Question

Two forces $3N$ and $2N$ are at an angle $\theta$ such that the resultant is $R$. The first force is now increased to $6N$ and the resultant become $2R$. The value of $\theta$ (in $degree$) is

Answer 1

Given,first force, $A=3N$
Second force, $B=2N$
And the resultant is $R$ when angle between them is $\theta$

$R=\sqrt{P^2+Q^2+2PQ\cos \theta }$

$R=\sqrt{3^2+2^2+2\left(3\right)\left(2\right)\cos \theta }$

$R=\sqrt{13+12\cos \theta }$

And when $A=6N$
the resultant becomes $2R$

$2R=\sqrt{6^2+2^2+2\left(6\right)\left(2\right)\cos \theta }$

$\Rightarrow 2\sqrt{13+12\cos \theta }=$

$\sqrt{6^2+2^2+2\left(6\right)\left(2\right)\cos \theta }$

$52+48\cos \theta =40+24\cos \theta$
$\Rightarrow \cos \theta =-\frac{1}{2}$
$\Rightarrow \theta ={120}^{\circ }$
Final answer: ${120}^{\circ }$