Question

A 150-kg ladder leans against a smooth wall, making an angle of 30 degrees with the floor. The centre of gravity of the ladder is one-third the way up from the bottom. How large a horizontal force must the floor provide if the ladder is not to slip?

Answer 1

the ladder is at equilibrium...so
Mg=N1
and f=N2
here N1 is the normal force exerted by the floor and N2 is the normal force exerted by the frictionless wall.
we will take the moments of forces about point B..
N1*0-f*0+mg*l/3cos30-N2*lsin30.........[l=the length of the ladder]
as the ladder has no angular movement...
so mg*l/3cos30-N2*lsin30=0
or N2=mg/3cot30
or N2=mg/sq.rt.3