When a force is applied to a body it will produce a tendency for the body to rotate about a point that is not on the line of action of the force. This tendency to rotate is sometimes called a torque, but most often it is called the moment of a force or simply the moment.
For example, consider a wrench used to unscrew the bolt. If a force is applied to the handle of the wrench it will tend to turn the bolt about point $O$. The magnitude of the moment is directly proportional to the magnitude of $\mathbf{F}$ and the perpendicular distance or moment arm $d$. The larger the force or the longer the moment arm, the greater the moment or turning effect.
Note that if the force $\mathbf{F}$ is applied at an angle $\theta \neq 90°$, then it will be more difficult to turn the bolt since the moment arm $d' = d\sin\theta$ will be smaller than $d$. If $\mathbf{F}$ is applied along the wrench, its moment arm will be zero since the line of action of $\mathbf{F}$ will intersect point $O$. As a result, the moment of $\mathbf{F}$ about $O$ is also zero and no turning can occur.
We can generalize the above discussion and consider the force $\mathbf{F}$ and point $O$ which lie in the shaded plane as shown.
The moment $\mathbf{M}_O$ about point $O$, or about an axis passing through $O$ and perpendicular to the plane, is a vector quantity since it has a specified magnitude and direction.
The magnitude of $\mathbf{M}_O$ is
$$ M_O = Fd $$
where $d$ is the moment arm or perpendicular distance from the axis at point $O$ to the line of action of the force. Units of moment magnitude consist of force times distance, $\text{N}·\text{m}$ or $\text{lb}·\text{ft}$ .
The direction of $\mathbf{M}_O$ is defined by its moment axis, which is perpendicular to the plane that contains the force $\mathbf{F}$ and its moment arm $d$.
The right-hand rule is used to establish the sense of direction of $\mathbf{M}_O$. According to this rule, the natural curl of the fingers of the right hand, as they are drawn towards the palm, represent the rotation, or if no movement is possible, there is a tendency for rotation caused by the moment. As this action is performed, the thumb of the right hand will give the directional sense of $\mathbf{M}_O$.
Notice that the moment vector is represented three-dimensionally by a curl around an arrow. In two dimensions this vector is represented only by the curl. Since in this case the moment will tend to cause a counterclockwise rotation, the moment vector is actually directed out of the page.
For two-dimensional problems, where all the forces lie within the x–y plane, the resultant moment $(M_R)_O$ about point O can be determined by finding the algebraic sum of the moments caused by all the forces in the system.
As a convention, we will generally consider positive moments as counterclockwise since they are directed along the positive $z$ axis (out of the page). Clockwise moments will be negative. Doing this, the directional sense of each moment can be represented by a plus or minus sign. Using this sign convention, with a symbolic curl to define the positive direction, the resultant moment is therefore
$$ \curvearrowleft +(M_R)_O = \Sigma Fd; \qquad (M_R)_O = F_1d_1 - F_2d_2 + F_3d_3 $$
If the numerical result of this sum is a positive scalar, $(M_R)_O$ will be a counterclockwise moment (out of the page); and if the result is negative, $(M_R)_O$ will be a clockwise moment (into the page).
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For each case illustrated in, determine the moment of the force about point $O$.
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Determine the resultant moment of the four forces acting on the rod about point $O$.
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