Question

Two forces $5$kg wt and $10$kg wt are acting with an inclination of ${120}^o$ between them. what is the angle the resultant makes with $10$kg wt?

Answer 1

Let P be the force of 5 kg weight and Q be the force of 10 kg weight.

By parallelogram law of vector addition, the resultant is,

$R^2=P^2+Q^2+2PQ\cos \theta$

$R^2=25+100+100\cos {120}^{\circ }$

$R=5\sqrt{3}weight$

Let $\alpha$ be the angle made by R with 10 kg weight.

$\tan \alpha =\frac{P\sin \theta }{P+Q\cos \theta }$

$\tan \alpha =\frac{5\sin 120}{5+10\cos 120}$

$\tan \alpha =\frac{1}{\sqrt{3}}$

$\alpha ={30}^{\circ }$