R) up. If it is pulled by a string, tension (T) acts along the string. If a lift is considered, the cable force acts up and weight acts down. Draw arrows from the particle, label them and make their directions unambiguous.

Why does this matter? Newton’s second law uses the resultant force, so an omitted force changes the algebra before any numerical work begins. The normal reaction often balances weight on a horizontal surface, but it is still a force and should be shown. “Smooth” means friction is absent. It does not mean normal reaction is absent.

Examiner test: candidates who draw forces on another object, or draw both sides of a Newton’s third-law pair on one diagram, usually cannot earn the diagram or method credit. Keep the question, “What acts on this body?” in view.

2. Newton’s first law and equilibrium

Newton’s first law says that a particle remains at rest, or moves with constant velocity in a straight line, unless acted on by an external resultant force. Therefore:

• (\sum F= 0) implies (a = 0).
• (a = 0) does not imply the particle is stationary. It may move at constant velocity.
• Balanced forces do not vanish individually; they have a zero vector sum.

For a particle in equilibrium, write separate equations in the horizontal and vertical directions where appropriate. In a simple lift travelling at constant speed, upward tension equals downward weight. In a towing question where the car and trailer travel at constant velocity, total driving force equals total resistance.

Memory sentence: zero resultant, zero acceleration, not necessarily zero speed. This distinction is frequently tested by an explanation question.

3. Newton’s second law: resultant force equals mass times acceleration

For constant mass, apply:

[
\sum F = ma
]

The symbol (\sum F) means the signed total of forces in one selected direction, not a single force chosen from the diagram. State a positive direction before substitution. For example, taking up as positive for a lift:

[
T - W = ma.
]

Taking down as positive would give (W-T=ma). Both are correct when the acceleration sign and final direction are consistent. SI units protect you: force in N, mass in kg, acceleration in m s(^{-2}). A newton is kg m s(^{-2}).

4. Mass, weight and lifts

Mass is a scalar measure of inertia, measured in kg. Weight is the gravitational force on that mass, measured in N:

[
W = mg.
]

Use (g=9.8 ext{ m s}^{-2}) unless a question states otherwise. In lift problems, calculate (W) before combining forces. If a lift accelerates upwards, cable tension exceeds weight. If it accelerates downwards, weight exceeds cable tension. If it travels down at a constant velocity, the forces balance, because the acceleration is zero.

A useful sense check is that an upward lift acceleration must be less than (T/m) only after weight has been subtracted. Candidates sometimes use (T=ma), accidentally treating the weight as if it did not exist.

5. Newton’s third law: pairs act on different bodies

Newton’s third law states that if body A exerts a force on body B, then body B exerts an equal-magnitude, opposite-direction force on body A. The two forces have the same type and act along the same line, but they act on different bodies.

For example, the Earth pulls a falling particle down by its weight; the particle pulls the Earth up with an equal gravitational force. The normal reaction on a block is not the third-law partner of the block’s weight. It is the contact force from the table on the block. This distinction earns explanation marks because it shows that candidates have identified the interacting bodies correctly.

6. Connected particles and smooth pulleys

A light, inextensible string over a smooth fixed pulley gives two key modelling results: the connected particles have the same magnitude of acceleration, and the tension is the same throughout that one string. A smooth horizontal plane means no friction; a smooth pulley changes the string’s direction but not its tension.

Use the high-reliability system-then-single method:

1. Draw separate free-body diagrams, but first consider both particles as one system. Choose the expected direction of motion as positive.
   2. Write one (\sum F$= ma) equation using the external forces only. Tension is internal to the two-particle system, so it cancels. This finds the common acceleration.$
   3. Isolate either one particle. Use (\sum F=ma) again to find tension.
2. Check the direction and scale. A hanging particle moving down has (T < mg); no connected system can accelerate faster than the isolated hanging particle would in free fall.

The WJEC 2.2.8 cases include a hanging particle connected either to another hanging particle or to a particle on a smooth horizontal plane. State the assumptions when invited: particles, light inextensible string, smooth fixed pulley, negligible air resistance and constant (g). Credit is given when these choices are used correctly in the modelling.

7. Perpendicular components and simple vectors

At this point in the course, retain horizontal and vertical directions separately. If a force is given as (\mathbf F=18\mathbf i-6\mathbf j) N on a 3 kg particle, then

[
\mathbf a=rac{\mathbf F}{m}=6\mathbf i-2\mathbf j ext{ m s}^{-2}.
]

Treat the (\mathbf i) and (\mathbf j) components independently. The later 2.4.9 material extends this to resolving angled forces, inclined planes and friction. In 2.2.8, clear component notation and correct units are more valuable than adding unneeded trigonometric lines.

Mathematical relationships and formula status

WJEC’s Appendix B lists (W=mg) and (F=ma) under Mechanics Forces and Equilibrium among formulae learners must use without them being supplied. Do not rely on a formula sheet for these relationships.

Practical applications and modelling

Mechanics is a simplified language for real systems. A lift question assumes a constant mass and a uniform cable force. A car-and-trailer question may be modelled as one particle when the pair has no relative motion: total mass is the sum of the masses, while coupling forces become internal. A pulley problem assumes a light, inextensible string and smooth pulley, so tension remains uniform. In reality, strings stretch, pulleys have friction and air resistance may matter. When an exam asks for limitations, name the assumption and explain its effect, for example: “The string may stretch, so the particles may not have exactly equal accelerations.”

There is no separate required practical for this mathematics specification point. The assessed practical skill is model construction: choosing assumptions, representing forces accurately and interpreting whether the calculated result is physically plausible.

Graph and data skills

Force questions frequently connect to 2.2.7 kinematics. On a velocity-time graph, the gradient is acceleration. For a fixed mass, a force-time graph can be interpreted through (a=F/m): a larger resultant force produces a larger acceleration. Do not infer velocity from a force alone without knowing time and initial conditions. When reading a graph, state the unit attached to the gradient or area before drawing a conclusion.