Resultant of a force system

By the end of this topic you will be able to

Resolve a force into perpendicular components.
Find the resultant of a system of concurrent coplanar forces.
State the direction of a resultant in the correct quadrant.

A force is a vector

A force has a size (magnitude), a direction and a point of application. Because it has direction, forces cannot be added like ordinary numbers. Two 10 N forces can add up to 20 N, to 0 N, or to anything in between, depending on the angle between them.

When several forces act at the same point (concurrent forces), they can all be replaced by a single force, the resultant $R$ , that has exactly the same effect on the body.

The component method

The most reliable way to add forces is to break each one into horizontal and vertical components, add the components, then put the total back together.

For a force $F$ at angle $\theta$ measured anticlockwise from the $+x$ axis:

F_x = F\cos\theta, \qquad F_y = F\sin\theta

Add all the $x$ -components and all the $y$ -components separately:

R_x = \sum F_x, \qquad R_y = \sum F_y

Then the magnitude and direction of the resultant are

R = \sqrt{R_x^2 + R_y^2}, \qquad \theta_R = \tan^{-1}\frac{R_y}{R_x}

Try it

Use the model below. Switch a force off by setting it to zero. Watch how the force polygon (the forces drawn tip to tail) closes on the resultant, and compare the readouts with your own component sums.

Interactive modelSet up to three forces. The red arrow is their resultant.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

Full screen

Worked example

Three forces act at a point: 100 N at $0^\circ$ , 80 N at $60^\circ$ and 50 N at $150^\circ$ . Find the resultant.

Components.

Force	$F_x$ (N)	$F_y$ (N)
100 N at $0^\circ$	$100\cos 0^\circ = 100.0$	$100\sin 0^\circ = 0$
80 N at $60^\circ$	$80\cos 60^\circ = 40.0$	$80\sin 60^\circ = 69.3$
50 N at $150^\circ$	$50\cos 150^\circ = -43.3$	$50\sin 150^\circ = 25.0$
Sum	$R_x = 96.7$	$R_y = 94.3$

Resultant. $R = \sqrt{96.7^2 + 94.3^2} = 135$ N, and $\theta = \tan^{-1}(94.3/96.7) = 44.3^\circ$ . Both components are positive, so the resultant is in the first quadrant.

Sense check. The three forces mostly point up and to the right, so a resultant of about 135 N at roughly $45^\circ$ is sensible.

Practice problems

Two forces of 30 N and 40 N act at a point at right angles to each other. Find the resultant and its angle from the 30 N force.

Show answer

$R = \sqrt{30^2+40^2} = 50$ N, at $\tan^{-1}(40/30) = 53.1^\circ$ from the 30 N force.
A force of 200 N acts at $40^\circ$ above the horizontal. Find its horizontal and vertical components.

Show answer

$F_x = 200\cos 40^\circ = 153.2$ N and $F_y = 200\sin 40^\circ = 128.6$ N.
Find the resultant of 50 N at $30^\circ$ and 70 N at $135^\circ$ .

Show answer

$R_x = 43.30 - 49.50 = -6.20$ N, $R_y = 25.00 + 49.50 = 74.50$ N. $R = 74.8$ N. $R_x$ is negative and $R_y$ positive, so the angle is $180^\circ - 85.2^\circ = 94.8^\circ$ (second quadrant).
What single extra force would hold the three forces in the worked example in equilibrium?

Show answer

135 N acting at $44.3^\circ + 180^\circ = 224.3^\circ$ (opposite to the resultant).

Key formulae

Components of a force: $F_x = F\cos\theta, \quad F_y = F\sin\theta$
Components of the resultant: $R_x = \sum F_x, \quad R_y = \sum F_y$
Magnitude of the resultant: $R = \sqrt{R_x^2 + R_y^2}$
Direction of the resultant: $\theta = \tan^{-1}\frac{R_y}{R_x}$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (35 min). 5 min: ask what happens when two 10 N forces pull on a ring (students often say 20 N); 10 min: components and the signs in each quadrant; 10 min: worked example, with students doing the table themselves; 10 min: let pairs use the model to find angles that give R = 0.

Activity. Ask students to set F1 = F2 = F3 = 50 N and find three directions that make the resultant zero (answer: $0^\circ$ , $120^\circ$ , $240^\circ$ , a closed equilateral polygon). This previews equilibrium.

Discussion. Why is the polygon rule the same as the component method?

Assessment idea. Give two forces and a required resultant direction; ask for the second force.