A 34 m long ladder weighing 10kg leans on a frictionless wall. Its...

Applying force balance in vertical direction

${N}_{1} = mg = 100N$

Applying torque balance at the bottom contact point,

$mg \times \dfrac{3}{2} = {N}_{2} \times 5 ⇒ {N}_{2} = {F}_{w} = 30N$

Applying force balance in horizontal direction

${f}_{s} = {N}_{2} = 30N$

Net reaction at the floor will be,

${F}_{f} = \sqrt{{{N}_{1}}^{2} + {{f}_{s}}^{2}} = \sqrt{{100}^{2} + {30}^{2}} = 10\sqrt{109}N$

Therefore, $\dfrac{{F}_{w}}{{F}_{f}} = \dfrac{30}{10\sqrt{109}} = \dfrac{3}{\sqrt{109}}$