Resolving forces and equilibrium of a particleEdexcel A-Level Maths: Revision notes

To resolve a force $F$ into two perpendicular components, use trigonometry. If the force makes an angle $\theta$ with a chosen direction, the component in that direction is $F\cos\theta$ and the component perpendicular to it is $F\sin\theta$. Choose the two directions to make the working short: along and perpendicular to a slope, or horizontal and vertical. Forces already in those directions need no resolving, so put as many forces on the axes as you can. Example: a weight $W$ on a line at angle $\theta$ to the vertical has components $W\cos\theta$ along the vertical and $W\sin\theta$ across it.

A particle is in equilibrium when the resultant force is zero, so the sum of the components in each of two perpendicular directions is zero. Resolve in two perpendicular directions and write two equations. Example: a weight of 40 N hangs from a horizontal string and a string at $30^\circ$ to the vertical. Vertically $T_1\cos30^\circ=40$, so $T_1=46.2$ N. Horizontally $T_2=T_1\sin30^\circ=23.1$ N. Three forces in equilibrium also form a closed triangle when drawn head to tail. Four or more forces need resolving.

For a particle of mass $m$ on a plane inclined at $\alpha$ to the horizontal, resolve parallel and perpendicular to the plane:

• weight component down the plane: $mg\sin\alpha$
• weight component into the plane: $mg\cos\alpha$. On a smooth plane the normal reaction is $R=mg\cos\alpha$ when nothing else acts perpendicular to the plane, and the particle slides with $a=g\sin\alpha$. Example: $\alpha=25^\circ$, $m=6$: $R=53.3$ N and $a=4.14\ \text{m s}^{-2}$. A horizontal force $P$ holding the particle at rest has the component $P\cos\alpha$ up the plane (when directed towards the plane), so $P\cos\alpha=mg\sin\alpha$ and $P=mg\tan\alpha$.

Write $F=ma$ along the direction of motion with the resolved components, and a zero-acceleration equation perpendicular to it. Example: 2 kg pulled up a smooth $30^\circ$ slope by 20 N parallel to the slope: $20-2g\sin30^\circ=2a$, so $a=5.1\ \text{m s}^{-2}$. Perpendicular: $R=2g\cos30^\circ=17.0$ N. If the string is cut, only $2g\sin30^\circ$ acts along the slope, giving a deceleration of $4.9\ \text{m s}^{-2}$ while it moves up. Use the suvat equations with that value of $a$ in each stage of the motion.

For a particle on a smooth slope joined over a smooth pulley to a hanging particle, write one equation of motion for each particle (the string is light and inextensible, so the tension and the acceleration are the same for both). Example: $A$ 4 kg on a $30^\circ$ slope, $B$ 3 kg hanging. $B$: $3g-T=3a$. $A$ (moving up): $T-4g\sin30^\circ=4a$. So $a=\frac{29.4-19.6}{7}=1.4$ and $T=25.2$ N. First decide which way the system moves by comparing $3g$ with $4g\sin30^\circ$. If $B$ lands, the string goes slack and $A$ decelerates at $g\sin30^\circ=4.9\ \text{m s}^{-2}$ while it continues up the slope.