Question

Two forces of magnitudes $P$ and $Q$ are inclined at an angle $\left(\theta \right)$, the magnitude of their resultant is $3Q$. When the inclination is changed to $(180-\theta )$, the magnitude of the resultant force becomes $Q$. The ratio of the forces $\left(\frac{P}{Q}\right)$ is $\frac{x}{y}$. Find $(x+y)$.

Answer 1

Given, magnitude of resultant of forces of magnitude $P$ and $Q$ is $3Q$ when angle between them is $\theta$,

$(3Q{)}^2=P^2+Q^2+2PQ\cos \theta$

$9Q^2=P^2+Q^2+2PQ\cos \theta \dots \left(i\right)$
And magnitude of resultant is $Q$ when angle between them is $(180-\theta )$
$Q^2=P^2+Q^2+2PQ\cos (180-\theta )$

$Q^2=P^2+Q^2-2PQ\cos \theta \dots \left(ii\right)$

By adding equation(i) and (ii)

$12Q^2=2P^2+2Q^2$

$8Q^2=2P^2$

$\frac{P^2}{Q^2}=\frac{8}{2}$

$\Rightarrow \frac{P}{Q}=2$
Compare it with $\frac{x}{y}$
$x=2,y=1$

so,$x+y=3$
Final answer:$3$