Question

Two forces P and Q, of magnitude $2F\text{and}3F$ respectively are at an angle $\theta$ with each other. If the force $Q$ is doubled, then their resultant also gets doubled. Then the angle $\theta$ is

• A. ${120}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is A ${120}^{\circ }$

Using Parallelogram law,

$R^2=P^2+Q^2+2PQ\cos \theta$
$4F^2+9F^2+12F^2\cos \theta =R^2$
$13F^2+12F^2\cos \theta =R^2$

When force $Q$ is doubled,
$4F^2+36F^2+24F^2\cos \theta =4R^2$
$4F^2+36F^2+24F^2\cos \theta =4(13F^2+12F^2\cos \theta )=52F^2+48F^2\cos \theta$
$\therefore \cos \theta =-\frac{12F^2}{24F^2}=-\frac{1}{2}\Rightarrow \theta ={120}^{\circ }$