Q. Two forces P and Q of magnitude 2F 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 is :

Solution:

$4F^2 + 9F^2 + 12F^2 \; \cos \; \theta = R^2$
$4F^2 + 36 \; F^2 + 24 \; F^2 \; \cos \; \theta = 4R^2$
$4F^2 + 36 \; F^2 + 24 \; F^2 \; \cos \; \theta$
$= 4(13F^2 + 12F^2 \cos \theta) = 52 \; F^2 + 48F^2 \cos \theta$
$\cos \theta = \frac{12 F^2}{24 F^2} = - \frac{1}{2}$

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