Question

The sum of the magnitudes of two forces acting at point is $18$ and the magnitude of their resultant is $12$. If the resultant is at ${90}^{\circ }$ with the force of smaller magnitude, what are the magnitudes of forces?

• A. $12,5$
• B. $14,4$
• C. $5,13$
• D. $10,8$

Answer 1

The correct option is C $5,13$

Let $P$ be the magnitude of smaller force and $Q$ be the magnitude of greater force, then according to the problem
$P+Q=18...\left(i\right)$

$R=\sqrt{P^2+Q^2+2PQ\cos \theta }=12...\left(ii\right)$
We know that the angle between resultant force vector and $P$ vector is,

$\tan \varphi =\frac{Q\sin \theta }{P+Q\cos \theta }=\tan {90}^{\circ }=\infty$
$\therefore P+Q\cos \theta =0...\left(iii\right)$

By solving $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$ we get
Putting the value $Q\cos \theta =-P$ in the equation
$P^2+Q^2+2PQ\cos \theta =144$, we get
$\Rightarrow Q^2-P^2=144,(Q-P)(Q+P)=144$
Solving for $P$ and $Q$ we get
$P=5$ and $Q=13$

Hence option C is the correct answer