Question

2. Two forces F1 and F2 act on a particle. The force F1 has magnitude 8 N and acts due east. The resultant of F1 and F2 is a force of magnitude 14 N acting in a direction whose bearing is 120q. Find (i) the magnitude of F2 , (4) (ii) the direction of F2 , giving your answer as a bearing to the nearest degree.

Answer 1

(i) $F_x=14\cdot\cos120°=-7\ (N)$

$F_y=14\cdot\sin120°=12.12\ (N)$

$F_{2y}=12.12\ (N)$ and $F_{2x}=-7-8=-15\ (N)$

$F_2=\sqrt{12.12^2+15^2}=19.3\ (N)$

(ii) $\tan\alpha=12.12/(-15)=-0.80829\to \alpha=141.05°$