Resolving Forces

AS · M1 · 13 min

Forces pointing in different directions cannot simply be added: $10\ \text{N}$ east and $10\ \text{N}$ north do not make $20\ \text{N}$ of anything. The fix is to resolve each force into two perpendicular components, add the components in each direction separately, and recombine. This one technique underpins almost every Paper 4 question on forces, from finding a resultant to equilibrium, inclined planes, connected particles and work done, so it is worth making it automatic.

Components of a force

Suppose a force $F$ acts at an angle $\theta$ above the horizontal. Its effect is exactly the same as two forces acting together: one horizontal and one vertical. These are its components.

The force F (solid) has the same effect as its two components (dashed): F cos θ along the direction it makes angle θ with, and F sin θ at right angles to that direction.

The force is the hypotenuse of a right-angled triangle, so ordinary trigonometry gives the two sides.

Components of a force

A force $F$ acting at angle $\theta$ to a given direction has

\text{component along that direction} = F\cos\theta, \qquad \text{component perpendicular to it} = F\sin\theta

The rule to remember: the component next to the angle (the adjacent side) uses cos; the component opposite the angle uses sin. Do not memorise "horizontal is cos": that is only true when the angle is measured from the horizontal. If the angle is measured from the vertical, the vertical component is $F\cos\theta$ and the horizontal one is $F\sin\theta$ .

Definition

Resolving a force in a direction means finding its component in that direction. The component of a force $F$ in a direction making angle $\theta$ with the force is $F\cos\theta$ .

Two useful special cases follow from $\cos 0^\circ = 1$ and $\cos 90^\circ = 0$ :

a force has its full value in its own direction;
a force has no component at right angles to itself.

The second fact is what makes resolving so powerful: if you resolve perpendicular to an unknown force, that force disappears from your equation.

Signs

Components are signed. Choose a positive direction for each axis (usually right and up, or up the slope and away from the slope) and give each component a $+$ or $-$ sign according to which way it points. A force of $20\ \text{N}$ pointing left has horizontal component $-20\ \text{N}$ .

Components of a single force

A force of $40\ \text{N}$ acts at $35^\circ$ above the horizontal. Find its horizontal and vertical components.

Solution

The angle is measured from the horizontal, so the horizontal component is adjacent:

\text{horizontal: } 40\cos 35^\circ = 32.8\ \text{N}, \qquad \text{vertical: } 40\sin 35^\circ = 22.9\ \text{N}

(Both to 3 significant figures. Keep the unrounded values, $32.766\ldots$ and $22.943\ldots$ , in your calculator if you need them later.)

Exact values you should know

Cambridge often gives angles through a trigonometric ratio rather than in degrees, so that the components come out exactly. The most common is " $\tan\alpha = \tfrac{3}{4}$ ", which describes a 3-4-5 triangle.

Ratios from Pythagorean triples

Given	$\sin$	$\cos$
$\tan\alpha = \tfrac{3}{4}$	$\tfrac{3}{5} = 0.6$	$\tfrac{4}{5} = 0.8$
$\tan\alpha = \tfrac{4}{3}$	$0.8$	$0.6$
$\tan\alpha = \tfrac{5}{12}$	$\tfrac{5}{13}$	$\tfrac{12}{13}$
$\tan\alpha = \tfrac{7}{24}$	$\tfrac{7}{25}$	$\tfrac{24}{25}$
$\sin\alpha = 0.6$	$0.6$	$0.8$

Also $\sin 30^\circ = \cos 60^\circ = \tfrac12$ , $\cos 30^\circ = \sin 60^\circ = \tfrac{\sqrt3}{2}$ , $\sin 45^\circ = \cos 45^\circ = \tfrac{\sqrt2}{2}$ .

To get $\sin$ and $\cos$ from $\tan\alpha = \tfrac{a}{b}$ , draw a right-angled triangle with opposite side $a$ and adjacent side $b$ , find the hypotenuse $\sqrt{a^2 + b^2}$ , and read off the ratios. Using exact ratios avoids rounding errors, and the mark scheme's answers will be built on them.

