Statics

Lesson 02

Moment -- or, a tendency to rotate

🟦  2.1  Introduction to moment

❏ The tendency to rotate

We have already learned that a force can be conceptualized as a tendency to translate.

In Statics, we must arrest the tendency to translate as well as the tendency to rotate.

A moment (of a force) can be thought of as a tendency to rotate about an axis.

That is, we can think of a moment is a tendency to rotate about an axis.

The introductory image shows a translational vector (↓) and a rotational vector (↺).

This may be the first time you have thought about rotation as a vector, because vectors are often depicted as straight arrows (↓). The curly arrow (↺) is indeed a vector, as it is used to convey both magnitude and direction (clockwise or counterclockwise).

❏ Translation vs. Rotation

❏ Moment (of a force)

We compute a moment by multiplying a force by a distance. 

Therefore, a moment has units of force times distance.

Inspect this flipbook for an introduction to the concept of moment. We see that a balanced cucumber does not create a moment while an off-center cucumber does create a moment.

For notation, we use the symbol M for moment. 

Graphically, we convey the idea of a moment with a curly arrow (↺ or ↻).

There are important rules that dictate when you must include or exclude the curly arrow from your diagram. These rules are critical, but it will take a while to get to them. They are explained in Section 2.13.

❏ Flipbook: An introduction to moment

🟦  2.2  Scalar expression for moment (of a force)

❏ The equation for moment (of a force)

The simplest scalar expression for moment is M = Fd. 

Verbally, we'd say "the moment about an axis is equal to a force multiplied by its perpendicular distance to that axis."

❏ 2D and 3D depictions and terminology

In 3D, the axis of rotation is a line (or really, a ray). In a 2D projection, the axis is a point, because the axis is perpendicular to your screen or paper. These are both illustrated here.

To properly visualize the moment, you must draw the 2D plane that is normal (perpendicular) to the axis of rotation.

We can use a subscript after the M to indicate the axis of rotation being studied. We can pick any point that lies on the axis of rotation and call it Point A. Therefore, we write MA  to indicate the moment about Point A.

❏ Is there a vector expression for moment (of a force)?

We will learn the vector formulation for moment (M = r x F) in Lesson 14.

❏ Didn't I already learn this in Physics?

You were introduced to moment in Physics, except that your professor likely called this concept torque. Engineers also use the word torque, but for us, it means something slightly different: the subtype of moment that twists a body. For the purpose of learning Statics, please use the term moment exclusively (at least for now).

❏ Moment about an axis (3D)



In a 3D drawing, we can dash in the axis of rotation. 

In 3D, that axis is shown as a line (or ray).


❏ Moment about a point (2D)



In a 2D drawing, we orient the drawing so that the axis is perpendicular to the plane of the screen or paper. 

Now, the axis of rotation appears as a point. It's really an axis. It just looks like a point. When we talk about taking moment about a point, we really mean about an axis.

🟦  2.3  Key terminology: rotational motion vs. a tendency to rotate

Let's use the analogy of a pushpin to learn more about moment -- a tendency to rotate.

Work carefully through this interactive animation. A square piece of cardboard rests on a corkboard. 

❏ Thought experiment #1:

A single pushpin connects the cardboard to the corkboard. We'll neglect any type of clamping or compression force from the pushpin, and we'll also neglect friction in the plane between the two materials.

We apply a force to the cardboard. That force, multiplied by its perpendicular distance to the axis of rotation creates a moment. Since there is no mechanism to arrest the motion, the cardboard rotates about the pushpin.

❏ Thought experiment #2:

We add a second pushpin and apply the same force. this time, the two pushpins deliver reaction forces to the system that counteract the cardboard's tendency to rotate. The state of static equilibrium is plainly evident.

❏ Key takeaways:

A moment may or may not cause rotational motion. Even when rotational motion is impeded, we will still compute moments. We can always think of a moment as a tendency to rotate, even when the system is in static equilibrium.

We can use the analogy of the pushpin whenever we compute a moment. We have to imagine the rotation of the body about that axis.

🟦  2.4  Moment arm

The equation M=Fd tells us that the magnitude of a moment is a function of the perpendicular distance between the force's line of action and the axis of rotation.

That perpendicular distance merits a special name: the moment arm. The larger the moment arm, the larger the moment.

Imagine that you want to use a crescent wrench to apply a moment that removes a hex-head bolt. Which wrench would you choose for the task: 1 or 2?

Since wrench 2 has a greater moment arm than wrench 1, it will deliver a larger moment to the hex-head bolt (assuming that the force P is held constant in this comparison). If you want to minimize your effort, choose wrench 2! If you're looking for an arm workout, choose wrench 1!

