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 $\cos \alpha =\frac{R}{2Q}$
$\Rightarrow P=-Q\left(4{\cos }^2\alpha -1\right)$
and $\cos \alpha =\frac{R}{2Q}$
$\Rightarrow P=-Q\left(\frac{R^2}{Q^2}-1\right)$
$\Rightarrow P=\frac{Q^2-R^2}{Q}$