Question

Q1.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

Explanation:

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?

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

Considering axis of rotation as shown:

$\sum( \tau) = 0$

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

=80x = 20(45) + 100(10)

=80x = 1900

=x = 23.75 cm

=L=50+x

=L=50+23.75

=L=73.75cm

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

Answer 2

Answer:

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