Resolving Forces and Inclined Planes

1 Resolving a Force into Components

In Year 1 forces acted along a line — up and down, or left and right. In Year 2 (DfE content area R · Forces and Newton's laws) forces act at any angle in a plane. The trick that makes every such problem manageable is to resolve: replace each force with two perpendicular components.

A force $F$ at angle $\theta$ to the horizontal

horizontal component $F\cos\theta$ · vertical component $F\sin\theta$
in vector form $\mathbf{F} = F\cos\theta\,\mathbf{i} + F\sin\theta\,\mathbf{j}$

"cos is across". The component adjacent to the angle you are given uses $\cos$; the component opposite it uses $\sin$. Always check which line the angle is measured from.

Combining components back into one force

If $\mathbf{R} = X\mathbf{i} + Y\mathbf{j}$: $|\mathbf{R}| = \sqrt{X^2 + Y^2}$, direction $\tan^{-1}\!\left(\dfrac{Y}{X}\right)$ from $\mathbf{i}$

Worked Example 1 — Components of a single force

A rope pulls with a force of $12$ N at $35^\circ$ above the horizontal. Find its horizontal and vertical components.

①Horizontal: $12\cos 35^\circ = 9.83$ N

②Vertical: $12\sin 35^\circ = 6.88$ N

③Check: $\sqrt{9.83^2 + 6.88^2} = 12.0$ ✓

Worked Example 2 — The resultant of three forces

Three forces act on a particle: $5$ N along $\mathbf{i}$, $8$ N at $60^\circ$ to $\mathbf{i}$, and $6$ N at $150^\circ$ to $\mathbf{i}$ (all measured anticlockwise). Find the magnitude and direction of the resultant.

①$X = 5 + 8\cos 60^\circ + 6\cos 150^\circ = 5 + 4 - 5.196 = 3.804$

②$Y = 0 + 8\sin 60^\circ + 6\sin 150^\circ = 6.928 + 3 = 9.928$

③$|\mathbf{R}| = \sqrt{3.804^2 + 9.928^2} = 10.6$ N

④Direction $\tan^{-1}\!\left(\dfrac{9.928}{3.804}\right) = 69.0^\circ$ anticlockwise from $\mathbf{i}$.

Make a table with one row per force and columns for $\mathbf{i}$ and $\mathbf{j}$. It keeps the signs under control.

2 Equilibrium of a Particle

A particle is in equilibrium when the resultant force on it is zero. With coplanar forces that gives two equations: the components in any two perpendicular directions each sum to zero. Two equations means you can find two unknowns.

Equilibrium of a particle

$\sum F_x = 0$ and $\sum F_y = 0$ (equivalently $\sum \mathbf{F} = \mathbf{0}$)

Worked Example 3 — A weight hanging from two strings

A particle of weight $20$ N hangs in equilibrium from two light strings making $30^\circ$ and $60^\circ$ with the horizontal. Find the tensions $T_1$ and $T_2$.

①Horizontal: $T_1\cos 30^\circ = T_2 \cos 60^\circ$, so $T_2 = \sqrt{3}\,T_1$.

②Vertical: $T_1 \sin 30^\circ + T_2 \sin 60^\circ = 20$.

③Substitute: $\tfrac12 T_1 + \sqrt{3}T_1 \cdot \tfrac{\sqrt3}{2} = 2T_1 = 20$.

④$T_1 = 10$ N and $T_2 = 10\sqrt{3} = 17.3$ N.

The steeper string carries the larger tension — a quick plausibility check.

Worked Example 4 — Finding a missing force

A particle is in equilibrium under forces $(3\mathbf{i} - 5\mathbf{j})$ N, $(-7\mathbf{i} + 2\mathbf{j})$ N and $\mathbf{P}$. Find the magnitude and direction of $\mathbf{P}$.

