A 34   m long ladder weighing 10   kg leans on a frictionless wall. Its feet rest on the floor 3…

Applying force balance in vertical direction

$N_1=mg=100 \mathrm{N}$

Applying torque balance at the bottom contact point,

$mg\times \frac{3}{2}=N_2\times 5\Rightarrow N_2=F_w=30 \mathrm{N}$

Applying force balance in horizontal direction

$f_s=N_2=30 \mathrm{N}$

Net reaction at the floor will be,

$F_f=\sqrt{{N_1}^2+{f_s}^2}=\sqrt{{100}^2+{30}^2}=10\sqrt{109} \mathrm{N}$

Therefore, $\frac{F_w}{F_f}=\frac{30}{10\sqrt{109}}=\frac{3}{\sqrt{109}}$