The syllabus also expects you to know $\sin\theta = \cos(90^\circ - \theta)$ , $\cos\theta = \sin(90^\circ - \theta)$ , $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta = 1$ . They appear naturally when you resolve: a force at $\theta$ to the horizontal is at $90^\circ - \theta$ to the vertical, so its vertical component can be written $F\cos(90^\circ - \theta) = F\sin\theta$ .

The resultant of several forces

The resultant of a set of forces is the single force that has the same effect as all of them together. To find it:

Finding a resultant

Choose two perpendicular directions, usually the $x$ - and $y$ -axes (or along and perpendicular to a slope).
Resolve every force in each direction, with signs. Add to get $X$ (total component in the $x$ -direction) and $Y$ (total component in the $y$ -direction).
The resultant has magnitude

R = \sqrt{X^2 + Y^2}

Sketch $X$ and $Y$ as the sides of a right-angled triangle to see which quadrant the resultant is in. Its angle with the $x$ -axis is $\tan^{-1}\left|\dfrac{Y}{X}\right|$ , measured from whichever side of the axis the sketch shows.
State the direction in words: " $15.8^\circ$ above the positive $x$ -axis", or "at $42.5^\circ$ to the negative $x$ -axis, above it".

Watch out

Your calculator's $\tan^{-1}$ only gives angles between $-90^\circ$ and $90^\circ$ . If $X$ is negative, the resultant points to the left and the calculator's angle is measured from the wrong side. Always sketch the components, use $\tan^{-1}\left|\tfrac{Y}{X}\right|$ to get the acute angle, and describe the direction from the sketch.

Resultant of three forces

Three coplanar forces act at a point: $25\ \text{N}$ along the positive $x$ -axis, $20\ \text{N}$ at $40^\circ$ above the negative $x$ -axis, and $12\ \text{N}$ at $20^\circ$ to the negative $y$ -axis, to the left of it. Find the magnitude and direction of the resultant.

Forces of 25 N along the positive x-axis, 20 N at 40° above the negative x-axis, and 12 N at 20° to the negative y-axis.

Solution

Resolve in the positive $x$ -direction (to the right). The $20\ \text{N}$ force makes $40^\circ$ with the negative $x$ -axis, so its horizontal component is $20\cos 40^\circ$ to the left. The $12\ \text{N}$ force makes $20^\circ$ with the vertical, so its horizontal component is $12\sin 20^\circ$ to the left.

X = 25 - 20\cos 40^\circ - 12\sin 20^\circ = 25 - 15.321 - 4.104 = 5.575

Resolve in the positive $y$ -direction (upwards):

Y = 20\sin 40^\circ - 12\cos 20^\circ = 12.856 - 11.276 = 1.579

Magnitude:

R = \sqrt{5.575^2 + 1.579^2} = 5.79\ \text{N}

Both $X$ and $Y$ are positive, so the resultant points up and to the right, at

\tan^{-1}\left(\frac{1.579}{5.575}\right) = 15.8^\circ \text{ above the positive } x\text{-axis}

Using exact ratios

Forces of $15\ \text{N}$ , $26\ \text{N}$ and $8\ \text{N}$ act at a point. The $15\ \text{N}$ force makes angle $\alpha$ above the positive $x$ -axis, where $\tan\alpha = \tfrac34$ . The $26\ \text{N}$ force makes angle $\beta$ above the negative $x$ -axis, where $\tan\beta = \tfrac{5}{12}$ . The $8\ \text{N}$ force acts along the negative $y$ -axis. Find the magnitude and direction of the resultant.

The 15 N force makes angle α with the positive x-axis, where tan α = 3/4. The 26 N force makes angle β with the negative x-axis, where tan β = 5/12. The 8 N force acts along the negative y-axis.

Solution

From the triples: $\sin\alpha = 0.6$ , $\cos\alpha = 0.8$ , $\sin\beta = \tfrac{5}{13}$ , $\cos\beta = \tfrac{12}{13}$ .

X = 15(0.8) - 26\left(\tfrac{12}{13}\right) = 12 - 24 = -12

Y = 15(0.6) + 26\left(\tfrac{5}{13}\right) - 8 = 9 + 10 - 8 = 11

R = \sqrt{(-12)^2 + 11^2} = \sqrt{265} = 16.3\ \text{N}

$X < 0$ and $Y > 0$ , so the resultant points up and to the left. The acute angle with the negative $x$ -axis is $\tan^{-1}\left(\tfrac{11}{12}\right) = 42.5^\circ$ . The resultant acts at $42.5^\circ$ above the negative $x$ -axis (equivalently $137.5^\circ$ anticlockwise from the positive $x$ -axis).