①$\mathbf{P} = -\big[(3 - 7)\mathbf{i} + (-5 + 2)\mathbf{j}\big] = 4\mathbf{i} + 3\mathbf{j}$

②$|\mathbf{P}| = \sqrt{16 + 9} = 5$ N

③Direction $\tan^{-1}\!\left(\tfrac34\right) = 36.9^\circ$ above $\mathbf{i}$.

3 Inclined Planes

On a slope, resolve parallel and perpendicular to the plane, not horizontally and vertically. Then the normal reaction and the motion each lie along one axis.

Particle of mass $m$ on a plane inclined at $\alpha$

component of weight down the slope: $mg\sin\alpha$
component of weight into the slope: $mg\cos\alpha$

$R = mg\cos\alpha$ only if nothing else pushes into or out of the plane. If there is a horizontal force, or a string at an angle to the slope, it has a perpendicular component too, and $R$ changes. Always resolve perpendicular to the plane from scratch.

Worked Example 5 — Sliding down a smooth slope

A particle of mass $4$ kg slides down a smooth plane inclined at $25^\circ$. Find its acceleration and the normal reaction.

①Parallel ($F = ma$): $4g\sin 25^\circ = 4a$, so $a = g\sin 25^\circ = 4.14$ m s$^{-2}$.

②Perpendicular (no acceleration): $R = 4g\cos 25^\circ = 35.5$ N.

On a smooth slope the acceleration $g\sin\alpha$ does not depend on the mass.

Worked Example 6 — Held by a horizontal force

A particle of mass $2$ kg is held at rest on a smooth plane inclined at $30^\circ$ by a horizontal force $P$. Find $P$ and the normal reaction.

①$P$ is horizontal, so it makes $30^\circ$ with the slope: component up the slope $P\cos 30^\circ$, component into the slope $P\sin 30^\circ$.

②Parallel: $P\cos 30^\circ = 2g\sin 30^\circ$, so $P = 2g\tan 30^\circ = 11.3$ N.

③Perpendicular: $R = 2g\cos 30^\circ + P\sin 30^\circ = 16.97 + 5.66 = 22.6$ N.

Here $R \neq mg\cos\alpha$, because $P$ also pushes the particle into the plane.

Worked Example 7 — Pulled up a smooth slope

A $3$ kg box is pulled up a smooth slope inclined at $20^\circ$ by a force of $30$ N parallel to the slope. Find its acceleration.

①Up the slope: $30 - 3g\sin 20^\circ = 3a$

②$30 - 10.06 = 3a$, so $a = 6.65$ m s$^{-2}$ up the slope.

4 Connected Particles on Slopes

The Year 1 pulley problems (see Forces and Newton's Laws) extend directly: write $F = ma$ for each particle separately, in its own direction of motion, then solve simultaneously.

Light inextensible string over a smooth pulley means: the tension is the same throughout the string, and both particles have the same speed and the same magnitude of acceleration.

Worked Example 8 — A slope and a hanging mass

Particle $A$ of mass $3$ kg lies on a smooth plane inclined at $30^\circ$. It is attached by a light inextensible string, passing over a smooth pulley at the top of the plane, to particle $B$ of mass $5$ kg hanging freely. The system is released from rest. Find the acceleration, the tension, and the force on the pulley.

①$B$'s weight $5g = 49$ N exceeds $A$'s component down the slope $3g\sin 30^\circ = 14.7$ N, so $B$ falls.

②$B$ (down): $5g - T = 5a$

③$A$ (up the slope): $T - 3g\sin 30^\circ = 3a$

④Add: $5g - 1.5g = 8a$, so $a = \dfrac{3.5g}{8} = 4.29$ m s$^{-2}$.

⑤$T = 5(g - a) = 5 \times 5.5125 = 27.6$ N.

⑥The two string sections leave the pulley along the slope (down at $30^\circ$ below horizontal) and vertically down, so the angle between them is $60^\circ$. The resultant is $2T\cos 30^\circ = 47.7$ N, along the bisector of the angle between the strings.

