question_answer

Three forces P, Q, and R acting on a particle are in equilibrium. If the angle between P and Q is double the angle between P and R, then P is equal to:

A) \[\frac{{{Q}^{2}}+{{R}^{2}}}{R}\]                              

B)                   \[\frac{{{Q}^{2}}-{{R}^{2}}}{Q}\]                             

C)                   \[\frac{{{Q}^{2}}-{{R}^{2}}}{R}\]                              

D)                   \[\frac{{{Q}^{2}}+{{R}^{2}}}{Q}\]

Correct Answer: B

Solution :

Let \[\alpha \] be the angle between P and R. Then angle between P and Q is \[2\alpha \] By Lamis theorem, we have \[\frac{P}{\sin (2\pi -3\alpha )}=\frac{Q}{\sin \alpha }=\frac{R}{\sin 2\alpha }\] \[\Rightarrow \]               \[\frac{P}{-\sin 3\alpha }=\frac{Q}{\sin \alpha }=\frac{R}{\sin 2\alpha }\] \[\Rightarrow \]               \[\frac{P}{-(3-4si{{n}^{2}}\alpha )}=\frac{Q}{1}=\frac{R}{2\cos \alpha }\] \[\Rightarrow \]\[P=-Q(3-4{{\sin }^{2}}\alpha )\]and \[\Rightarrow \] \[\cos \alpha =\frac{R}{2Q}\] \[\Rightarrow \]\[P=-Q(4co{{s}^{2}}\alpha -1)\]and \[\cos \alpha =\frac{R}{2Q}\] \[\Rightarrow \]               \[P=-Q\left( \frac{{{R}^{2}}}{{{Q}^{2}}}-1 \right)\] \[\Rightarrow \]               \[P=\frac{{{Q}^{2}}-{{R}^{2}}}{Q}\]