Statics

Lesson 3

Free-Body Diagrams and the Equations of Equilibrium

🟦 3.1 Introduction to Free-Body Diagrams (FBDs)

🟦 3.2 How to create FBDs: FOLLOW THE RULES!

🟦 3.3 Visual communication and drawing

🟦 3.4 Introduction to the equations of equilibrium (E.o.E)

🟦 3.5 Example problems for evaluating static equilibrium

🟦 3.6 Example problem for solving for an unknown force (or moment)

🟦 3.7 Example: a collinear force system (a stack of books)

🟦 3.8 Example: a concurrent force system (three cables pull on a ring)

🟦 3.9 Example: using moment equilibrium to solve for unknowns

🟦 3.10 Pulley sheaves, pins, and cables

🟦 3.11 Qualitative equilibrium

➜ Practice Problems

🟦 3.1 Introduction to Free-Body Diagrams (FBDs)

The free-body diagram or FBD is the most important concept in all of Statics.

We begin with the problem statement, or context. We will typically call this the loading diagram.

In order to create a FBD, start with the loading diagram, and then fully free the body from the context.

Remove supports, connections, elements, and members until you have isolated whatever it is you want to study.

Everything that is removed (cut, disassembled, not shown, etc.) is replaced with its effect on the body.

The effect will be a force and/or a moment.

Inspect this flipbook very carefully.

❏ Flipbook: a series of free-body diagrams

It contains 1 loading diagram (image 1) and 5 FBDs (images 2 through 6).

Pay special attention to slide 6, the FBD of the monkey. Notice that the normal force was not drawn coincident with the center (centroid) of the monkey's mass. It is offset a bit to the right. This is because the FBD of the monkey is not symmetric. Don't worry about this too much right now. The key point is that each vector must be drawn in the exactly correct location -- and it is not necessarily at the center (centroid) of the object.

In Physics, your professors would lump all of the mass of an object (a baseball, a planet, a box on a ramp, etc.) into a single point. Graphically they would draw that as a solid dot, and then apply forces to that dot. This may have been called "particle equilibrium." In Statics, only the simplest systems can be modeled as a particle. We have to draw the actual geometry of the object or assembly of objects.

FBDs are trickier than students think, and they are the foundation of the course.

❏ Sample FBD in static equilibrium

🟦 3.2 How to create FBDs: FOLLOW THE RULES!

Once you learn the rules for creating FBDs, you can apply them to any solid mechanics problem in this class and beyond. The sooner you learn them, the easier it will be to master Statics!

❏ Determine what you want in the FBD.

- Figure out what unknown the problem is asking for you to solve. The problem statement is usually a loading diagram.
- The FBD must expose that unknown as an action on the body.
- In order to expose a force or moment on a body, you MUST remove / detach / disassemble the element that transfers the force to the body.

For example, say that you apply equal and opposite forces (F1 and F2) to two wood blocks, as shown. You're asked to solve for the normal force that exists at surface c-c.

Inspect the three FBDs below carefully. Two allow for you to solve for N. The other one does not.

❏ Solve for the force at surface c-c

Forces F1 and F2 are both compressive and share the same line of action (d-d).

FBD of Blocks A and B as a system

This is a FBD in static equilibrium. It represents the system of Blocks A and B together.

It doesn't help us solve the normal force at c-c.

Since we didn't pull the blocks apart at c-c, that force cannot be drawn on this FBD. (If you were to draw it on this drawing, it would be a concept error.)

This FBD is not useful to us.

❏ FBD of Blocks A and B as a system

This FBD would only allow you to state that F1 = F2.

FBD of Block A alone

This is also a FBD in static equilibrium.

In this FBD, we have isolated Block A. Block A feels "push" forces on both sides.

Since we removed Block B from the diagram, we must replace it with its effect.

The effect of Block B on Block A is the leftward push force, N.

This FBD will allow us to solve for N.

❏ FBD of Block A alone

This FBD shows that N = F1.

FBD of Block B alone.

This FBD is again in static equilibrium.

We have isolated Block B. Block B "feels" push forces on both sides.

Since we removed Block A from the diagram, we have replaced it with its effect (the rightward push force N).

This FBD is just as useful as the prior one, if our goal is just to solve for N.

❏ FBD of Block B alone

This FBD shows that N = F2.

❏ While learning, don't be afraid to personify the FBD.

We usually think about personification in literature courses. It's when an inanimate object is written in a way that it is feeling human emotions. In mechanics, the FBD shows us what each body "feels" or experiences. It's like the body is the protagonist of the story. The body in the FBD does not know what is happening in the distance - it only "feels" contact forces applied by adjacent bodies.

❏ Be ready for some ambiguity and flexibility with respect to modeling self-weight of the body

If a weight (body force, W = mg) is given to you in the problem statement, it's a sign that you should include it in the FBD. We do this when the weight of the body is large enough to impact our answer in a meaningful way. In Statics, it is common to neglect self-weight, since it's often negligible compared to the loads into the system. In the example above, the self-weight of Blocks A and B has been intentionally excluded from the analysis, and that is OK. Engineers are modeling reality. Our models are imprecise and imperfect, but that is OK, because engineers think in terms of safety, not precision.

