Forces as vectors and resolving forcesEdexcel International A Level Maths: Revision notes

A force has both a magnitude (measured in newtons, N) and a direction, so it is a vector. In two dimensions a force is written in component form as $\mathbf{F}=(a\mathbf{i}+b\mathbf{j})$ N, or as a column vector, where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors (often east and north, or horizontal and vertical). The magnitude is $|\mathbf{F}|=\sqrt{a^2+b^2}$. If $\theta$ is the angle between $\mathbf{F}$ and $\mathbf{i}$, then $\tan\theta=\frac{b}{a}$ (take care with the quadrant). Forces obey the rules of vector addition.

To resolve a force is to replace it with two perpendicular components that together have the same effect. For a force of magnitude $F$ at angle $\theta$ to a chosen direction: $\text{component along the direction}=F\cos\theta,\qquad\text{component perpendicular to it}=F\sin\theta.$ A 20 N force at $60^\circ$ above the horizontal has horizontal component $20\cos60^\circ=10$ N and vertical component $20\sin60^\circ=10\sqrt{3}\approx17.3$ N. In vector form, $\mathbf{F}=10\mathbf{i}+10\sqrt{3}\mathbf{j}$.

You may resolve in any two perpendicular directions. For forces on a slope inclined at $\alpha$ to the horizontal it is usually easiest to resolve parallel and perpendicular to the plane. A weight $W$ acting vertically downwards has component $W\sin\alpha$ down the slope and $W\cos\alpha$ into the slope. For a weight of 30 N on a $30^\circ$ slope: down the slope $30\sin30^\circ=15$ N; into the slope $30\cos30^\circ=26.0$ N. Always state the direction in which you are resolving, and take one direction as positive.

The resultant is the single force with the same effect as all the forces together. Add forces component by component: $\mathbf{R}=\mathbf{F}_1+\mathbf{F}_2+\ldots$ Worked example. $\mathbf{F}_1=(3\mathbf{i}+4\mathbf{j})$ N and $\mathbf{F}_2=(-7\mathbf{i}+2\mathbf{j})$ N give $\mathbf{R}=(-4\mathbf{i}+6\mathbf{j})$ N. Then $|\mathbf{R}|=\sqrt{16+36}=\sqrt{52}=7.21$ N, and the angle with $\mathbf{j}$ is $\tan^{-1}\left(\frac{4}{6}\right)=33.7^\circ$. For forces given by magnitude and angle, resolve each into $\mathbf{i}$ and $\mathbf{j}$ components first, then add.

If the resultant is known, an unknown force is found by subtraction: $\mathbf{T}=\mathbf{R}-\mathbf{P}-\mathbf{Q}$. Equate $\mathbf{i}$ components and $\mathbf{j}$ components separately to find unknown constants. A force $a\mathbf{i}+b\mathbf{j}$ with $a<0$ points partly west (or left), and with $b<0$ partly south (or down), so describe the direction with a clear reference, such as '$14.6^\circ$ west of south'.