WAEC 2011 · Paper 2 · Q18

(a)
Forces $F_1 = (3\text{ N}, 210^\circ)$ and $F_2 = (4\text{ N}, 120^\circ)$ act on a particle of mass $7\text{ kg}$ which is at rest. Calculate the: (i) acceleration of the particle; (ii) velocity of the particle after 3 seconds.
Separate values with commas, e.g. 3, −2
(b)
$F_1 = (2\mathbf i + 3\mathbf j)\text{ N}$ , $F_2 = (-5\mathbf j)\text{ N}$ and $F_3 = (6\mathbf i - 4\mathbf j)\text{ N}$ act on a body. Find the magnitude and direction of the fourth force that will keep the body in equilibrium.

Worked solution (try it first)

$F_1$ : $3\sin210^\circ = -1.5$ east and $3\cos210^\circ = -2.598$ north.
$F_2$ : $4\sin120^\circ = 3.464$ east and $4\cos120^\circ = -2$ north.
Resultant: $1.964\mathbf i - 4.598\mathbf j$ , of size $\sqrt{3.857 + 21.143} = \sqrt{25}$
$= 5\text{ N}$ .
(A check: the bearings differ by $90^\circ$ , so $\sqrt{3^2 + 4^2} = 5$ .)
$a = \dfrac{F}{m}$
$= \dfrac57$
$\approx 0.71\text{ m s}^{-2}$ .

From rest, $v = at$
$= \frac57 \times 3$
$= \frac{15}{7}$
$\approx 2.14\text{ m s}^{-1}$ .

$F_1 + F_2 + F_3 = (2 + 0 + 6)\mathbf i + (3 - 5 - 4)\mathbf j$
$= 8\mathbf i - 6\mathbf j$ .
The fourth force cancels it: $-8\mathbf i + 6\mathbf j$ .
Its size is $\sqrt{64 + 36} = 10\text{ N}$ .
It points 8 west and 6 north: $\tan^{-1}\frac86 \approx 53^\circ$ west of north, a bearing of about $307^\circ$ .