Question

Three forces $\mathrm{P},\mathrm{Q},\mathrm{R}$ are acting at a point in a plane. The angle between $\mathrm{P},\mathrm{Q}$ and $\mathrm{Q},\mathrm{R}$ are $150°$ and $120°$ respectively, then for equilibrium; forces $\mathrm{P},\mathrm{Q},\mathrm{R}$ are in the ratio

• A. $1:2:3$
• B. $1:2:3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• C. $3:2:1$
• D. $3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}:2:1$

Answer 1

The correct option is D

$3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}:2:1$

The explanation for the correct answer.

Find the required ratio?

Given: Angle between $\mathrm{P}$ and $\mathrm{Q}$ is $150°$ and an angle between $\mathrm{Q}$ and $\mathrm{R}$ is $120°$

Angle between $\mathrm{P}$ and $\mathrm{R}$$=360°-(150°+120°)=90°$

Using Lami's Theorem$\frac{\mathrm{A}}{\mathrm{sin\alpha }}=\frac{\mathrm{B}}{\mathrm{sin\beta }}=\frac{\mathrm{C}}{\mathrm{sin\gamma }}$, where $\mathrm{A},\mathrm{B},\mathrm{C}$ are forces and $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$ are angles.

$$\begin{gathered} \therefore \frac{\mathrm{P}}{\sin 120°}=\frac{\mathrm{Q}}{\sin 90°}=\frac{\mathrm{R}}{\sin 150°} \\ \frac{\mathrm{P}}{{\displaystyle \raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{\mathrm{Q}}{1}=\frac{\mathrm{R}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ \frac{\mathrm{P}}{\sqrt{3}}=\frac{\mathrm{Q}}{2}=\frac{\mathrm{R}}{1} \end{gathered}$$

Forces $\mathrm{P},\mathrm{Q},\mathrm{R}$ are in the ratio $\sqrt{3}:2:1$

Hence option (d) is the correct answer.