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Q.
The resultant of two forces $3P$ and $2P$ is $R$. If the first force is doubled, then the resultant is also doubled. The angle between the two forces is
Motion in a Plane
Solution:
$R^{2}=(3 P)^{2}+(2 P)^{2}+2(3 P)(2 P) \cos\, \theta$
$\Rightarrow R^{2}=13 P^{2}+12 P^{2} \cos \theta\, \dots(1)$
When the first force is doubled, the resultant is doubled,
so, $(2 R)^{2}=(6 P)^{2}+(2 P)^{2}+2(6 P)(2 P) \cos \,\theta$
$\Rightarrow 4 R^{2}=36 P^{2}+4 P^{2}+24 P^{2} \cos \,\theta$
$\Rightarrow R^{2}=10 P^{2}+6 P^{2} \cos \,\theta\, \dots(2)$
Equating (1) & (2), we get
$13 P^{2}+12 P^{2} \cos\, \theta=10 P^{2}+6 P^{2} \cos \,\theta$
$3 P^{2}=-6 P^{2} \cos \,\theta$
so, $\cos\, \theta=-\frac{1}{2}$
$\Rightarrow \theta=120^{\circ}$