# A beam of length 300 cm has two blocks of 30 kg and 25 kg placed at either end such that it is balanced on a pivot . What is the distance of the pivot from the two blocks.

### 1 Answer | Add Yours

The beam of length 3 m has two blocks placed at either end. The mass of the blocks is 30 kg and 25 kg. Each of the blocks is pulled down by the Earth with a force equal to the product m*g where m is the mass of the block. This force results in a torque about the pivot.

For the beam to remain balanced, the torque due to each of the blocks should be equal in magnitude. As they are placed on opposite ends, the direction of torque is also opposite that makes the net torque on the beam about the pivot equal to 0.

Torque = `F*D*sin theta ` where F is the force applied at a distance D from the pivot point in a direction `theta` .

Let the distance of the 30 kg block from the pivot be L, the distance of the 25 kg block is (3 - L). Equating the torque due to the two blocks gives: 30*L*sin 90 = 25*(3 - L)*sin 90

=> 30L = 75 - 25L

=> L = 75/55 = 15/11

(3 - L) = 18/11

The block with mass 39 kg is 15/11 m from the pivot and the block with mass 25 kg is 18/11 m from the pivot.