### Moments/torque (Biomechanics)

#### Moments/Torque

The turning effect of the force on the body on which it is acting is measured by the moment of a force, also known as torque.

The moment of a force depends on the magnitude of the force and the distance from the axis of rotation.

The moment of a force about a point is (the magnitude of the force) $\times$ (the perpendicular distance of the line of action of the force from the point).

Suppose that this perpendicular distance is denoted $d_{p}$.

When there are several forces acting on a body the moments about a point can be added so long as a positive direction (clockwise or anticlockwise) is specified and is considered for each moment.

For further examples see the moments page of the Mechanics section.

#### Worked Example: Finding the resultant torque

###### Example

Suppose that two individuals weighing $250 \ \mathrm{N}$ and $220 \ \mathrm{N}$ sit on opposite sides of a seesaw. The heavier individual is $2 \ \mathrm{m}$ away from the seesaw's axis of rotation, whereas the other is $2.2 \ \mathrm{m}$ away. Which end of the seesaw will drop?

###### Solution

We know that the seesaw will rotate in the direction of the resultant torque at its axis of rotation. To find the resultant torque, $T_r$, we sum the torques created by each individual, which can be found using $T = F \times d_p$.

Let the torque created by the heavier individual be $T_1$, and $T_2$ for the other individual. $T_1$ and $T_2$ are in opposite directions (clockwise and anticlockwise) so we need to minus one from the other.

So we have \begin{align} T_r & = T_1 - T_2, \\ & = (250 \mathrm{N} \times 2 \mathrm{m}) - (220 \mathrm{N} \times 2.2 \mathrm{m}), \\ & = 16 \mathrm{N m}. \end{align} The resultant torque is in a positive direction, therefore the heavier individual's end of the seesaw will fall.

#### Worked Example: Finding the resultant torque when distance given is not perpendicular

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