Answer
The vertical distance between hands and feet must be $~~0.88~m$
Work Step by Step
In part (a), we found that the least horizontal pull by the hands and push by the feet that will keep the climber stable is $~~F = 337~N$
We can consider the torque about a rotation axis at the climber's center of mass.
The two friction forces at the hands and feet are directed upward, and there is a counterclockwise torque from these friction forces.
The rock walls exert a force $F$ on the hands and the feet, and there is a clockwise torque from these forces.
Let $h_1$ be the vertical distance of the feet below the center of mass.
Let $h_2$ be the vertical distance of the hands above the center of mass.
Note that $h_1+h_2 = h$
We can find $h$:
$\tau_{net} = 0$
$(d)(F~\mu_1)+(d+w)(F~\mu_2)-(h_1)(F)-(h_2)(F) = 0$
$(h_1)(F)+(h_2)(F) = (d)(F~\mu_1)+(d+w)(F~\mu_2)$
$(h_1)+(h_2) = (d)~(\mu_1)+(d+w)~(\mu_2)$
$h = (0.40~m)(0.40)+(0.40~m+0.20~m)(1.2)$
$h = 0.88~m$
The vertical distance between hands and feet must be $~~0.88~m$
