WAEC 2012 · Paper 2 · Q17

Forces $(30 \text{ N}, 030^\circ)$ and $(40 \text{ N}, 060^\circ)$ act on a body of mass 20 kg initially at rest on a smooth horizontal floor. Calculate the:

(a)
magnitude of the resultant force;
(b)
direction of the resultant force;
(c)
acceleration of the body.

Worked solution (try it first)

Bearings are measured clockwise from north, so a force $(F, \theta)$ has east part $F\sin\theta$ and north part $F\cos\theta$ .
East: $30\sin30^\circ + 40\sin60^\circ = 15 + 20\sqrt3$
$= 49.641$ N.
North: $30\cos30^\circ + 40\cos60^\circ = 15\sqrt3 + 20$
$= 45.981$ N.

(a)

Magnitude: $\sqrt{49.641^2 + 45.981^2} = 67.66$ N.

(b)

Angle east of north: $\tan^{-1}\dfrac{49.641}{45.981} = 47.2^\circ$ , so the resultant acts on a bearing of $047^\circ$ .

(c)

$F = ma$ : $a = \dfrac{67.66}{20} = 3.383$ m s $^{-2}$ .

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