a uniform rod of length 1.5m is placed over a wedge at 0.5m from one end .a force of 100 N is applied at its one end near the wedge to keep it horizontal .find weight of rod and reaction of wedge

Explanation:

The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end.  The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.

Sum the torques about the wedge (it may help to draw a diagram first).  Take counterclockwise to be positive.

∑τ = Iα

W (0.25 m) − (100 N) (0.50 m) = 0

W = 200 N

Sum the forces in the y direction.

∑F = ma

F − 100 N − 200 N = 0

F = 300 N

because if the tool that you are using is not balanced it there would be false mass.

490 n

f=125n a=2.5m/s2