Question

The sum of the magnitudes of two forces is $18\text{N}$ and the magnitude of their resultant is $12\text{N}$. If the resultant makes an angle of ${90}^{\circ }$ with the smaller force, then find the magnitude of the forces.

• A. $5\text{N},13\text{N}$
• B. $6\text{N},12\text{N}$
• C. $10\text{N},8\text{N}$
• D. $\text{None of these}$

Answer 1

The correct option is A $5\text{N},13\text{N}$

Let us say the magnitudes of the two vectors are $P$ and $Q$, where $P<Q$
As per the question, $P+Q=18\Rightarrow P=18-Q\dots \dots \left(1\right)$
and, resultant $R=12=\sqrt{P^2+Q^2+2PQ\cos \theta }\dots \dots \left(2\right)$

If $\varphi$ is the angle between $P$ and $R$,
$\tan \varphi =\frac{Q\sin \theta }{P+Q\cos \theta }$
Given $\varphi ={90}^{\circ }$
$\Rightarrow \frac{Q\sin \theta }{P+Q\cos \theta }=\tan {90}^{\circ }=\infty$
$\Rightarrow P+Q\cos \theta =0\dots \dots \left(3\right)$

From eq. (1) & (3), we have
$Q(1-\cos \theta )=18\dots \dots \left(4\right)$
Also from eq (1) and (2), we have
$R^2=P^2+Q^2+2PQ\cos \theta$
$\Rightarrow R^2=P^2+Q^2+2PQ+2PQ\cos \theta -2PQ$
$\Rightarrow (P+Q{)}^2-R^2=2PQ(1-\cos \theta )$
$\Rightarrow 2PQ(1-\cos \theta )={18}^2-{12}^2=180$
or, $PQ(1-\cos \theta )=90\dots \dots \left(5\right)$

Dividing (5) by (4), we get
$P=5\text{units}$ and $Q=13\text{units}$
(from (1))