Question

Three forces $\overset{\rightarrow}{P},\overset{\rightarrow}{Q}\mspace{6mu}\overset{\rightarrow}{R}$are acting at a point in a plane. The angles between $\overset{\rightarrow}{P}$and $\overset{\rightarrow}{Q}$and$\overset{\rightarrow}{Q}$and$\overset{\rightarrow}{R}$ are 150° and 120⁰ respectively, then for equilibrium, forces P, Q, R are in the ratio

• A. $1:2:\sqrt{3}$
• B. $1:2:3$
• C. 3:2:1
• D. $\sqrt{3}:2:1$

Answer 1

Answer: (D)

$\sqrt{3}:2:1$

Answer 2

Clearly, the angle between P and R is

$360{^\circ} - (150{^\circ} + 120{^\circ}) = 90{^\circ}$. By Lami's theorem,$\frac{P}{\sin 120{^\circ}} = \frac{Q}{\sin 90{^\circ}} = \frac{R}{\sin 150{^\circ}} \Rightarrow \frac{P}{\sqrt{3}/2} = \frac{Q}{1} = \frac{R}{1/2} \Rightarrow \frac{P}{\sqrt{3}} = \frac{Q}{2} = \frac{R}{1}$