Answer
The force on each hand is $196N$ and on each foot is $96N$
Work Step by Step
There are 3 forces acting on the person here:
- The person's weight $\vec{W}$ located at his center of gravity and pointing downward.
- The total normal force acting on both feet $\sum \vec{N_f}$, pointing upward.
- The total normal force acting on both hands $\sum \vec{N_h}$, pointing upward.
Because the person is in equilibrium, $$\sum N_f+\sum N_h-W=0$$ $$\sum N_f+\sum N_h=584N (1)$$
Also, a person in equilibrium means the net external torque acting on the person is zero. If we choose the rotation axis to be at the person's feet, then:
- $\sum N_f$ produces no torque, because the rotation axis passes through this force.
- The lever arm between $W$ and the axis is $0.84m$, and $W$ produces a clockwise rotation, so $\tau_W\lt0$. Therefore, $\tau_W=-584N\times0.84m=-490.56N.m$
- The lever arm between $\sum N_h$ and the axis is $0.84m+0.41m=1.25m$, and $\sum N_h$ produces a counterclockwise rotation, so $\tau_{N_h}\gt0$. Therefore, $\tau_{N_h}=1.25\sum N_h$
$$\tau_{N_h}+\tau_W=0$$ $$1.25\sum N_h=490.56$$ $$\sum N_h=392N$$
Plugging this into (1), we have $\sum N_f=192N$
The normal force is equally distributed between 2 hands and 2 feet, so the force on each hand and each feet is $$N_h=\frac{392}{2}=196N$$ $$N_f=\frac{192}{2}=96N$$