Decide the direction of motion first by comparing the driving forces. If you guess wrong on a smooth system you just get a negative acceleration — but once friction is involved (see Friction) a wrong guess gives nonsense, because friction's direction depends on the motion.

5 Quick Reference

Components

$F\cos\theta$ adjacent to $\theta$, $F\sin\theta$ opposite.

Resultant

$\sqrt{X^2+Y^2}$ at $\tan^{-1}(Y/X)$.

Equilibrium

$\sum F_x = 0$, $\sum F_y = 0$.

Down a slope

$mg\sin\alpha$.

Into a slope

$mg\cos\alpha$.

Smooth slope

Released from rest: $a = g\sin\alpha$.

Normal reaction

Resolve perpendicular every time — include all forces.

Pulleys

Same $T$ and same $a$ for both particles.

Two strings at $\phi$

Force on pulley $2T\cos(\phi/2)$.

6 Practice Questions

Use $g = 9.8$ m s$^{-2}$ and give answers to 3 significant figures where they are not exact.

Question 1

Find the horizontal and vertical components of a $50$ N force acting at $40^\circ$ above the horizontal.

▶ Show solution

Horizontal: $50\cos 40^\circ = 38.3$ N. Vertical: $50\sin 40^\circ = 32.1$ N.

Question 2

Find the magnitude and direction of the resultant of $(4\mathbf{i} + 3\mathbf{j})$ N and $(2\mathbf{i} - 7\mathbf{j})$ N.

▶ Show solution

Resultant $= 6\mathbf{i} - 4\mathbf{j}$.

Magnitude $\sqrt{36 + 16} = \sqrt{52} = 7.21$ N.

Direction $\tan^{-1}\!\left(\tfrac46\right) = 33.7^\circ$ below $\mathbf{i}$.

Question 3

A particle of weight $W$ hangs from two light strings making $40^\circ$ and $50^\circ$ with the horizontal. The tension in the first string is $15$ N. Find the other tension and $W$.

▶ Show solution

Horizontal: $15\cos 40^\circ = T_2\cos 50^\circ$, so $T_2 = \dfrac{15\cos 40^\circ}{\cos 50^\circ} = 17.9$ N.

Vertical: $W = 15\sin 40^\circ + T_2\sin 50^\circ = 9.64 + 13.69 = 23.3$ N.

Question 4

A particle is released from rest on a smooth plane inclined at $15^\circ$. Find its speed after it has moved $3$ m down the plane.

▶ Show solution

$a = g\sin 15^\circ = 2.536$ m s$^{-2}$.

$v^2 = u^2 + 2as = 0 + 2(2.536)(3) = 15.22$, so $v = 3.90$ m s$^{-1}$.

Question 5

A particle of mass $5$ kg is held at rest on a smooth plane inclined at $35^\circ$ by a force $P$ acting up the line of greatest slope. Find $P$ and the normal reaction.

▶ Show solution

Parallel: $P = 5g\sin 35^\circ = 28.1$ N.

Perpendicular: $R = 5g\cos 35^\circ = 40.1$ N.

Question 6

A box of mass $2$ kg on a smooth slope inclined at $20^\circ$ is pushed up the slope by a horizontal force of $15$ N. Find its acceleration and the normal reaction.

▶ Show solution

The horizontal force makes $20^\circ$ with the slope.

Parallel: $15\cos 20^\circ - 2g\sin 20^\circ = 2a \Rightarrow 14.10 - 6.70 = 2a \Rightarrow a = 3.70$ m s$^{-2}$.

Perpendicular: $R = 2g\cos 20^\circ + 15\sin 20^\circ = 18.42 + 5.13 = 23.5$ N.

Question 7

A particle is in equilibrium under three forces: $12$ N along $\mathbf{i}$, $10$ N at $120^\circ$ anticlockwise from $\mathbf{i}$, and a third force $\mathbf{F}$. Find the magnitude of $\mathbf{F}$ and the angle it makes with $\mathbf{i}$.

