Solve this:
Q. A uniform ladder of length 3.25 m and weighing 250 N is placed against a smooth vertical wall with its lower end 1.25 m from the wall. If coefficient of friction between the ladder and the floor is 0.3, then, find the frictional force acting on the ladder at the point of contact between ladder and floor.

Dear student,


Let Ff be the frictional force.Rf=250NBC=(3.25)2-(1.25)2=3mTaking moments about B and equating them,Ff×3=(Rf×1.25)-(250×0.625)=156.2NFf=156.23=52.1N

 
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