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) $