Answer
The force from the floor on the person's hands is $392~N$
The force from the floor on the person's feet is $274~N$
Work Step by Step
We can consider the torques about a rotation axis at the person's feet.
The person's weight exerts a counterclockwise torque about this axis.
The force from the floor on the hands $F_h$ exerts a clockwise torque about this axis.
Since the system is in equilibrium, the clockwise torque is equal in magnitude to the counterclockwise torque. We can find the force from the floor on the person's hands $F_h$:
$\tau_{cw} = \tau_{ccw}$
$(1.70~m)~(F_h) = (1.00~m)(68~kg)(9.80~m/s^2)$
$F_h = \frac{(1.00~m)(68~kg)(9.80~m/s^2)}{1.70~m}$
$F_h = 392~N$
The force from the floor on the person's hands is $392~N$
Since the system is in equilibrium, the downward force is equal in magnitude to the sum of the upward forces. We can find the force from the floor on the person's feet $F_f$:
$F_h+F_f = mg$
$F_f = mg-F_h$
$F_f = (68~kg)(9.80~m/s^2)-(392~N)$
$F_f = 274~N$
The force from the floor on the person's feet is $274~N$
