WAEC 2011 · Paper 2 · Q8

(a)
A particle of mass $400\text{ g}$ is moving under the action of two forces $F_1 = (35\text{ N}, 210^\circ)$ and $F_2 = (35\sqrt3\text{ N}, 300^\circ)$ and a resistance of $40\text{ N}$ . Find the magnitude of the: (i) resultant of $F_1$ and $F_2$ ; (ii) resultant force acting on the particle.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(i)

$F_1$ : $35\sin210^\circ = -17.5$ east and $35\cos210^\circ = -30.31$ north.
$F_2$ : $35\sqrt3\sin300^\circ = -52.5$ east and $35\sqrt3\cos300^\circ = 30.31$ north.
Add: $F_1 + F_2 = -70\mathbf i + 0\mathbf j$ , so the resultant is $70\text{ N}$ (due west).

(ii)

The resistance acts against the motion, so the resultant force is $70 - 40 = 30\text{ N}$ .

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