Question

A body of mass $5kg$ is suspended by string making angle ${60}^o$ & ${30}^o$ with horizontal then:
(a) $T_1=25N$ (b) $T_2=25N$
(c) $T_1=25\sqrt{3}N$ (d) $T_2=25\sqrt{3}N$

• A. a, b
• B. a, d
• C. c, d
• D. b, c

Answer 1

The correct option is B a, d

As the mass is at rest the resultant of forces acting on it are equal to zero

so forces in vertical direction are

$T_{1}sin{30}^o+T_{2}sin{60}^o-mg=0$

$\frac{T_1}{2}+\frac{\sqrt{3}{T}_2}{2}=5g$

$T_1+\sqrt{3}{T}_2=100$

similarly in horizontal direction

$T_{1}cos{30}^o-T_{2}cos{60}^o=0$

$\frac{T_1}{T_2}=\frac{1}{\sqrt{3}}$

solving above equations will give us

$T_1+\sqrt{3}\times \sqrt{3}{T}_1=100$

$T_1=25N$

$T_2=25\sqrt{3}N$