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 $P$
$=OM=R\cos \theta =Q$ (given)

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