Working backwards: finding an unknown force

Often the resultant is given and a force is unknown. Resolving still gives one equation per direction, so two directions let you find two unknowns.

A very common pattern ends with two equations of the form

P\cos\theta = a, \qquad P\sin\theta = b

Square and add, using $\sin^2\theta + \cos^2\theta = 1$ , to get $P$ ; divide to get $\theta$ :

Key result

P = \sqrt{a^2 + b^2}, \qquad \tan\theta = \frac{b}{a}

Finding a force and its direction

Two forces act at a point $O$ . One has magnitude $30\ \text{N}$ and acts along the negative $y$ -axis. The other has magnitude $P\ \text{N}$ and acts at angle $\theta$ above the positive $x$ -axis. The resultant of the two forces has magnitude $20\ \text{N}$ and acts along the positive $x$ -axis. Find $P$ and $\theta$ .

Solution

Resolve in each direction, setting the total equal to the corresponding component of the resultant.

$x$ -direction: $P\cos\theta = 20$

$y$ -direction: $P\sin\theta - 30 = 0$ , so $P\sin\theta = 30$

Square and add:

P^2(\cos^2\theta + \sin^2\theta) = 20^2 + 30^2 = 1300 \quad\Rightarrow\quad P = \sqrt{1300} = 36.1

Divide:

\tan\theta = \frac{30}{20} = 1.5 \quad\Rightarrow\quad \theta = 56.3^\circ

Resultant in a given direction (exam standard)

Four coplanar forces act at a point: $40\ \text{N}$ along the positive $x$ -axis, $10\ \text{N}$ along the positive $y$ -axis, $F\ \text{N}$ at $60^\circ$ above the negative $x$ -axis, and $30\ \text{N}$ at $30^\circ$ below the negative $x$ -axis. The resultant of the four forces acts along the positive $y$ -axis. Find $F$ and the magnitude of the resultant.

Forces of 40 N along the positive x-axis, 10 N along the positive y-axis, F N at 60° above the negative x-axis, and 30 N at 30° below the negative x-axis.

Solution

The resultant is along the $y$ -axis, so its $x$ -component is zero.

40 - F\cos 60^\circ - 30\cos 30^\circ = 0

0.5F = 40 - 25.981 = 14.019 \quad\Rightarrow\quad F = 28.04 = 28.0

The resultant is the total $y$ -component:

Y = 10 + F\sin 60^\circ - 30\sin 30^\circ = 10 + 24.282 - 15 = 19.28

The resultant is $19.3\ \text{N}$ along the positive $y$ -axis. Keep $F = 28.038\ldots$ unrounded when finding $Y$ . Using the rounded $F = 28.0$ gives $Y = 19.249$ , which rounds to $19.2$ and would lose the accuracy mark: a clear illustration of why you never round an intermediate value.

Resolving on an inclined plane

On a slope, the natural directions are parallel to the plane and perpendicular to it, because the normal reaction and friction already lie along these. Only the weight (and any horizontal or vertical applied force) needs resolving.

On a plane inclined at α, the weight mg has component mg sin α down the plane and mg cos α into the plane. The angle between the weight and the normal to the plane is also α.

Weight on a plane inclined at α

\text{down the plane: } mg\sin\alpha, \qquad \text{into the plane: } mg\cos\alpha

Why the angle is $\alpha$ . The weight is perpendicular to the horizontal and the normal is perpendicular to the plane. Turning both lines through $90^\circ$ does not change the angle between them, so the angle between the weight and the normal equals the angle between the horizontal and the plane, $\alpha$ . The component into the plane is adjacent to this angle, so it is $mg\cos\alpha$ .

A quick check: as $\alpha \to 0$ (flat ground), the component down the plane should vanish and the component into the plane should become the full weight. $\sin 0 = 0$ and $\cos 0 = 1$ , so the formulae pass.

