Question

Find the tension in OB and AB in the given figure. Also, calculate the tension in OB when just after the string AB is burnt.

Answer 1

To find:

• Tension in OB and AB in the given figure

• Tension in OB after AB is burnt.

Calculation:

Applying Lami's Theorem :

$\therefore \: \dfrac{T_{OB}}{ \sin( {90}^{ \circ} ) } = \dfrac{T_{AB}}{ \sin( {90}^{ \circ} + {90}^{ \circ} - \theta) } = \dfrac{mg}{ \sin( {90}^{ \circ} + \theta ) }$

$= > \: \dfrac{T_{OB}}{ \sin( {90}^{ \circ} ) } = \dfrac{T_{AB}}{ \sin( {180}^{ \circ} - \theta) } = \dfrac{mg}{ \cos( \theta ) }$

$= > \: \dfrac{T_{OB}}{ \sin( {90}^{ \circ} ) } = \dfrac{T_{AB}}{ \sin( \theta) } = \dfrac{mg}{ \cos( \theta ) }$

$= > \: \dfrac{T_{OB}}{1 } = \dfrac{T_{AB}}{ \sin( \theta) } = \dfrac{mg}{ \cos( \theta ) }$

$= > \: T_{OB} = \dfrac{T_{AB}}{ \sin( \theta) } = \dfrac{mg}{ \cos( \theta ) }$

So,

$1) \: T_{OB} = \dfrac{mg}{ \cos( \theta) }$

$2) \: T_{AB} = \dfrac{mg \sin( \theta) }{ \cos( \theta) } = mg \tan( \theta)$

After AB is burnt:

$\therefore \: T_{OB} - mg \cos( \theta) = \dfrac{m {v}^{2} }{l}$

$= > \: T_{OB} - mg \cos( \theta) = \dfrac{m {(0)}^{2} }{l}$

$= > \: T_{OB} - mg \cos( \theta) = 0$

$= > \: T_{OB} = mg \cos( \theta)$

Hope It Helps.