Question

Three forces $P, Q$, and $R$ acting on a particle are in equilibrium. If the angle between $P$ and $Q$ is double the angle between $P$ and $R$, then $P$ is equal to :

• A. $\frac{Q^{2}+R^{2}}{R}$
• B. $\frac{Q^{2}-R^{2}}{Q}$
• C. $\frac{Q^{2}-R^{2}}{R}$
• D. $\frac{Q^{2}+R^{2}}{Q}$

Answer 1

Answer: (B)

Let $\alpha$ be the angle between $P$ and $R$.
Then angle between $P$ and $Q$ is $2 \alpha$

By Lami's theorem, we have
$\frac{P}{\sin (2 \pi-3 \alpha)}=\frac{Q}{\sin \alpha}=\frac{R}{\sin 2 \alpha}$
$\Rightarrow \frac{P}{-\sin 3 \alpha}=\frac{Q}{\sin \alpha}=\frac{R}{\sin 2 \alpha}$
$\Rightarrow \frac{P}{-\left(3-4 \sin ^{2} \alpha\right)}=\frac{Q}{1}=\frac{R}{2 \cos \alpha}$
$\Rightarrow P=-Q\left(3-4 \sin ^{2} \alpha\right)$
and $\cos \alpha=\frac{R}{2 Q}$
$\Rightarrow P=-Q\left(4 \cos ^{2} \alpha-1\right)$
and $\cos \alpha=\frac{R}{2 Q}$
$\Rightarrow P=-Q\left(\frac{R^{2}}{Q^{2}}-1\right)$
$\Rightarrow P=\frac{Q^{2}-R^{2}}{Q}$