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.
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?
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.
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