Question

The resultant of two forces P and Q acting at an angle a is equal to $(2m+1)\sqrt{P^2+Q^2}.$ When the forces act at an angle $({90}^{}-\alpha )$, the resultant is $(2m-1)\sqrt{P^2+Q^2}.$ Then value of $tan\alpha$ is:

Answer 1

Solution

In case I :

$|\overrightarrow{P}+\overrightarrow{Q}|=\sqrt{P^2+Q^2+2P\theta cos\theta }$

$\Rightarrow \left(2MH\right)\sqrt{P^2+Q^2}=\sqrt{P^2+Q^2+2PQcos\theta }$

$\Rightarrow (2M+1{)}^2(P^2+Q^2)=P^2+Q^2+2PQcos\theta$

$\Rightarrow \frac{(4m^2+4m)}{2\left(PQ\right)}(P^2+Q^2)=cos\theta$

In case II

$|P^2+Q^2|=\sqrt{P^2+Q^2+2PQsin\theta }$

$\Rightarrow (2m-1)\sqrt{P^2+Q^2}=\sqrt{P^2+Q^2+2PQsin\theta }$

$\Rightarrow (2m-1{)}^2(P^2+Q^2)=P^2+Q^2+2PQsin\theta$

$\Rightarrow \frac{(4m^2-4m)}{2\left(PQ\right)}(P^2+Q^2)=sin\theta$

So, $tan\theta =\frac{sin\theta }{cos\theta }=\frac{4m^2-4m}{\left(2PQ\right)}\frac{\left(2PQ\right)}{(4m^2+4m)}\frac{(P^2+Q^2)}{(P^2+Q^2)}$

$\Rightarrow tan\theta =\frac{\left(4m\right)}{\left(4m\right)}\frac{(m-1)}{(m+1)}=\frac{m-1}{m+1}$

Here $Q=\alpha$ So, $tan\alpha =\frac{m-1}{m+1}$