▶ Show solution

Sum of the two known forces: $\mathbf{i}$: $12 + 10\cos 120^\circ = 7$; $\mathbf{j}$: $10\sin 120^\circ = 8.660$.

So $\mathbf{F} = -7\mathbf{i} - 8.660\mathbf{j}$.

$|\mathbf{F}| = \sqrt{49 + 75} = \sqrt{124} = 11.1$ N.

$\tan^{-1}\!\left(\tfrac{8.660}{7}\right) = 51.1^\circ$, and $\mathbf{F}$ is in the third quadrant, so it acts at $180^\circ + 51.1^\circ = 231^\circ$ anticlockwise from $\mathbf{i}$.

Question 8

Particle $A$ ($4$ kg) is on a smooth plane inclined at $40^\circ$, connected by a light inextensible string over a smooth pulley at the top to particle $B$ ($3$ kg) hanging freely. Determine which way the system moves, and find the acceleration and tension.

▶ Show solution

Compare: $3g = 29.4$ N against $4g\sin 40^\circ = 25.2$ N. So $B$ descends.

$B$: $3g - T = 3a$. $A$: $T - 4g\sin 40^\circ = 4a$.

Add: $3g - 4g\sin 40^\circ = 7a \Rightarrow a = \dfrac{4.203}{7} = 0.600$ m s$^{-2}$.

$T = 3(g - a) = 27.6$ N.

Question 9

A particle of mass $3$ kg rests on a smooth plane inclined at $\alpha$, where $\tan\alpha = \tfrac34$. It is held in equilibrium by a force $P$ acting at $30^\circ$ above the line of greatest slope. Find $P$ and the normal reaction.

▶ Show solution

From $\tan\alpha = \tfrac34$: $\sin\alpha = 0.6$, $\cos\alpha = 0.8$.

Parallel: $P\cos 30^\circ = 3g(0.6) = 17.64 \Rightarrow P = 20.4$ N.

Perpendicular: $R + P\sin 30^\circ = 3g(0.8) \Rightarrow R = 23.52 - 10.18 = 13.3$ N.

Here $P$ pulls partly away from the plane, so $R$ is less than $mg\cos\alpha$.

Question 10

A wedge has two smooth faces inclined at $30^\circ$ and $60^\circ$ to the horizontal, meeting at a smooth pulley at the top. Particle $A$ ($4$ kg) is on the $30^\circ$ face and particle $B$ ($3$ kg) on the $60^\circ$ face, joined by a light inextensible string over the pulley. The system is released from rest.

(a) Show that $B$ moves down its face. (b) Find the acceleration and the tension. (c) Find the speed of $B$ after $1.5$ s and the distance it has moved. (d) Find the magnitude of the force exerted on the pulley by the string.

▶ Show solution

(a) Component of $B$'s weight down its face: $3g\sin 60^\circ = 25.5$ N. For $A$: $4g\sin 30^\circ = 19.6$ N. $25.5 > 19.6$, so $B$ moves down.

(b) $B$: $3g\sin 60^\circ - T = 3a$. $A$: $T - 4g\sin 30^\circ = 4a$.

Add: $25.46 - 19.6 = 7a \Rightarrow a = 0.837$ m s$^{-2}$.

$T = 4a + 19.6 = 22.9$ N.

(c) $v = 0 + 0.837 \times 1.5 = 1.26$ m s$^{-1}$; $s = \tfrac12(0.837)(1.5)^2 = 0.942$ m.

(d) The strings run down the two faces, so the angle between them is $180^\circ - 30^\circ - 60^\circ = 90^\circ$.

Resultant $= \sqrt{T^2 + T^2} = T\sqrt{2} = 32.5$ N.

Resolving Forces and Inclined Planes · A Level Maths Year 2 · Created with MathJax