Answer
The magnitude of the force at point A from the calf muscle is $~~2700~N$
Work Step by Step
We can consider the net torque about a rotation axis at point B.
Since the foot is in equilibrium, $\tau_{net} = 0$
Note that the floor pushes up at point P with a force of $900~N$
We can find the force $F_A$ at point A from the calf muscle:
$\tau_{net} = 0$
$(15~cm)(900~N)-(5.0~cm)~F_A = 0$
$(5.0~cm)~F_A = (15~cm)(900~N)$
$F_A = \frac{(15~cm)(900~N)}{5.0~cm}$
$F_A = 2700~N$
The magnitude of the force at point A from the calf muscle is $~~2700~N$.
