Question

If the resultant of two forces P and Q acting on a particle is at right angle to Q, then the angle between the forces is

• A. ${\cos }^{-1}\left(\frac{P}{Q}\right)$
• B. $\pi -{\cos }^{-1}\left(\frac{Q}{P}\right)$
• C. $\pi -{\cos }^{-1}\left(\frac{P}{Q}\right)$
• D. None of these

Answer 1

Answer: (B)

$\tan {90}^o=\frac{P\sin \alpha }{Q+P\cos \alpha }$

$\Rightarrow$ $Q+P\cos \alpha =0$
$\Rightarrow$ $\cos \alpha =-\frac{Q}{P}$
$\Rightarrow$ $\alpha ={\cos }^{-1}\left(-\frac{Q}{P}\right)$
$\Rightarrow$ $\alpha =\pi -{\cos }^{-1}\left(\frac{Q}{P}\right)$