Question

If the resultant of two forces of magnitudes P and Q acting at a point at an angle of $60^{o}$ is $\sqrt{7}Q,$ then P/Q is

• A. 1
• B. $\frac{3}{2}$
• C. 2
• D. 4

Answer 1

Answer: (C)

2

Answer 2

$R^{2} = P^{2} + Q^{2} + 2PQ\cos\theta$

⇒ $(\sqrt{7}Q)^{2} = P^{2} + Q^{2} + 2PQ\cos 60{^\circ}$

⇒ $7Q^{2} = P^{2} + Q + PQ$ ⇒ $P^{2} + PQ - 6Q^{2} = 0$

⇒ $P^{2} + 3PQ - 2PQ - 6Q^{2} = 0$

⇒ $P(P + 3Q) - 2Q(P + 3Q) = 0$

⇒ $(P - 2Q)(P + 3Q) = 0$

⇒ $P - 2Q = 0$ or $P + 3Q = 0$

From $P - 2Q = 0$ ⇒ $\frac{P}{Q} = 2$.