Statics

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.

M = the moment of a force about an axis

F = the force

d = the perpendicular distance between the force and the axis

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

❏ A key equation to learn

❏ 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.

❏ Moment about an axis (3D)

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

❏ Moment about a point (2D)

In a 2D drawing, we orient the view so that the axis of rotation is perpendicular to the plane of the screen or paper. It looks like a point, but it's really an axis. When we talk about taking moment about a point, we really mean about an axis.

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

Yes, there is! We will learn the vector formulation for moment (M = r x F) in Lesson 14. Stay tuned.

❏ Wasn't this covered in Physics?

You were introduced to the idea of 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: a subcategory of moment for offset forces that tend to twist a body. For the purpose of learning Statics, please use the term moment exclusively.

🟦 2.3 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 the 3D interactive. A square piece of cardboard is secured to a corkboard with a pushpin.

❏ 3D interactive: the pushpin analogy

The 3D interactives can take a little time to load (it's mostly a function of Wifi speed). Please be patient: they're worth the wait!

❏ Follow-up #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. We have rotational motion.

❏ Follow-up #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. There is a tendency to rotate, but that tendency is counteracted by the forces exerted by the pushpins.

❏ Key takeaway:

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.

The analogy of the pushpin is useful whenever we compute a moment. We simply imagine the rotation of the body about the pushpin's 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).

Want to minimize your effort? Choose wrench 2!
Want a good 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 the opposite. We want to impede motion and maintain static equilibrium. In this type of scenario, we wish to minimize the moment arm, as much as possible, through our engineering 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.

❏ 3D interactive: practice with moment arms

❏ Flipbook: Practice with moment arms

Here's a key takeaway from that activity: when a force is coincident to a point, it does not cause a moment about any axis that goes through that point.

🟦 2.5 Sign convention for moment (very important!!!)

❏ 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). In your calculations:

never ever give F a sign
never ever give d a sign

Since F and d do not bear signs, our sign convention for M is done by inspection. We must visualize the body's tendency to rotate about the axis we're investigating.

A moment that tends to rotate a body counterclockwise (CCW or ↺) is considered positive.
A moment that tends to rotate a body clockwise (CW or ↻) is considered negative.

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

❏ CCW positive and CW negative

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

As an example, consider the five images below. Each is a mini-flipbook with 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, and work these as many times as needed to master the concept.

❏ 3D sign convention for moments

Another way to work through signs of moments is to use the right hand. Curl the fingers of your right hand in the direction of the body's tendency to rotate.

When you're making a "thumbs up" gesture, the moment is positive.
The "thumbs down" gesture is a negative moment.

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

Explore the 3D interactive to learn about the R.H.R. and get a preview of double-arrow notation. The 2D image summarizes key concepts from the interactive.

❏ 3D interactive: thumbs up convention

❏ "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 compatible and interchangeable in 2D scalar problems. When we get to 3D problems, we'll see that the "thumbs up / thumbs down" method will be more useful than assessing CW vs. CCW rotation.

Practice the "thumbs up / thumbs down" method in the five mini flipbooks 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 dimensions to our piece of cardboard and 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, under the assumption that the material doesn't fail.

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. Use a 4-step process:

determine the moment arm
multiply distance by force
write down the product, including units
lastly, determine the sign by inspection

🟦 2.7 Moment summation and Varignon's theorem

The process of determining the net moment about an axis due to all of the forces on the body is called moment summation. More commonly, people say that they are summing moments.

When we need to calculate a moment (of a force or a system of forces), we can either use the force vector/s given to us, or we can break them into component forces.

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

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 of each moment algebraically (as it's possible that one component could cause a clockwise rotation while the other might cause a counterclockwise rotation).

❏ Application of Varignon's Theorem

🟦 2.8 Three strategies for moment summation

Let's go through three strategies that can be used to simplify moment summations.

The premise behind all three strategies is identical: we wish to calculate the moment of force F1 (magnitude 2 kips) about point A. We know the inclination of F1 and its position relative to A.

The answer is the same regardless of the strategy: the moment about A is –4.66 kip·inches (alternatively expressed as 4.66 kip·inches ↻).

