Engineering Statics: Open and Interactive

Section4.1Moment of Force

A moment of force, or torque, is a measure of the tendency of that force to rotate a body about a selected point or axis, called the moment center. This tendency increases with the magnitude of the force, and also with the distance between the line-of-action of the force and the moment center.
Moments are vector quantities, so they have magnitude and direction and obey all the rules of vector arithmetic, even dot and cross products, described in Chapter 2.

Subsection4.1.1Magnitude of a Moment

The magnitude of a moment describes how hard it turns, in the same way that the magnitude of a force describes how hard it pushes or pulls. When you turn a door knob, the magnitude of the moment is how hard you twist the knob.
The magnitude of a moment is found by multiplying the magnitude of the force by the distance between the line of action of the force and the center of rotation, as illustrated in Figure 4.1.1. This distance called the moment arm or the perpendicular distance, $$d_{\perp}\text{.}$$
$$M =F d_{\perp}\text{.}\tag{4.1.1}$$
Since the moment is the product of the force’s magnitude and the perpendicular distance, the closer the point is to the force’s line of action, the smaller the moment. If the point lies upon the force’s line of action, then the moment arm is zero, making the moment zero as well. But notice that a force may be slid along its line-of-action without changing the moment, because neither the magnitude of the force nor the moment arm changes. The magnitude of a moment is a positive quantity regardless of whether it produces a clockwise or counter-clockwise tendency.
Moments are the product of a force with a distance, so they have units of [force] $$\times$$ [distance] such as N-m or ft-lb.

Example4.1.2.

A mobile crane is used to raise a $$\lb{500}$$ load of building materials. Boom $$AC$$ is $$\ft{40}$$ long and the tension of the boom pendant $$AB$$ is $$\lb{1500}\text{.}$$
Use the definition of the moment (4.1.1) to determine the magnitude and direction of:
1. $$M_1\text{,}$$ the moment about point $$C$$ caused by the load, $$W\text{.}$$
2. $$M_2\text{,}$$ the moment about point $$C$$ caused by tension $$T\text{.}$$
\begin{align*} M_1 \amp = \ftlb{17,885} \circlearrowright\\ M_2 \amp = \ftlb{19,590} \circlearrowleft \end{align*}
Solution.
1. Given: $$W = \lb{500}\text{,}$$ $$T = \lb{1500}\text{,}$$ boom length $$AC = \ft{40}\text{.}$$
2. Sketch right triangles containing the perpendicular distances $$d_1$$ and $$d_2$$ from point $$C$$ to the lines of action of $$W$$ and $$T\text{.}$$
3. Find angles in the right triangles.
\begin{align*} \alpha \amp = \tan^{-1} \left(1/2\right) = \ang{26.6}\\ \theta \amp = \alpha - \ang{7.5} = \ang{19.1} \end{align*}
4. Use trigonometry to find the perpendicular distances.
\begin{gather*} d_1 = 40 \cos \alpha = \ft{35.8}\\ d_2 = 40 \sin \theta = \ft{13.06} \end{gather*}
5. Apply the definition of the moment (4.1.1), and solve for the moments.
\begin{align*} M_1 \amp = W d_1 = (\lb{500})(\ft{35.8}) =\ftlb{17,885}\\ M_2 \amp = T d_2 = (\lb{1500})(\ft{13.06}) = \ftlb{19,590} \end{align*}
By inspection we see that the tension produces a counterclockwise rotation and the load produces a clockwise rotation. Counterclockwise moment $$M_2$$ is greater than clockwise moment $$M_1$$ because the boom pendant must also support the weight of the boom, which we have not accounted for.
As we will see in the next section, counterclockwise moments indicate a positive scalar moment in two dimensions, and a vector that points in the $$+\khat$$ direction in three.

Example4.1.3.

The bent column $$ABC$$ is firmly fixed to the ground at $$A\text{.}$$
Determine the magnitude and direction of the moment that a $$\N{400}$$ force applied at $$C$$ produces about point $$A\text{.}$$
\begin{equation*} M_A = \Nm{2956} \end{equation*}
Solution.
This problem is similar to the previous example, but the geometry required to find the perpendicular distance is slightly more difficult.
1. Use the law of sines and law of cosines to find distance $$d$$ from $$A$$ to $$C$$ and the angle $$\phi$$ in triangle $$ABC\text{.}$$
\begin{align*} d^2 \amp= 5.5^2 + 3^2 - 2(3)(5.5) \cos \ang{120}\\ d \amp = \m{2.47}\\ \\ \frac{d}{\sin \ang{120}} \amp = \frac{\m{5.5}}{\sin \phi} \implies \phi = \ang{39.6} \end{align*}
2. Then, find the angle $$\theta$$ in the right triangle and solve for the perpendicular distance.
\begin{align*} \theta \amp= \ang{180} - \ang{30} - \ang{50}- \phi = \ang{60.36}\\ d_\perp \amp = d\ \sin \theta = \m{6.49} \end{align*}
3. Finally, use (4.1.1) to find the magnitude of $$M_A$$ created by $$\N{400}$$ force $$F\text{.}$$
\begin{equation*} M_A = F\ d_\perp = \Nm{2596} \end{equation*}
This force produces a clockwise rotation of the column, so the corresponding scalar moment is negative and the moment expressed as a vector is
\begin{equation*} \vec{M}_A = \Nm{2596} \circlearrowright \end{equation*}

Subsection4.1.2Direction of a Moment

In two-dimensional problems, moment vectors are perpendicular to the page and moments simplify to clockwise or counter-clockwise rotation about a point, called the point of interest or moment center. The point of interest can be visualized as the point where the axis of rotation pierces the page. Since these moment vectors point into the third dimension, they are usually drawn as circular arrows in the plane of the page.
Consider the two moments acting on a horizontal surface, shown below. To determine the sense of these moments, i.e. whether the vectors point up or down, apply the right hand rule. Imagine grasping the axis of rotation with your right hand, with your fingers curling in the direction of rotation; your thumb then points in the direction of the moment vector. Using this technique, we see that counter-clockwise moments point up, or out of the page, while the clockwise moments point down or into the page. In other words, counter-clockwise moment vectors point towards the positive $$z$$ direction and clockwise moments point in the $$-z$$ direction.
Pause your reading now to apply the right-hand rule yourself to determine the direction of the two moments in Figure 4.1.4
In three dimensions, moment vectors are more difficult to visualize, because they can point in any direction in 3-D space, however they still point along an axis of rotation with sense determined by the right hand rule.
As discussed in Subsection 2.8.1, there are several alternative methods for applying the right-hand rule, however they all define a direction that is mutually perpendicular to two other vectors. To find the direction of a moment, the first of these vectors is a position vector which points from the point of interest to a point on the force’s line of action, and is designated $$\vec{r}\text{.}$$ The second vector is the force, $$\vec{F}\text{.}$$
With the three-finger method, you align your right-hand index finger with vector $$\vec{r}$$ and your middle finger with the force $$\vec{F}\text{;}$$ your thumb will point in the direction of the moment vector. Alternately, you can align your thumb with $$\vec{r}$$ and your index finger with $$\vec{F}\text{;}$$ your middle finger will point in the direction of the moment vector.
With the point-and-curl method, place your right hand flat and point your fingertips in the direction of $$\vec{r}\text{.}$$ Rotate your hand until the force $$\vec{F}$$ is perpendicular to the back of your hand and can rotate your fingers. In this position, your thumb defines the direction of the moment vector and also the axis of rotation.
Any of these techniques may be used to find the direction of a moment. They all produce the same result so you don’t need to learn them all, but make sure you have at least one method you can use accurately and consistently.