Question

If the resultant of two forces of magnitudes P and Q acting a point at an angle of ${60}^{0}issqrt7Q$ then P/Q is

Answer 1

The resultant of two vector P and Q is given by

$R=\sqrt{P^2+Q^2+2AQcos\theta }$

$\Rightarrow \sqrt{7Q}=\sqrt{P^2+Q^2+2PQ\times \frac{1}{2}}$

$[\because cos{60}^{\circ }=1/2]$

dividing by Q, we have

$\sqrt{7}=\sqrt{(\frac{P}{Q}{)}^2+1(P/Q)}$

$\Rightarrow 7(\frac{P}{Q}{)}^2+P/Q-6=0$

$\Rightarrow (P/Q+3)(P/Q-2)=0$

$\Rightarrow P/Q=2$ $\because P/Q\ne -3$