Weight components on a slope

A box of mass $5\ \text{kg}$ rests on a plane inclined at $25^\circ$ to the horizontal. Find the components of its weight parallel and perpendicular to the plane.

Solution

Weight $= 5 \times 10 = 50\ \text{N}$ .

\text{parallel (down the plane): } 50\sin 25^\circ = 21.1\ \text{N}

\text{perpendicular (into the plane): } 50\cos 25^\circ = 45.3\ \text{N}

A force applied horizontally to a particle on a slope also has to be resolved. A horizontal force $P$ pushing a particle towards the slope makes angle $\alpha$ with the plane, so it has component $P\cos\alpha$ up the plane and $P\sin\alpha$ into the plane. This case is developed in Equilibrium on inclined planes.

Common mistakes

Where resolving goes wrong

Sin and cos the wrong way round. Decide from the diagram which component is adjacent to the marked angle. If the angle is given from the vertical, the vertical component is the cos one.
Forgetting signs. A force pointing left or down has a negative component in the usual directions. Write each term with its sign before adding.
Calculator in radians. Paper 4 angles are in degrees. A wildly wrong answer usually means the calculator mode is wrong.
Rounding too early. Carry at least 4 or 5 significant figures through the working and round only the final answer to 3 s.f.
Misreading the direction of the resultant. $\tan^{-1}$ does not know which quadrant you are in. Sketch the components.

Exam technique

Exam tip

Questions on resultants are usually worth 4 to 6 marks: typically one mark for each resolved equation, one for the magnitude, one for the direction. Show both resolved totals ( $X = \ldots$ , $Y = \ldots$ ) explicitly so that method marks are available even if the arithmetic slips.
A direction must be stated unambiguously: give the angle and what it is measured from (" $56.3^\circ$ above the positive $x$ -axis", " $32^\circ$ to the left of the upward vertical"). An angle on its own may lose the mark.
When a question gives $\tan\alpha = \tfrac34$ or similar, use the exact values $0.6$ and $0.8$ , not a rounded angle of $36.9^\circ$ . The final answers are often exact integers.
"Calculations are always required": scale drawings are not accepted.

Summary

A force $F$ at angle $\theta$ to a direction has component $F\cos\theta$ in that direction and $F\sin\theta$ perpendicular to it.
A force has no component perpendicular to itself, so resolving perpendicular to an unknown force eliminates it.
The resultant of several forces has components $X = \sum(\text{x-components})$ and $Y = \sum(\text{y-components})$ ; magnitude $\sqrt{X^2 + Y^2}$ , direction from a sketch and $\tan^{-1}|Y/X|$ .
$P\cos\theta = a$ and $P\sin\theta = b$ give $P = \sqrt{a^2 + b^2}$ and $\tan\theta = b/a$ .
On a plane inclined at $\alpha$ : weight component $mg\sin\alpha$ down the plane and $mg\cos\alpha$ into the plane.
Learn the 3-4-5, 5-12-13 and 7-24-25 ratios; use exact values whenever the question gives them.

Practice

Question

A force of $50\ \text{N}$ acts at $28^\circ$ above the horizontal. Find its horizontal and vertical components.
Two forces of magnitudes $9\ \text{N}$ and $12\ \text{N}$ act at a point at right angles to each other. Find the magnitude of their resultant and the angle it makes with the $9\ \text{N}$ force.
A particle of mass $4\ \text{kg}$ is on a plane inclined at $20^\circ$ to the horizontal. Find the components of its weight parallel and perpendicular to the plane.
Forces of $8\ \text{N}$ along the positive $x$ -axis, $6\ \text{N}$ along the positive $y$ -axis and $10\ \text{N}$ at $45^\circ$ below the negative $x$ -axis act at a point. Find the magnitude and direction of the resultant.
Three forces act at a point: $P\ \text{N}$ at angle $\theta$ above the positive $x$ -axis, $18\ \text{N}$ along the negative $x$ -axis and $7\ \text{N}$ along the negative $y$ -axis. Their resultant is $10\ \text{N}$ along the positive $y$ -axis. Find $P$ and $\theta$ .
A force of $50\ \text{N}$ acts at angle $\alpha$ above the positive $x$ -axis, where $\tan\alpha = \tfrac{7}{24}$ . Forces of $20\ \text{N}$ along the negative $x$ -axis and $30\ \text{N}$ along the negative $y$ -axis also act. Find the exact components of the resultant, its magnitude, and its direction.
Two forces, each of magnitude $10\ \text{N}$ , act at a point with an angle of $60^\circ$ between them. Find the magnitude of their resultant.
A force of $20\ \text{N}$ acts along the positive $x$ -axis and a force of $P\ \text{N}$ acts at angle $\theta$ above the negative $x$ -axis. The resultant has magnitude $25\ \text{N}$ and is perpendicular to the $20\ \text{N}$ force. Find $P$ and $\theta$ .
Two forces of magnitudes $6\ \text{N}$ and $10\ \text{N}$ act at a point, with angle $\theta$ between them. Their resultant has magnitude $14\ \text{N}$ . Find $\theta$ , and find the angle between the resultant and the $6\ \text{N}$ force.

