Question

Three light strings are connected at the point $P$. A weight $W$ is suspended from one of the strings. End $A$ of string $AP$ and end $B$ of string $PB$ are fixed as shown. In equilibrium, $PB$ is horizontal and $PA$ makes an angle of ${60}^{\circ }$ with the horizontal. If the tension in $PB$ is $30N$, then the tension in $PA$ and weight $W$ are respectively given by

• A. $60N,30N$
• B. $\frac{60}{\sqrt{3}}N,\frac{30}{\sqrt{3}}N$
• C. $60N,30\sqrt{3}N$
• D. $60\sqrt{3}N,30\sqrt{3}N$

Answer 1

The correct option is C $60N,30\sqrt{3}N$


Here $W=mg$
For equilibrium of block of weight $W$,
$W=T_2$

For horizontal equilibrium at point $P$, we can say
$T_1\cos {60}^{\circ }=30N$
$⟹T_1=60N$

For vertical equilibrium at point $P$, we can say
$T_1\sin {60}^{\circ }=T_2=W$
$⟹W=60\times \frac{\sqrt{3}}{2}=30\sqrt{3}N$