Question

A block of mass 6 kg is kept in equilibrium with the
help of strings as shown in figure. The tension in string
PQ is (g = 10 m/s2)

Answer 1

Given:

A block of mass 6 kg is kept in equilibrium with the help of strings as shown in figure.

To find:

Tension in string PQ.

Calculation:

Let tension in PR be T1 and that of PQ be T2.

Applying Lami's Theorem:

$\rm{ \therefore \: \dfrac{T1}{ \sin( {90}^{ \circ} ) } = \dfrac{T2}{ \sin( {90}^{ \circ} + {30 }^{ \circ} ) } = \dfrac{6g}{ \sin( {90}^{ \circ} + {60}^{ \circ} ) } }$

$\rm{ = > \: \dfrac{T1}{ \sin( {90}^{ \circ} ) } = \dfrac{T2}{ \cos( {30 }^{ \circ} ) } = \dfrac{6g}{ \cos( {60}^{ \circ} ) } }$

$\rm{ = > \: \dfrac{T1}{ 1 } = \dfrac{T2}{ \cos( {30 }^{ \circ} ) } = \dfrac{6g}{ \cos( {60}^{ \circ} ) } }$

$\rm{ = > \: T1 = \dfrac{T2}{ \cos( {30 }^{ \circ} ) } = \dfrac{6g}{ \cos( {60}^{ \circ} ) } }$

Considering the second two values:

$\rm{ = > \: \dfrac{T2}{ \cos( {30 }^{ \circ} ) } = \dfrac{6g}{ \cos( {60}^{ \circ} ) } }$

$\rm{ = > \: \dfrac{T2}{ \cos( {30 }^{ \circ} ) } = \dfrac{60}{ \cos( {60}^{ \circ} ) } }$

$\rm{ = > \: \dfrac{T2}{ ( \frac{ \sqrt{3} }{2} )} = \dfrac{60}{ ( \frac{1}{2} )} }$

$\rm{ = > \: \dfrac{T2}{ \sqrt{3} } = 60 }$

$\rm{ = > \: T2 = 60 \sqrt{3} \: N }$

So, final answer:

$\boxed{ \bf{ \: T2 = 60 \sqrt{3} \: N }}$