A weightless ladder 20 m  long rests against a frictionless wall at an angle of 60° from the horizontal. A 150 kg man is 4 m from the top of the ladder. A horizontal force is needed to keep it from slipping. Choose the correct magnitude of the force from the following

AB is the ladder, Let F be the horizontal force and W is the weight of man.    
Let ${\mathrm{N}}_1 \mathrm{and} {\mathrm{N}}_2$ be normal reactions of ground and wall, respectively, then for vertical equilibrium 
$\left.W=N_1-\cdots -\cdots 1\right)$

For horizontal equilibrium    

$\left.N_2=F .......2\right)$

$N_2\left(AB\sin 60°\right)-W\left(AC\cos 60°\right)=0$

$\begin{array}{l}F\left[20\times \frac{\sqrt{3}}{2}\right]-W\left[16\times \frac{1}{2}\right]=0\\ \Rightarrow F=\frac{8W\times 2}{20\sqrt{3}}=\frac{4W}{5\sqrt{3}}=\frac{150\times 4}{5\sqrt{3}}\text{ kg-wt}\\ =40\sqrt{3}=40\times 1.73=69.2\text{ kg-wt }\end{array}$