Question

A body of mass $60 \,kg$ 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

Answer 1

Answer: (D)

103.9 gN

Answer 2

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