Question

A uniform rod of 100 cm of weight 100 gf, is balanced at 60 cm mark, when a weight of 20 gf is suspended at the 15 cm mark. To balance the uniform rod a weight of 80 gf should be suspend at​

Answer 1

Given:

A uniform rod of 100 cm of weight 100 gf, is balanced at 60 cm mark, when a weight of 20 gf is suspended at the 15 cm mark.

To find:

Where should 80gf force be suspended to balance the rod?

Calculation:

First of all, refer to the diagram.

• Since the rod is in rotational equilibrium, the net torque on the rod will be zero.

Considering axis of rotation as shown:

$\sum( \tau) = 0$

$\rm \implies \: 20(60 - 15) + 100(60 - 50) - 80(x) = 0$

$\rm \implies \: 80x = 20(45) + 100(10)$

$\rm \implies \: 80x = 1900$

$\rm \implies \: x = 23.75 \: cm$

$\rm \implies \: L = 50 + x$

$\rm \implies \: L = 50 + 23.75$

$\rm \implies \: L = 73.75 \: cm$

So, 80gf weight has to be given at 73.75 mark on the rod.