Forces 90N and 120N act in the directions 120° and 240° respectively. Find the resultant of these forces.
\(-45(2i + \sqrt{2}j)\)
\(60(\sqrt{3}i + 7j)\)
\(30(7i + \sqrt{3}j)\)
\(-15(7i + \sqrt{3}j)\)
Correct answer is D
\(F = F\cos\theta + F\sin\theta\)
\(\implies 90N = 90\cos 120° + 90\sin 120°\)
\(120N = 120 \cos 240° + 120 \sin 240°\)
\(R = F_{1} + F_{2} \)
= \((90 \cos 120 + 120 \cos 240)i + (90\sin 120 + 120 \sin 240)j\)
= \(90(-0.5) + 120(-0.5))i + (90(\frac{\sqrt{3}}{2}) + (120(-\frac{\sqrt{3}}{2}))j\)
= \(-105i - 15\sqrt{3}j = -15(7i + \sqrt{3}j)\)