Every Paper 4 question starts the same way: draw the forces, split each one into two perpendicular components, and add the components. Master resolving and the rest of mechanics follows.
A force \(F\) at angle \(\theta\) to a direction has component \(F\cos\theta\) along that direction and \(F\sin\theta\) perpendicular to it. To find a resultant, add all the components in two perpendicular directions to get \(X\) and \(Y\); then \(R = \sqrt{X^2 + Y^2}\) and \(\tan\alpha = \dfrac{Y}{X}\).
Fig. 1: a 20 N force at 35° to the horizontal splits into 16.4 N horizontally and 11.5 N vertically (drawn to scale)
Section 01
Drawing the forces.
Before any calculation, draw a force diagram: the particle as a dot, every force as an arrow starting from it, each labelled. Paper 4 treats bodies as particles, so all the forces act at one point.
Weight \(W = mg\), vertically down, with \(g = 10\ \text{m s}^{-2}\) in this component. A 3 kg mass weighs 30 N.
Normal reaction \(R\): from a surface, perpendicular to that surface.
Friction \(F\): from a rough surface, along the surface, opposing motion (or the tendency to move).
Tension \(T\): in a string, pulling away from the particle along the string. A rod can also push: that is a thrust.
Driving force and resistance: given in the question (for example air resistance); never invent forces that are not stated.
Force on A by B
Newton's third law partner
Ground pushes up on a box (normal reaction)
Box pushes down on the ground, same size
Earth pulls the box down (weight)
Box pulls the Earth up, same size
Examiner trap
Weight and normal reaction are NOT a third-law pair: they act on the same body and are different kinds of force. A third-law pair always acts on two different bodies and is the same type of force.
Section 02
Resolving into components.
A force is a vector: it has a size and a direction. Any force can be replaced by two perpendicular components that together have exactly the same effect. Choose the directions to suit the problem: horizontal and vertical, or parallel and perpendicular to a slope.
Worked example Horizontal and vertical components
A force of 50 N acts at \(30^\circ\) above the horizontal. Find its horizontal and vertical components.
The horizontal direction is next to the \(30^\circ\) angle, so use cos: \(50\cos 30^\circ = 43.3\) N.
The vertical direction is across from the angle, so use sin: \(50\sin 30^\circ = 25\) N.
Answer: 43.3 N horizontally and 25 N vertically upwards.
Worked example Components of weight on a slope
A particle of mass 4 kg rests on a plane inclined at \(25^\circ\) to the horizontal. Find the components of its weight parallel to and perpendicular to the plane.
The angle between the weight and the perpendicular to the plane is also \(25^\circ\) (both lines are turned through the slope angle).
Perpendicular to the plane (next to the angle): \(40\cos 25^\circ = 36.3\) N.
Down the plane (across from the angle): \(40\sin 25^\circ = 16.9\) N.
Answer: 16.9 N down the plane and 36.3 N into the plane.
Your turn
A horizontal force of 15 N acts on a particle on a plane inclined at \(30^\circ\). Find its components parallel to and perpendicular to the plane.
The force makes \(30^\circ\) with the plane, so parallel component \(= 15\cos 30^\circ = 13.0\) N and perpendicular component \(= 15\sin 30^\circ = 7.5\) N.
Examiner trap
Do not learn "horizontal is cos". The rule is: the component next to the angle you are given uses cos. If the angle is measured from the vertical, the vertical component is the cos one.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
A force of 30 N acts at \(60^\circ\) to the horizontal. What is its horizontal component?
Quick check+10 XP
A book rests on a table. Which force is the Newton's third law partner of the book's weight?
Quick check+10 XP
Forces of 5 N and 12 N act at right angles. What is the magnitude of the resultant?
Quick check+10 XP
A particle of mass 3 kg is on a plane inclined at \(30^\circ\). What is the component of its weight down the plane?
Section 03
Finding the resultant.
The resultant is the single force with the same effect as all the forces together. Resolve every force in two perpendicular directions, add with signs, then rebuild one force with Pythagoras and inverse tan. Calculation is always required; a scale drawing earns nothing.
Worked example Two perpendicular forces
Forces of 6 N due east and 8 N due north act on a particle. Find the magnitude and direction of the resultant.
The components are already perpendicular: \(X = 6\), \(Y = 8\).
\(R = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\) N.
\(\tan\alpha = \dfrac{8}{6}\), so \(\alpha = 53.1^\circ\) north of east.
Answer: 10 N at \(53.1^\circ\) to the 6 N force.
Worked example Three forces
Three forces act at a point: 20 N horizontally to the right, 15 N at \(40^\circ\) above the horizontal to the left, and 10 N vertically down. Find the resultant.
Right as positive: \(X = 20 - 15\cos 40^\circ = 20 - 11.49 = 8.509\).
Up as positive: \(Y = 15\sin 40^\circ - 10 = 9.642 - 10 = -0.358\).
\(R = \sqrt{8.509^2 + 0.358^2} = 8.52\) N.
\(\tan\alpha = \dfrac{0.358}{8.509}\), so \(\alpha = 2.4^\circ\); \(Y\) is negative, so the resultant points below the horizontal.
Answer: 8.52 N at \(2.4^\circ\) below the horizontal, to the right.
Worked example Two forces not at right angles
Forces of 10 N and 14 N act at a point with an angle of \(50^\circ\) between them. Find the magnitude of the resultant.
Keep the signs. A force pointing left or down gives a NEGATIVE component; adding its size instead is the most common way to lose every mark in a resultant question.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
Forces of 8 N along the positive \(x\)-axis, 6 N along the positive \(y\)-axis and 5 N at \(45^\circ\) below the negative \(x\)-axis act at a point. Find the magnitude of the resultant.
Exam style+20 XP
A force of \(P\) N acts at \(30^\circ\) above the horizontal and a force of 12 N acts horizontally in the opposite direction. Their resultant is vertical. Find \(P\).
Exam style+20 XP
A horizontal force of 15 N acts on a particle on a plane inclined at \(30^\circ\) to the horizontal, pushing it towards the plane. Find the component of the force parallel to the plane.
Exam style+20 XP
Two forces of 10 N and 14 N act at a point, with \(50^\circ\) between their directions. Find the magnitude of their resultant.
Skill unlocked
Forces resolved.
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