In a dynamics problem such as this, the objective is to cause motion. In this scenario, the bigger the moment arm, the better the design.

In a statics problem, our objective is very different. We want to impede motion and maintain static equilibrium. In this type of scenario, we wish to minimize the moment arm, as possible, through smart design decisions.

❏ Compare moment arms of two wrenches

Let's practice computing moment arms. First, browse the 3D interactive to get a sense for the geometry. Then, work through the 2D flipbook until you can compute moment arms accurately.

❏ Flipbook: Practice with moment arms

One key takeaway embedded in that activity is the fact that when a force is coincident to a point, it does not cause a moment about that point.

🟦  2.5  Sign convention for moment

❏ 2D sign convention for moments

The magnitude of a moment is computed with M=Fd. 

In this equation, F and d are both magnitudes (neither positive nor negative).

We now need to establish a sign convention for moment. Since F and d do not bear signs, we determine the sign of M by inspection -- we must visualize the tendency to rotate:

Fair warning: this is the first Statics concept that predicts student success in this course. Students that don't learn this sign convention well now tend to struggle with the entire course and beyond.

❏ CCW positive and CW negative

❏ Mastering the sign convention for moments: CW vs. CCW

For example, consider the five examples below. Each is a mini-flipbook containing two images. You can think of the yellow square object as a piece of cardboard if you like. Determine whether the moment about pushpin A (point A) is positive or negative. Check your work on each problem; work these as many times as needed to master this important concept.

Do you remember the Ace of Spades homework problem with the toothpicks glued to it? That analogy correlates to the exercise above. The pink vectors represent people pushing and pulling on the Ace of Spades.

❏ 3D sign convention for moments

Another way to work through signs of moments is to use your right hand. You curl the fingers of your right hand in the direction of the tendency to rotate. When you're making a "thumbs up" gesture, the moment is positive. "Thumbs down" is a negative moment.

There are a few caveats to this type of thinking: it only works when x is rightward (→) and y is upward (↑). We can call this approach the right-hand-rule (R.H.R.) for determining the sign of a moment.

Explore the 3D interactive to learn about the R.H.R. You'll also get a preview of double-arrow notation. The 2D image gives you bullet points for the key concepts from the interactive.

❏ "Thumbs up" and "thumbs down" logic

❏ Mastering the sign convention for moments: thumbs up vs. thumbs down

The "thumbs up vs. thumbs down" method for assessing the sign of a moment is an alternative to thinking CW vs. CCW. The two methods are 100% compatible or interchangeable. Just be forewarned that when we get to 3D problems, the "thumbs up / thumbs down" method will be more useful than thinking CW vs. CCW.

Practice the "thumbs up / thumbs down" method in the five mini flip-books below. Each mini flipbook has two slides.

Did you notice that when a force is coincident with a point, the moment arm is zero, and therefore the moment is zero? That's the second time this has been emphasized, so it's probably an important takeaway from this lesson. Basically, if you're trying to rotate a rigid body about an axis, you better apply a force that is not coincident to that axis.

🟦  2.6  Putting it all together

Let's add some discrete values to our piece of cardboard. And this time we anchor it with five pushpins: A, B, C, D, and E.

Is the cardboard in static equilibrium or in a state of motion? With 5 pushpins, it's definitely stationary.

Of course, we can still compute moments, as moments are defined as the tendency to rotate.

In these five mini-flipbooks, calculate the moment of the 4N force about A, B, C, D, and E. Follow these steps:

🟦  2.7  Varignon's theorem

When we need to calculate the moment about a point (an axis), we can either use the force vector/s given to us, or we can break them into components.

Varignon's theorem (published by French mathematician Pierre Varignon in 1687) states that the moment of a resultant force about a point is equal to the summation of the moments of the component forces.

The same concept holds true whenever you want to find the moment (tendency to rotate) about an axis due to a combination of applied forces.

The process of determining the net moment about an axis due to all of the forces on the body is called moment summation, or more commonly, summing moments.

Here's how I'd paraphrase Varignon's theorem:

When you want to sum moments about a point (an axis), you can use the vector/s given to you in the problem, or break them into components. Just make sure that you assign the sign algebraically (as it's possible that one component could cause a clockwise rotation and the other cause a counterclockwise rotation). 

❏ Application of Varignon's Theorem

In this series of images, we see some strategies commonly used to simplify moment summations. The premise is that one force is applied (F1) and that we wish to calculate the moment about A. 

