A light weightless bar is provided at it centre and weight of 5n and 10n placed 3m and 2m, respectively from the pivot on one side are balanced by weight 20n weight on the other side how far in the 20

We can solve this problem using the principle of moments, which states that the sum of the moments of the forces acting on an object about any point is equal to zero if the object is in equilibrium. First, let us find the total moment of the 5n and 10n weights about the pivot point. We can calculate this by multiplying the force by the perpendicular distance from the force to the pivot: total moment = (5n x 3m) + (10n x 2m) = 25nm Next, we can set up an equation for the moment of the 20n weight about the pivot point: 20n x d = 25nm where d is the distance of the 20n weight from the pivot point. Solving for d, we get: d = (25nm) / (20n) = 1.25m Therefore, the 20n weight is located 1.25m from the pivot point.