Question

A block of mass 3m is suspended by a meter scale rod of mass m as shown in the figure. If the tension in string A is equal to kmg in equilibrium, then the value of k will be :

Answer 1

Given:

A block of mass 3m is suspended by a meter scale rod of mass m as shown in the figure. The tension in string A is equal to kmg in equilibrium.

To find:

Value of k

Calculation:

The rod is at rotational equilibrium , such that total torque will be zero.

Considering instantaneous axis of rotation to be at point B:

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

$=>\tau1+\tau2+\tau3=0$

$= > T_{A}(100) - 3mg(25) - mg(50) = 0$

$= > T_{A}(100) - 75mg - 50mg = 0$

$= > T_{A}(100) - 125mg = 0$

$= > T_{A}(100) = 125mg$

$= > kmg(100) = 125mg$

$= > \: 100k = 125$

$= > k = \dfrac{125}{100}$

$= > k = \dfrac{5}{4}$

So, final answer is:

$\boxed{\bf{ k = \dfrac{5}{4} }}$