Question

Two forces $P$ and $Q$ act such that the resultant $R$ is equal to the magnitude of $P$. If $P$ is doubled, what is the angle made by the new resultant with the force $Q$?

• A. $0^{\circ }$
• B. ${30}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D ${90}^{\circ }$


$R=\sqrt{P^2+Q^2+2PQ\cos \theta }$
where $\theta$ is the angle between $P$ and $Q$
According to question,
$P^2=P^2+Q^2+2PQ\cos \theta$
$⟹Q^2+2PQ\cos \theta =0$
$⟹Q(Q+2P\cos \theta )=0$...(1)
If $\alpha$ is the angle made by the new resultant with $Q$ then,
$\tan \alpha =\frac{2P\sin \theta }{Q+2P\cos \theta }$
Using equ. (1), we can say
$\tan \alpha =\frac{2P\sin \theta }{0}$
$⟹\alpha ={90}^0$