Resolving forces and resultantsAQA A-Level Maths: Revision notes

A force is a vector with magnitude and direction. It can be written as $F\mathbf{i}+G\mathbf{j}$ or as a magnitude and angle. To resolve a force of magnitude $F$ at angle $\theta$ to a chosen direction, take two perpendicular directions: $\text{along: }F\cos\theta,\qquad \text{perpendicular: }F\sin\theta.$ The angle $\theta$ must be the angle between the force and the direction used for the cosine. For a $20$ N force at $60^\circ$ above the horizontal: horizontal $=10$ N, vertical $=17.3$ N. On a plane inclined at $\alpha$ to the horizontal, the weight $mg$ resolves into $mg\sin\alpha$ down the plane and $mg\cos\alpha$ into the plane (perpendicular).

The resultant of several forces is the single force with the same effect: the vector sum. In component form add the $\mathbf{i}$ parts and the $\mathbf{j}$ parts separately: $(4\mathbf{i}-3\mathbf{j})+(-7\mathbf{i}+5\mathbf{j})=-3\mathbf{i}+2\mathbf{j}.$ For a resultant $X\mathbf{i}+Y\mathbf{j}$: $\text{magnitude}=\sqrt{X^2+Y^2},\qquad \tan\theta=\left|\frac{Y}{X}\right|,$ where $\theta$ is the angle with the $\mathbf{i}$ direction (check the quadrant from the signs). For forces that are not given as vectors, resolve each into horizontal and vertical components first, then add.

A particle is in equilibrium when the resultant force is zero, so it is at rest or moving at constant velocity. In components: $\sum F_x=0,\qquad \sum F_y=0.$ To solve: draw a clear force diagram, choose two perpendicular directions, resolve every force and write two equations. For a lamp of weight $117.6$ N hung from strings at $30^\circ$ and $60^\circ$ to the ceiling: horizontally $T_1\cos30^\circ=T_2\cos60^\circ$, vertically $T_1\sin30^\circ+T_2\sin60^\circ=117.6$, giving $T_1=58.8$ N and $T_2=102$ N. With three forces in equilibrium, any one force equals the negative of the sum of the other two: $\mathbf{R}=-(\mathbf{P}+\mathbf{Q})$. Modelling words: light means negligible mass, smooth means no friction, inextensible means constant length.

Newton's second law is $\mathbf{F}=m\mathbf{a}$: the resultant force equals mass times acceleration, in the same direction. When forces are not along the direction of motion, resolve them and apply the law in each direction separately. For a block on a smooth slope of angle $\alpha$: perpendicular to the plane there is no acceleration, so $R=mg\cos\alpha$; down the plane $mg\sin\alpha=ma$, so $a=g\sin\alpha$. For $\alpha=30^\circ$, $a=4.9\ \text{m s}^{-2}$, independent of the mass. Use $W=mg$ for weight with $g=9.8\ \text{m s}^{-2}$. Once the acceleration is known, the constant acceleration equations give velocity and displacement.

If the forces are vectors, so are the resultant and acceleration, and the vector equations of motion apply: $\mathbf{F}=m\mathbf{a},\quad \mathbf{v}=\mathbf{u}+\mathbf{a}t,\quad \mathbf{r}=\mathbf{u}t+\tfrac12\mathbf{a}t^2.$ Example: forces $(6\mathbf{i}+4\mathbf{j})$ N and $(-2\mathbf{i}+8\mathbf{j})$ N act on a $2$ kg particle starting from rest at $O$. Resultant $=4\mathbf{i}+12\mathbf{j}$, so $\mathbf{a}=2\mathbf{i}+6\mathbf{j}$ and $|\mathbf{a}|=6.32$. After $3$ s, $\mathbf{r}=\frac12(2\mathbf{i}+6\mathbf{j})(9)=9\mathbf{i}+27\mathbf{j}$ and the distance from $O$ is $\sqrt{810}=28.5$ m. If the resultant force is constant, the particle moves in a straight line only when it starts at rest or $\mathbf{u}$ is parallel to $\mathbf{a}$.