# Biophysics Problem 21

A man is standing as shown. His upper body weighs $1000\; N$ (i.e. $100\; kg$ or $220 \;lbs$)! and has a centre of gravity $C$ about $0.6 \;m$ from an effective pivot point $O.$ Suppose the muscles in his back which maintain the position are a perpendicular distance of $6 \;cm$ from the pivot point. What force are they exerting?

#### First Step

First, let's determine the lengths. We know that:

$OC = 0.6\;m \\ OA = 0.06\;m$

And OB can be determined by:

$OB = OC \times \cos50^\circ \\ = 0.6 \times 0.643 \\ = 0.39\; m$

Now you are ready to solve the problem by taking moments about $O.$

Try to calculate $F$ now.

#### Calculations

The $1000\;N$ force exerts a clockwise torque about $O.$

$F$ exerts a counter-clockwise torque of equal magnitude about $O.$

Therefore,

$F \times OA = 1000 \times OB \\ F = 6430\; N$