Answers

Horizontal $50\cos 28^\circ = 44.1\ \text{N}$ ; vertical $50\sin 28^\circ = 23.5\ \text{N}$ .
$R = \sqrt{9^2 + 12^2} = 15\ \text{N}$ , at $\tan^{-1}\left(\tfrac{12}{9}\right) = 53.1^\circ$ to the $9\ \text{N}$ force.
Weight $40\ \text{N}$ . Parallel: $40\sin 20^\circ = 13.7\ \text{N}$ down the plane. Perpendicular: $40\cos 20^\circ = 37.6\ \text{N}$ into the plane.
$X = 8 - 10\cos 45^\circ = 0.929$ ; $Y = 6 - 10\sin 45^\circ = -1.071$ . $R = \sqrt{0.929^2 + 1.071^2} = 1.42\ \text{N}$ . $X > 0$ , $Y < 0$ : the resultant points down and to the right, at $\tan^{-1}\left(\tfrac{1.071}{0.929}\right) = 49.1^\circ$ below the positive $x$ -axis.
$x$ : $P\cos\theta - 18 = 0$ ; $y$ : $P\sin\theta - 7 = 10$ . So $P\cos\theta = 18$ , $P\sin\theta = 17$ , giving $P = \sqrt{613} = 24.8$ and $\theta = \tan^{-1}\left(\tfrac{17}{18}\right) = 43.4^\circ$ .
$\sin\alpha = \tfrac{7}{25}$ , $\cos\alpha = \tfrac{24}{25}$ , so the $50\ \text{N}$ force has components $48$ and $14$ . $X = 48 - 20 = 28$ , $Y = 14 - 30 = -16$ . $R = \sqrt{1040} = 32.2\ \text{N}$ , at $\tan^{-1}\left(\tfrac{16}{28}\right) = 29.7^\circ$ below the positive $x$ -axis.
Take one force along the $x$ -axis. $X = 10 + 10\cos 60^\circ = 15$ , $Y = 10\sin 60^\circ = 8.660$ . $R = \sqrt{15^2 + 8.660^2} = 17.3\ \text{N}$ (exactly $10\sqrt3$ ), along the bisector of the angle between the forces.
The resultant is along the $y$ -axis. $x$ : $20 - P\cos\theta = 0$ ; $y$ : $P\sin\theta = 25$ . So $P = \sqrt{20^2 + 25^2} = \sqrt{1025} = 32.0$ and $\theta = \tan^{-1}\left(\tfrac{25}{20}\right) = 51.3^\circ$ .
Take the $6\ \text{N}$ force along the $x$ -axis. $X = 6 + 10\cos\theta$ , $Y = 10\sin\theta$ . Then

X^2 + Y^2 = 36 + 120\cos\theta + 100\cos^2\theta + 100\sin^2\theta = 136 + 120\cos\theta = 196

so $\cos\theta = 0.5$ and $\theta = 60^\circ$ . Then $X = 11$ , $Y = 8.660$ , and the resultant makes $\tan^{-1}\left(\tfrac{8.660}{11}\right) = 38.2^\circ$ with the $6\ \text{N}$ force.

Components of a force

Signs

Exact values you should know

The resultant of several forces

Working backwards: finding an unknown force

Resolving on an inclined plane

Common mistakes

Exam technique

Summary

Practice

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