In the figure a $54$ kg rock climber is in a lie$-$back climb along a fissure, with hands pulling on one side of the fissure and feet pressed against the opposite side. The fissure has width w $=0\mathrm{.}30$ m$,$ and the center of mass of the climber is a horizontal distance d $=0\mathrm{.}30$ m from the fissure. The coefficient of static friction between hands and rock is $\backslash$mu $1=0\mathrm{.}40,$ and between boots and rock it is $\backslash$mu $2=1\mathrm{.}10.$ The climber adjusts the vertical distance h between hands and feet until the $($identical$)$ pull by the hands and push by the feet is the least that keeps him from slipping down the fissure. $($He is on the verge of sliding.$)$

$($a$)$ What is the least horizontal pull by the hands and push by the feet that will keep the climber stable? $($b$)$ What is the value of h$?$