Question

The sum of magnitudes of two forces acting at a point is $16N$ and their resultant $8\sqrt{3}N$ is at ${90}^o$ with the force of smaller magnitude. The two forces (in $N$) are

• A. $11,5$
• B. $9,7$
• C. $6,10$
• D. $4,12$
• E. $2,14$

Answer 1

The correct option is E $2,14$

According to question,
$P+Q=16\Rightarrow P=16-Q$
and resultant
$R=\sqrt{P^2+Q^2+2PQ\cos \theta }$
$8\sqrt{3}=\sqrt{P^2+Q^2+2PQ\cos \theta }$
$\tan {90}^o=\frac{\sin \theta }{P+Q\cos \theta }$
$P+Q\cos \theta =0$
$\cos \theta =-\frac{P}{Q}$
$\therefore 8\sqrt{3}=\sqrt{P^2+Q^2+2PQ(-\frac{P}{Q})}$
$=\sqrt{P^2+Q^2-2P^2}$
$8\sqrt{3}=\sqrt{Q^2-P^2}$
$Q^2-P^2=192$
$Q^2-{(Q-16)}^2=192$
$Q^2-Q^2-256+32Q=192$
$32Q=192+256$
$Q=14$
$\therefore P=2$