Question

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

• A. $130.9gN$
• B. $60gN$
• C. $50gN$
• D. $103.9gN$

Answer 1

The correct option is D $103.9gN$

Horizontal component = $T_1\cos {30}^o-T=0$

Vertical component = $T_1\sin {30}^o-mg=0$

$\therefore \frac{T_1\cos {30}^o}{T_1\sin {30}^o}=\frac{T}{mg}$

$T=mg\cot {30}^o$

$=60\times 10\times 1.732$

$=103.9gN$