Question

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

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

Answer 1

Answer: (E)

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