Question

The resultant of two forces P and Q is of magnitude P. If P be doubled, the resultant will be Inclined to Q at an angle

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

Answer 1

Answer: (D)

${90}^o$

$\text{Step 1: Resultant of two forces}$

Let $\theta$ be the angle between $P$ ad $Q$

$R=[P^2+Q^2+2PQ\cos \theta {]}^{1/2}$

$\Rightarrow R^2=P^2+Q^2+2PQ\cos \theta$

$\Rightarrow P^2=P^2+Q^2+2PQ\cos \theta$ $[\because R=P]$

$\Rightarrow Q^2+2PQ\cos \theta =0$

$\Rightarrow Q+2P\cos \theta =0$ $....\left(1\right)$

$\text{Step 2: Angle between new resultant and Q[Ref. fig.]}$

If P is doubled then $P^{{}^{\prime }}=2P$

Angle between $2P$ and $Q$ will still be $\theta$

Let $\alpha$ be the angle between new resultant $R^{{}^{\prime }}$ and $Q$

Since $ODAC$ is parallelogram $OD=AC=2P$

$\tan \alpha =\frac{BC}{OA+AB}=\frac{2P\sin \theta }{Q+2P\cos \theta }$

Using equation (1)

$\tan \alpha =\infty \to \alpha ={90}^o$

Hence Option D is correct.