Question

Two forces P and Q have a resultant R and the resolved part of R in the direction of P is of magnitude Q. Then the angle between the forces is:

• A. $ {{\sin }^{-1}}{{\left( \frac{P}{2Q} \right)}^{1/2}} $
• B. $ 2{{\sin }^{-1}}{{\left( \frac{P}{2Q} \right)}^{1/2}} $
• C. $ {{\cos }^{-1}}{{\left( \frac{P}{2Q} \right)}^{1/2}} $
• D. $ 2{{\cos }^{-1}}{{\left( \frac{P}{2Q} \right)}^{1/2}} $

Answer 1

Answer: (B)

Let $\alpha$ be the angle between the force and their resultant $R$ makes an angle $\theta$ with the direction of $P$.
Then Resolved part of $R$ in the direction of $P$
$=O M=R \cos \theta=Q$ (given)

$\therefore Q=O M=P+Q \cos \alpha$
$\Rightarrow Q(1-\cos \alpha)=P$
$\Rightarrow Q\left[2 \sin ^{2}\left(\frac{\alpha}{2}\right)\right]=P$
$\Rightarrow \sin ^{2}\left(\frac{\alpha}{2}\right)=\frac{P}{2 Q}$
$\Rightarrow \sin \left(\frac{\alpha}{2}\right)=\left(\frac{P}{2 Q}\right)^{1 / 2}$
$\Rightarrow \left(\frac{\alpha}{2}\right)=\sin ^{-1}\left(\frac{P}{2 Q}\right)^{1 / 2}$
Hence, $\alpha=2 \sin ^{-1}\left(\frac{P}{2 Q}\right)^{1 / 2}$