❏ Think hard about Third Law Pairs.

The only time that you will have both N3L pairs in a FBD is if you are modeling a particle at the interface between two objects. There won't be a practical purpose for this type of FBD in Statics. If you think of contact forces as pushes or pulls, it will help you deduce which N3L pair force should be drawn on each FBD.

❏ The line of action for each vector is critically important.

Remember from Lesson 02 that moment summation equations are a function of the perpendicular distance between the point of interest and the vector's line of action. Incorrect lines of action for vectors will yield incorrect moment summations.

❏ The point of application for each vector is (mostly) important.

In general, we like to draw pushes and pulls properly, especially at the beginning of a Statics course. As we become more sophisticated, we'll learn some tricks that will break this rule. Please stay flexible; there are few absolutes in life.

❏ Remember the difference between loading diagrams and FBDs.

They're similar, but also different. Loading diagrams show the context; you can NOT run equations on a loading diagram. The FBD frees the body from the context and is an analytical tool we use to model the problem numerically.

❏ Your calculations in Statics must be based on a FBD.

In Statics, you don't write any equations until you have drawn the FBD. Different FBDs mean different equations, so it's impossible to assess whether or not you understand Statics concepts unless there is both a FBD and a set of equations.

🟦 3.3 Visual communication and drawing

❏ Drawing is a mode of thinking

Statics is certainly based in principles from physics (mostly Newton's Third Law), but its strongest connection may actually be to geometry. If you think back to your high school geometry class, you'll (hopefully) remember doing a lot of drawings.

The time you spend drawing Statics FBDs is time well spent. The act of making good quality drawings, wherein the geometry and the vectors are drawn to scale, will help you think through the problem.

In Statics, the FBDs, other diagrams, figures, and plots are just as important as reaching a correct numeric answer.

❏ A sample loading diagram

We will solve this problem later this semester. This is a loading diagram of an excavator.

❏ Coach's corner

Draw the geometry of the FBD to scale. If you have graph paper, great. If not, try to draw the ratios of the lengths, and angular measurements reasonably accurately.
Overcome any perceptions you may have that you "can't draw" or that "engineers are good at math, not drawing." A positive attitude and growth mindset helps.
Take advantage of this opportunity to hone your graphical communication skills.
Adopt the habit of drawing the pictures/diagrams nice and large. Your calculations are secondary, and will complement the information shown in the figure.

❏ Another sample loading diagram

We will also solve this problem later this semester. This is a 2D planar truss.

🟦 3.4 Introduction to the equations of equilibrium (E.o.E)

For the first part of this course, we will focus on Statics problems that can be modeled in a 2D plane. These can be called planar problems.

We will work on 3D problems beginning in Lesson 13.

Here is how we could express the three (2D, or planar) equations of equilibrium (E.o.E.) in casual conversation:

(1) The sum of the forces in the x-direction must equal zero. (The net x-direction force is zero.)

(2) The sum of the forces in the y-direction must equal zero. (The net y-direction force is zero.)

(3) The sum of the moments about any z-axis must equal zero. (The net moment is zero.)

This defines the state of static equilibrium.

❏ Key points (for a 2D problem):

The E.o.E. can only be applied to a FBD. They cannot be applied to a loading diagram.

If all three planar E.o.E. are satisfied, then the body is in static equilibrium, and is studied in Statics.

If any planar E.o.E. is not satisfied, then the body is in motion, and is studied in Dynamics.

net force = translational motion
net moment = rotational motion
a net force and moment = the body is both translating and rotating

Generally, in a Statics course, we use the E.o.E. to solve for unknown forces (and/or moments) that impede motion and enable the state of static equilibrium. These unknowns can be thought of as reactions. We will learn about reactions in Lesson 04.

🟦 3.5 Example problems for evaluating static equilibrium

❏ Scenario #1

Let's explore the idea of static equilibrium with an example. Let's take a small block of wood as an example.

We can determine whether or not there's a tendency to translate by checking for a net force in the x-direction and/or y-direction.

We also need to determine whether or not there's a tendency to rotate. If a body is in static equilibrium, then we can sum moments about any point, and the sum of the moments will equal zero.

Point E looks appealing. All four vectors are coincident with Point E and therefore cause no moment about E. Therefore, the sum of the moments about E is equal to zero, meaning that there is no tendency to rotate.

To check our work, we can spot check another point and run another moment summation equation. Let's sum moments about A:

That equation also shows no tendency for the body to rotate. If the body is in static equilibrium, then you can sum moments about any point (axis) and calculate zero.

❏ FBD of Scenario #1

❏ Scenario #2

In this scenario, the horizontal 4N force has a line of action that coincides the block's center (C).

By inspection, the forces are balanced in the y-direction: +8N - 8N = 0.

