Question

A body of mass $60kg$ suspended by means of three strings $P,Q$ and $R$ as shown in the figure is in equilibrium. The tension in the string $P$ is

• A. 130.9 gN
• B. 60 gN
• C. 50 gN
• D. 103.9 gN
• E. 100 gN

Answer 1

Answer: (D)

The free body diagram of mass M is shown in figure. Taking component of forces,
Taking component of forces
$R\cos \theta =Mg$
$R\cos 60^{\circ }=Mg(i)$
$andR\sin 60^{\circ }=T(ii)$
By Eqs. (i) and (ii), we get
$\therefore \tan 60^{\circ }=\frac{T}{Mg}$
$orT=Mg\tan 60^{\circ }$
$orT=60\times g\times \sqrt{3}=103.9gN$