Question

A uniform beam of mass m is pivoted at one end and held in equilibrium at an angle q by a horizontal wire, as shown above. The tension T in the wire is

Answer 1

Given:

A uniform beam of mass m is pivoted at one end and held in equilibrium at an angle $\theta$ by the horizontal wire.

To find:

Tension in the wire

Calculation:

Since the rod is in rotational equilibrium hence the torque due to tension and the torque due to weight would be equal and opposite.

$\therefore \: \sum( \tau) = 0$

$= > \: T \bigg\{l \sin( \theta) \bigg \} - mg \bigg \{\dfrac{l}{2} \cos( \theta) \bigg \} = 0$

$= > \: T \bigg\{ \cancel{l} \sin( \theta) \bigg \} = mg \bigg \{\dfrac{ \cancel{l}}{2} \cos( \theta) \bigg \}$

$= > \: T = \dfrac{1}{2} mg \times \dfrac{ \cos( \theta) }{ \sin( \theta) }$

$= > \: T = \dfrac{1}{2} mg\cot( \theta)$

Hence final answer is :

$\boxed{ \bold{\: T = \dfrac{1}{2} mg\cot( \theta)}}$