Let's say that the magnitude of F1 is 2 kips.

We could use F1 times d1, but we'd have to do a little geometry to calculate d1. (Calculate d1 yourself. The answer is 2.330 inches.)

The moment about A is -4.66 kip•inches (which can also be expressed as 4.66 kip•inches ↻).

Alternatively, we could break F1 into components F2 and F3. 

The bounding box has been dashed in to help you visualize the relationship between source vector F1 (you could call it the resultant force if you like) and components F2 and F3. 

This technique is a great strategy for this particular example, since d2 and d3 are given in the problem.

Just watch the signs, since the moment caused by the vertical component is positive and the moment caused by the horizontal component is negative.

Do this calculation on your own. You should also get an answer of -4.66 kip•inches. Again, you can express this as 4.66 kip•inches ↻. 

When expressing answers, be sure not to use a double negative. If you were to write –4.66 kip•inches ↻, your answer could be interpreted incorrectly (as in a negative clockwise value could be erroneously interpreted as a counterclockwise or positive moment).

Another strategy is to draw a line between the vector's point of application and the point (axis) of interest. That line defines a new coordinate system. 

Draw a bounding box around vector F1 in the new coordinate system and resolve the vector into component forces F4 and F5.

Note that F1 = (F2 + F3) = (F4 + F5). Also note that since F1 was drawn as a pull force, then the other vectors should also create a pull force (arrows pointing away from the body).

This technique works a little better when we bring the power of vector notation to play in Lessons 13-15. (We can dot a force vector with a unit vector to project it to a certain axis.) For now, we'll do these calculations manually in 2D space.

Try this one on your own. It's easy to calculate d4 with the Pythagorean Theorem. To figure out F4, the key is to draw and compute angles correctly. (F4 has a magnitude of 0.7278 kips).

Of course, you'll get the same answer as above (4.66 kip•inches ↻).

These problems may be challenging for you if it's been a while since you thought about geometry: complementary angles, supplementary angles, angle congruency, etc. Unfortunately, you can't rely pattern recognition or plugging into generic equations to compute moments in a 2D scalar problem. Your moment summation equation will need to be custom-derived in every problem. The good news is that you can use the strategies above to manipulate force vectors and ultimately work smarter (faster) and not harder!

🟦  2.8  Putting it all together: example problem for moment summation

Below are sample calculations for summing moments as a result of multiple forces in a system.

Instead of using pushpins to visualize axes of rotation, I'm using nails in this example.

Work both examples independently and spot check your work with the solution. Remember to assess signs by inspection.

What is the moment about A due to applied forces P1, P2, and P3? (Neglect the forces caused by the nail reaction at B.)

Now, calculate the moment about B due to these same three forces. (Again, neglect any forces that may be transferred from nail A into the system.)

The fact that both answers are the same is a coincidence only. You can prove this to yourself by summing moments about a couple other points.

🟦  2.9  Applied moments

We already know that we can apply a force to a body. The word applied means we're talking about an input into the system. It can also be called a load.

We can also apply a moment to a body.

Let's say that you need to turn the dial of a thermostat. You grip the knob with your right index finger and thumb, clamp them together while also using friction to rotate the knob.

As you see in the image, the combination of clamping (normal force) and friction force (shear force, tangential in this particular example) exerted by your fingers can be a complex distribution of vectors.

As an alternative to drawing the forces, we can simply draw their net effect with a curly vector: the applied moment (tendency to rotate). In this particular example, rotation is not impeded by other forces in the system, so the dial turns.

❏ Applying a moment to a body

The left image shows the forces exerted by the person to turn the dial. 

The right image shows the couple moment, as well as the dial in the rotated position (from cold to hot).

We are transitioning into the 2nd major concept of this lesson.

The first concept was moment summation, which is a mathematical operation. A force offset from a point will cause both a tendency to translate and a tendency to rotate.

The second concept is an applied moment, which represents the tendency to rotate without a tendency to translate.

Make sure you understand the difference between these two concepts.

🟦  2.10  Force couples and couple moments

Applied moments are a subcategory of what we call couple moments.

The confusing thing is that not many people call a couple moment a couple moment. People usually just call a couple moment a moment. The terminology always creates enormous confusion for Statics students learning this for the first time.

In this lesson I will be intentional with terminology to aid your learning. 

❏ Categories of couple moments

We start with the force couple. Whenever you have two forces that are parallel, equal in magnitude, and opposite in direction, you can call them a force couple. They team up to cause rotation.

For instance, lay your phone or calculator on the desk. If you want to rotate the object about its centroid, apply a force couple per the instructions above.

