# Biophysics Problem 19

The force \(F\) acts 3/4 of the way along the boom of length \(20\; m.\) What is the torque (moment) of the force about the point \(O\)?

There are two ways this problem could be solved. Let's use the method where we break the force down into component perpendicular and parallel to the boom. The moment of \(F\) about \(O\) will be equal to the sum of the moments of the components about the same point.

Since the parallel component will pass through \(O\), the torque about \(O\) will be zero regardless of the magnitude of the parallel components.

All that you have to calculate is the moment of the perpendicular component about the point \(O.\)

Let's start by calculating the perpendicular component itself:

\(\text{Perpendicular component} = F \times \cos 20^\circ \\ = 100 \times 0.342 \\ = 34.2\;N\)

This component acts \(15\;m\; (\text{3/4 of the boom length from)} \;O.\) So the magnitude of the torque is:

\(\text{Torque} = 34.2 \times 15\\ = 513\; Nm\)

That's all. I hope you remembered that the direction of the torque is \(+\), or into the page.