Resultant Force Formula: R = √(F₁² + F₂² + 2F₁F₂ cos θ)

The resultant force formula for two forces $F_1$ and $F_2$ acting at an angle θ to each other is $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta}$ . Pull a box with 3 N east and 4 N north, at 90° apart, and the effect is a single 5 N pull toward the north east.

The resultant is the one force that could replace all the others and produce the same acceleration. It comes from the parallelogram law of vector addition. For three or more forces, the component method below is faster and less error prone. The resultant on a body is its net force, so this page pairs with the net force formula and the wider laws of motion section.

Quick facts

Two forces
$R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta}$
Direction
$\tan\alpha = \frac{F_2\sin\theta}{F_1 + F_2\cos\theta}$ , α measured from F₁
At right angles
$R = \sqrt{F_1^2 + F_2^2}$
Range
$|F_1 - F_2| \le R \le F_1 + F_2$
Many forces
$R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$
SI unit
newton (N)

Parallelogram law behind the formula

Draw $F_1$ and $F_2$ from the same point and complete the parallelogram. The diagonal from that point is the resultant. Applying the cosine rule to the triangle formed gives $R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta$ . The plus sign appears because θ is the angle between the forces, not the inside angle of the triangle, which is 180° − θ.

The direction follows from the same diagram: $\tan\alpha = \frac{F_2\sin\theta}{F_1 + F_2\cos\theta}$ , where α is the angle between R and $F_1$ .

Special cases to know by heart

Resultant of two forces at key angles
Angle θ	Resultant R	Comment
0°	$F_1 + F_2$	Same direction, maximum
60°	$\sqrt{F_1^2 + F_2^2 + F_1F_2}$	cos 60° = 0.5
90°	$\sqrt{F_1^2 + F_2^2}$	Pythagoras
120°, equal forces F	$F$	Same size as each force
180°	$\|F_1 - F_2\|$	Opposite directions, minimum

Two equal forces F at angle θ give $R = 2F\cos\frac{\theta}{2}$ , bisecting the angle. That shortcut saves time in multiple choice questions.

Worked examples on the resultant force formula

Example 1: forces at right angles

Forces of 3.0 N and 4.0 N act on a point at 90° to each other. Find the resultant and its direction.

$R = \sqrt{3.0^2 + 4.0^2} = \sqrt{25} = 5.0\ \text{N}$

$\tan\alpha = \frac{4.0}{3.0} \Rightarrow \alpha = 53.1^\circ$ from the 3.0 N force.

Example 2: equal forces at 60°

Two 10 N forces act at 60° to each other. Find the resultant.

$R = \sqrt{100 + 100 + 2 \times 10 \times 10 \times 0.5} = \sqrt{300} = 17.3\ \text{N}$

Shortcut check: $2 \times 10 \times \cos 30^\circ = 17.3\ \text{N}$ . It points along the bisector, 30° from each force.

Example 3: three forces by components

A ring is pulled by 20 N east, 30 N north and 12 N west. Find the resultant.

$\sum F_x = 20 - 12 = 8\ \text{N}$ , $\sum F_y = 30\ \text{N}$ .

$R = \sqrt{8^2 + 30^2} = \sqrt{964} = 31.0\ \text{N}$

Direction: $\tan\alpha = \frac{30}{8} \Rightarrow \alpha = 75.1^\circ$ north of east.

Component method for any number of forces

Resolve each force: $F_x = F\cos\phi$ and $F_y = F\sin\phi$ , with φ measured from the positive x axis.
Add all x parts, keeping signs. Do the same for y.
Combine: $R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$ .
Find the direction with $\tan\alpha = \frac{\sum F_y}{\sum F_x}$ , then check the quadrant from the signs.

If both sums are zero, the resultant is zero and the body is in equilibrium. Then the acceleration is zero and, from the force formula, nothing about its velocity changes.

Resultant and equilibrium

The force needed to hold a body still against a resultant R is the equilibrant: the same size, opposite direction. For three forces in equilibrium, Lami's theorem gives $\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}$ , where each angle is the one between the other two forces. Hanging signs held by two cords, solved on the tension force formula page, are a typical use.

Mistakes with resultants

Adding sizes directly when the forces are at an angle.
Using the inner triangle angle in the cosine rule and getting a minus sign.
Forgetting to check that R lies between the difference and the sum of the two forces.
Reading the direction from tan α without checking the quadrant.

Quick test: two forces of 6 N and 8 N can give any resultant from 2 N to 14 N, never 15 N.

Frequently asked questions

What is the formula for resultant force?

For two forces at angle θ, R = √(F₁² + F₂² + 2F₁F₂ cos θ). For many forces, add the x and y components and combine them with Pythagoras.

What is the resultant of two forces at right angles?

Use Pythagoras: R = √(F₁² + F₂²). For 6 N and 8 N it is 10 N.

When is the resultant maximum and minimum?

Maximum, F₁ + F₂, when the forces point the same way. Minimum, |F₁ − F₂|, when they point in opposite directions.

Is resultant force the same as net force?

Yes, for the forces on one body. Each term names the single combined force, and that combination sets the acceleration through F = ma.