A single force causes a tendency to translate, as well as a tendency to rotate.

A force couple only causes rotation. There is no net force, and no tendency to translate. That's why they are special.

❏ Definition of a force couple

Let's define some symbols:

F = the magnitude of either force in the couple

d = the perpendicular distance between the two forces

We can convert the force couple (e.g. ↓↑) into an equivalent couple moment. (↺), with this simple equation:

M = Fd

where M = the couple moment

❏ Definition of a couple moment

It's common to represent force couples as couple moments in statics problems. 

When you see an applied moment, imagine someone grabbing a knob in the system and forcefully turning it. If it's helpful, you can always convert an applied moment into any arbitrary force couple, as shown in the diagram.

❏ Equivalencies

🟦  2.11  Example problem: moment summation with applied moments

Here is an example problem that shows you how to include couple moments (concept 2) in your moment summation operation (concept 1).

Track your units! Each term in the moment summation must have units of force times distance.

Couple moments have units of force times distance built in.

🟦  2.12  Are moments subcategorized as forces?

Forces and moments have different units. We think of them as different concepts:

But sometimes we think of moments as a subcategory of forces. Bear with me here. Let's say you are doing some structural analysis in a commercial Finite Element Analysis software and you want to apply a couple moment to your analytical model. You see an icon for "apply forces" but you don't see one for "apply moments." Frustrated and confused, you open up the "apply forces" dialogue box to discover that there's a way to apply couple moments in that location.

Here's the explanation.

Since every moment can be converted into two forces that are parallel, equal in magnitude, and opposite in direction, we sometimes include them under the umbrella of "forces."

For this reason, sometimes in this text you'll see me write "Forces (and moments)" as a parenthetical note. Whenever you see that, just be reminded that couple moments (commonly simply called moments) are equivalent to two forces that are equal in magnitude, opposite in direction, and parallel.

🟦  2.13  When to include / exclude the curly arrow for moment in a drawing

There are very specific rules that dictate when you must include (and must exclude) the curly arrow symbol for moment in a Statics drawing:

❏  Concept 1 (the idea that moment is the tendency to rotate caused by an off-center force):

❏  Concept 2 (a couple moment):

You've got this. You can do it. 🩷

➜ Practice Problems

Scrambled answers (without units) to all problems EXCEPT problem 8:

-56.6     -40     -35     -9     -7     -6     -2.74     5     10     28.5     91.5     92     180     240     575


Problem 1.

A big kid (65#) and a little kid (50#) are trying to balance on a seesaw.

You sum moments about A to determine whether the seesaw rotates clockwise, counterclockwise, or remains horizontal (in static equilibrium, with no net moment)

Problem 2.

A dad (exerting a force of 800N) tries to balance on a seesaw with his three kids (200N, 300N, and 400N).

Same question as above:

Problem 3.

You are working in a restaurant, and investigating a swinging door that separates the kitchen from the dining area.

The door rotates about hinges that are (approximately) aligned with the z-axis.

What is the moment of the 4 pound force about the axis of the hinges? (It's positive, by the way, becaues when you use the right-hand-rule your right thumb points in the positive z-direction, but you don't need to worry about that in this lesson.)

Problem 4.

A yellow object lies on a table. Four forces are applied.

Maybe this object is a piece of cardboard, maybe it's a phone, maybe it's sticky note pad, etc. It doesn't matter.

You need to sum moments about A. What is the total moment about A.

Practice both sign conventions. Express your answer with positive/negative signs first. Then, re-write the answer with curly arrows (↺ or ↻). Look at the example in section 2.11 for an example.

ΣMA = ?

Problem 5.

It's the exact same object, but this time we want to sum moments about B.

ΣMB = ?

Problem 6.

And now... sum moments about C.

ΣMC = ?

Problem 7.

A force is applied at A to a horse-shoe shaped object.

What is the moment of the force at B? ΣMB = ?

Solve for this two ways:

Problem 8.

A 10 kN force is applied to an amorphous blob as shown.

Solve for the moment about A (ΣMA = ?) three ways, as described in Section 2.7:

Problem 9.

Rigid body ABC is subjected to two forces.

The line of action of the 100N force is defined by BC.

What is the summation of the moments about A?

ΣMA = ?

Problem 10.

Rigid body ABC is subjected to two forces.

The line of action of the 58k force is defined by BC.

What is the summation of the moments about A?

ΣMA = ?

Problem 11.

Convert each force couple into a couple moment.

Problem 12

Four forces and one couple moment have been applied to a rigid body.