question_answer

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)         \[2{{\cos }^{-1}}{{\left( \frac{P}{2Q} \right)}^{1/2}}\]

Correct Answer: B

Solution :

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, \[\alpha =2{{\sin }^{-1}}{{\left( \frac{P}{2Q} \right)}^{1/2}}\]