# 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, 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.

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

\(F\) exerts a counter-clockwise torque of equal magnitude about \(O.\)

Therefore,

\( F \times OA = 1000 \times OB \\ F = 6430\; N\)