Resultant forces and dynamics in a planeEdexcel A-Level Maths: Revision notes

A force has both a magnitude (in newtons) and a direction, so it is a vector. In 2D, forces are given in magnitude-direction form: for example 12 N on a bearing of $060^\circ$. Bearings are measured clockwise from north and written with three digits. To add forces, add them as vectors: the resultant is the single force with the same effect as all of them together. Forces acting in the same line add or subtract. Forces at an angle need components or a triangle. The resultant of 8 N east and 6 N north is $\sqrt{8^2+6^2}=10$ N, at a bearing of $\tan^{-1}\left(\frac{8}{6}\right)=053^\circ$.

Choose perpendicular directions, usually east and north. A force $F$ on a bearing $\theta$ has components:

• east: $F\sin\theta$
• north: $F\cos\theta$. This formula works for every bearing, including those over $90^\circ$, because the signs come out correctly. For $F=6$ N at $210^\circ$: east $=6\sin210^\circ=-3$ N and north $=6\cos210^\circ=-5.20$ N (so it points south-west). Add the east components and add the north components to get the components of the resultant $R_E$ and $R_N$.

From the components $R_E$ and $R_N$: magnitude $R=\sqrt{R_E^2+R_N^2}$ and direction $\tan\alpha=\frac{|R_E|}{|R_N|}$ for the angle from north, then adjust for the quadrant to get the bearing. For two forces $P$ and $Q$ with angle $\phi$ between their lines of action (both drawn from the same point) the parallelogram gives $R^2=P^2+Q^2+2PQ\cos\phi$. Equivalently, draw the triangle of forces head to tail: the interior angle is $180^\circ-\phi$, and the cosine rule then has $-2PQ\cos(180^\circ-\phi)$. Example: 20 N and 15 N with $60^\circ$ between them: $R^2=400+225+2(20)(15)\cos60^\circ=925$, so $R=30.4$ N.

Newton's second law is a vector equation: $\mathbf{F}=m\mathbf{a}$. The acceleration is in the direction of the resultant force, with magnitude $a=\frac{R}{m}$. In components: $a_E=\frac{R_E}{m}$ and $a_N=\frac{R_N}{m}$. A body that starts at rest moves in a straight line in the direction of the resultant while the forces are constant. Example: 4 kg particle, resultant 10 N: $a=2.5\ \text{m s}^{-2}$ and $v=2.5t$. Use suvat separately in each component direction, then combine: speed $=\sqrt{v_E^2+v_N^2}$.

If a particle is in equilibrium under several forces, the resultant is zero. If one force is unknown, it must be equal and opposite to the resultant of the others. Example: 12 N on $060^\circ$ and 5 N on $150^\circ$ have a resultant of 13 N on $083^\circ$ (to the nearest degree), so the force needed for equilibrium is 13 N on $263^\circ$. When a force is removed, the resultant of the others is equal and opposite to the force removed, so it gives the new acceleration directly.