Question

Three forces P, Q & R are acting at a point in a plane. The angle between P and Q, Q and R are ${150}^o$ and ${120}^o$ respectively. Then for equilibrium, forces P, Q & R are in the ratio :

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

Answer 1

The correct option is B $\sqrt{3}:1:2$

$Accordingtoquestion.............forequilibrium:\frac{P}{\sin p}=\frac{Q}{\sin q}=\frac{R}{\sin r}where,p=AnglebetweenQ\&R={120}^{0}r=AnglebetweenP\&Q={150}^{0}q=AnglebetweenP\&R={360}^0-\left({120}^0+{150}^0\right)={90}^{0}hence,P:Q:R=\sin p:\sin q:\sin r=\sin {120}^0:\sin {90}^0:sin{150}^0=\frac{\sqrt{3}}{2}:\frac{1}{2}:\frac{1}{1}\therefore \sqrt{3}:1:2sothattheoptionisB.$