❏ Strategy 1: direct perpendicular distance

Here, we simply multiply F1 times moment arm d1, but have to do some geometric analysis in order to calculate d1 (see below).

❏ Strategy 2: use x- and y- components

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.

Watch your signs: the moment caused by the vertical component is positive and the moment caused by the horizontal component is negative. The solution is below.

❏ Strategy 3: use the force component perpendicular to the moment arm

Another strategy is to construct 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. Here is the solution:

These problems may be challenging if your middle school / high school geometry skills are rusty. Unfortunately, you can't rely on pattern recognition or plugging into generic equations for these types of problems. You'll need to custom-derive your moment summation equation for each problem.

🟦 2.9 Example: the moment summation operation

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, this examples features nails.

Work both examples independently and spot check your work with the solution, below.

What is the moment about A due to applied forces P1, P2, and P3?

Now, calculate the moment about B due to these same three forces.

The fact that both answers are the same is a coincidence only.

🟦 2.10 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.

🟦 2.11 Force couples and couple moments

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

People usually just call a couple moment a moment. The terminology always creates enormous confusion for Statics students learning this for the first time. Statics students tend to mix up moment summations with couple moments.

❏ Categories of couple moments

A force couple is defined as two forces that are parallel, equal in magnitude, and opposite in direction. They team up to cause rotation without translation.

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

A single force causes a tendency to translate, as well as a tendency to rotate about any axis off its line of action.

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

🟦 2.12 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).

Write out 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.

🟦 2.13 Are moments subcategorized as forces?

Forces and moments have different units, with different concepts:

a force is a tendency to translate
a moment is a tendency to rotate

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 why. Since every couple moment can be converted into a force couple, we sometimes include them under the umbrella of "forces."

For this reason, sometimes in this text you'll see the term "Forces (and moments)". Whenever you see that, remember that couple moments (commonly simply called moments) are equivalent to a force couple.

🟦 2.14 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 moment summation operation):

This is an OPERATION.
We never draw the curly arrow.
We certainly think about the curly arrow, and students often rotate their finger in the air to think through the sign convention.
But we don't actually draw the curly arrow symbol.
It's drawn in this lesson, but only as a teaching tool. I don't know how to teach this concept in an online textbook without drawing the curly arrow symbol.
Including a curly arrow on a Statics drawing for the moment summation operation is a concept error.

❏ Concept 2 (the presence of a couple moment):

We always draw the curly arrow.
It is a part of the system behavior.
It can't be omitted.
Omitting this curly arrow on a Statics drawing is a concept error.

➜ Practice Problems

Scrambled answers (without units):

-56.6 -40 -35 -9 -7 -6 -2.75 5 8.91 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)

What is the sum of the moments about A? ΣMA = ?
Does the seesaw rotate about A? If so, which way?

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:

What is the sum of the moments about A? ΣMA = ?
Does the seesaw rotate about A? If so, which way?

Problem 3.

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

Part (a). You need to sum moments about A. What is the total moment (tendency to rotate) about A? The proper notation is: ΣMA = ...

Part (b). What is the total moment (tendency to rotate) about B? ΣMB = ...

Part (c). What is the total moment (tendency to rotate) about C? ΣMC = ...

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

Problem 4.

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

Part (a). You need to solve the moment (tendency to rotate) about B. This can be expressed as ΣMB = ?. Solve for the moment about B by breaking the 8 kip force into its x- and y- components.

Part (b). Work the problem again. This time, calculate the calculate the perpendicular distance between the force and point B. ΣMB = ?

Problem 5.

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? ΣMhinges = ?

By the way, this moment is would be classified as positive, because when you use the right-hand-rule your right thumb points in the positive z-direction. This will be important later in the course.

Problem 6.

Convert each force couple into a couple moment.

Ma = ?
Mb = ?
Mc = ?
What can you infer about the difference in notation when you compare the prompts for Problems 1-5 to this problem?

Problem 7.

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:

using x and y components of the force that are oriented to the plane of this screen
solving for the perpendicular distance between the 10 kN force and point A
resolving the 10 kN force into components parallel to and perpendicular to line AB