a uniform object ladder of length 3.5 m and weighing 250n is placed against a smoth vertical wall with its lower end 1.25 m from the wall . if coefficent of friction between the ladder and floror is 0.3 , then the frictional force acting on the ladder at that point of contact betweenh ladder and floor

Dear student 

W is the weight of the ladder, acting downwards on the center of mass of the ladder. N is the normal reaction from the ground on the ladder and f is the friction offered by the ground to the ladder.
Consider the topmost point of the ladder as our reference point.
Now, we assume that the ladder is at rest. So, the torques due to different forces will cancel out each other. Considering our reference point, the torque due to the weight of ladder is W x (1.25)/2
The torque due to normal reaction from the ground is N x 1.25
The torque due to friction is 3(by pythagoras theorem) x f.
Now, there is no force acting in vertical direction on the ladder other than its weight. So, the normal reaction, N=W.
Now, by equating the clockwise and anticlockwise torques, we get f=52.0833333N (nearly equal to 52.1N).
I hope this helps you.
Feel free to ask if you don't get it.