question_answer

Two forces \[P\] and \[Q\] acting at a point have a resultant\[R\]. The resolved of \[R\] in the direction of \[P\] is a magnitude\[Q\]. Then, the angle between the forces and the magnitude of the resultant are

A) \[\theta ={{\cos }^{-1}}\left( \frac{P-Q}{Q} \right),\,\,R=\sqrt{{{P}^{2}}-{{Q}^{2}}+2PQ}\]

B) \[\theta ={{\cos }^{-1}}\left( \frac{Q-P}{Q} \right),\,\,R=\sqrt{{{Q}^{2}}-{{P}^{2}}+2PQ}\]

C) \[\theta ={{\cos }^{-1}}\left( \frac{P-Q}{P} \right),\,\,R=\sqrt{2{{P}^{2}}-{{Q}^{2}}+2PQ}\]

D)  None of the above

Correct Answer: B

Solution :

Let \[\theta \] be the angle between the forces \[P\] and \[Q\], then                 \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta \]           ... (i) Let the resultant \[R\] make an angle \[\alpha \] with the direction of force \[P\] and \[\beta \] with the direction of force\[Q\]. Then,       \[\sin \alpha =\frac{Q\sin \theta }{R},\,\,\cos \alpha =\frac{P+Q\cos \theta }{R}\] Given, resolved part of \[R\] in the direction of                 \[P=Q\] \[\Rightarrow \]               \[P+Q\cos \theta =Q\] \[\Rightarrow \]               \[\cos \theta =\frac{Q-P}{Q}\] \[\Rightarrow \]               \[\theta ={{\cos }^{-1}}\left( \frac{Q-P}{Q} \right)\] which is the required angle. Putting\[\cos \theta =\frac{Q-P}{Q}\]in Eq. (i), we get                 \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2P(Q-P)\] \[\Rightarrow \]               \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ-2{{P}^{2}}\] \[\Rightarrow \]               \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\] \[\Rightarrow \]               \[R=\sqrt{{{Q}^{2}}-{{P}^{2}}+2PQ}\]