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  4. The square of resultant of two equal forces is three times their product. Angle between the forces is

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Q. The square of resultant of two equal forces is three times their product. Angle between the forces is

AMUAMU 2008

Solution:

Let $\theta$ be the angle between vectors $\vec{ P }$ and $\vec{ Q }$
whose resultant is $\vec{ R }$.
Here, $P=Q$ and $R^{2}=3 P Q=3 P^{2}$
As $ R^{2}=P^{2}+Q^{2}+2 P Q \cos \theta$
$\therefore 3 P^{2}=P^{2}+P^{2}+2 P^{2} \cos \theta$
or $ 3 P^{2}-2 P^{2}=2 P^{2} \cos \theta$
or $ P^{2}=2 P^{2} \cos \theta$
or $ 1=2 \cos \theta$
$\therefore \cos \theta=\frac{1}{2}$,
thus, $\cos \theta=\cos 60^{\circ}$
or $ \theta=60^{\circ}=\frac{\pi}{3}$