A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the beam to be balanced, a 2.9-kg mass must be placed on the left-end, and a 8.00-kg mass must be placed at another location on the beam. How far from the left-end is the location of the 8.00-kg mass

x ’= 1,735 m,  measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

           x_{cm} = 1.2 -1

          x_ {cm} = 0.2 m

          Σ τ = 0

          w₁ 1.2 + mg 0.2 - W₂ x = 0

          x =

          x =

let's calculate

          x = 2.9 1.2 + 4 0.2 / 8

          x = 0.535 m

measured from the pivot point

measured from the far left is

           x’= 1,2 + x

           x'=  1.2 + 0.535

           x ’= 1,735 m

momentum of soft ball three times less than baskey ball