But in the x-direction, there is a net 4N force in the positive (rightwards) direction. This block is in translational motion; the tendency to translate is not arrested.

Is the block also in rotational motion? For this check, we sum moments about the block's center (C). All three forces in the system are coincident with point C. This means that the net moment on the block is zero, and there is no rotation.

❏ FBD of Scenario #2

❏ Scenario #3

This scenario is nearly identical to the prior one. The only change is that we have moved the 4N force to a different point of application.

We already know from Scenario #2 that the block is in translational motion rightward.

Sum moments about any point to determine the tendency to rotate. Again, we sum moments about the block's center (C):

We conclude that this block is both translating rightwards and rotating clockwise.

❏ FBD of Scenario #3

🟦 3.6 Example problem for solving for an unknown force (or moment)

Let's say that a ruler lays on a desk. We neglect friction. Three people apply forces to the ruler, as shown, at A, B, and D. The forces share the same line of action.

Your job is apply a force at C that puts the system in static equilibrium.

This example problem is trivial in content (you can determine that a 1# leftward force is needed by inspection, by subtracting 4# from 5#).

This problem is not here for the content. It's here to serve as an example of an important problem-solving concept.

❏ A ruler lays flatwise on a table and forces are applied

In this text, as a teaching tool, I sometimes dash an unknown vector and give it arrowheads at both ends.

Please don't do this as part of your problem-solving process.

It's simply a teaching tool used when I want to specify the line of action of an unknown vector while discussing the direction with students. For instance, I might say: "can we deduce whether Fc is leftwards or rightwards?"

❏ A double-headed arrow is simply a placeholder

In your problem-solving process, on your FBD, you must assume a direction for an unknown force (or moment) vector.

if the sign in your answer is positive, it CONFIRMS your assumption
if the sign in your answer is negative, it REVERSES your assumption

In this drawing, you'll see that someone has (incorrectly) assumed that the unknown force at C is rightwards.

It is OK to incorrectly assume the direction (or sense) of an unknown force (or moment).

❏ On your FBD, you must make an assumption for the direction of the unknown force (or moment)

Here is how to properly set up the analysis to solve for Fc, based on the rightwards assumption.

Inspect the calculation carefully and emulate it when you solve problems.

At the end of the problem, if you discover that your assumed direction is incorrect, do not revise the FBD to reflect the correct direction of the vector. When someone views your work, they will think that you made a concept error. The example calculation provided here is the best one I know of; please adopt and emulate this reasoning throughout the course.

❏ Closely emulate this calculation example

🟦 3.7 Example: a collinear force system (a stack of books)

Imagine a stack of three books (A, B, and C) on a table. They are piled up concentrically (symmetrically).

In this flipbook, you'll see various FBDs of one or more books.

Since the stack of books is supported by a table, and therefore in static equilibrium, we can sum forces in the vertical direction in order to solve for unknown forces.

Specifically, we are interested in solving for the normal forces (N) transferred between the books.

We will use one of the equations of equilibrium (the summation of forces in the vertical direction must equal zero) to solve for these unknowns.

Work this problem carefully, following each step from the flipbook, and drawing each FBD.

The final image in the flipbook illustrates a new idea called the exploded FBD. This is when you draw each body in the system as its own FBD. The advantage is that you can see the N3L pairs reverse direction (or sense) between adjacent bodies. It's a great learning tool.

❏ Flipbook: A stack of books is supported by a table

Concept question #1: did the FBDs in the flipbook follow THE RULES?

Yes. Remember: you can only draw a vector on a FBD when the body causing that force is not depicted as part of the free-body.

Concept question #2: was this a 2D problem or a 3D problem?

We modeled the problem in 2D by visualizing a side view of the stack of books. Whenever we can simplify the 3D world into a 2D projection (or elevation), we choose to do that. The flipbook illustrated a 2D or planar model for the stack of books.

Reminder: if you were asked to depict the normal force between books B and C and tried to draw that on a FBD that consisted of the entire stack of books (A, B, and C), it would be a concept error. In order to reveal the force transferred between bodies B and C, you must disassemble the stack at that interface to reveal the normal force between B and C.

This problem can be modeled as a collinear force system. That means that all of the forces in the system share the same line of action. Collinear problems are the easiest problems to solve, because you only have to use one E.o.E. to solve unknowns. You solved these previously in your Physics class. The example problem in Section 3.6 was also a collinear force system.

🟦 3.8 Example: a concurrent force system (three cables pull on a ring)

In a concurrent force system, all of the vectors in the FBD have lines of action that intersect at the same point in space.

For instance, inspect the photo of a web of suspended cables in Rome, Italy.

Most of these cables are not taut, but imagine that you were to build a model of this system out of rope or bungee cords. In order to make the system of cables taut, you would need to tension them significantly.

At each point in the web where multiple cables connect, we can perform a concurrent force analysis.

We can use two E.o.E. to solve these types of problems:

sum of forces in the x-direction equals zero
sum of forces in the y-direction equals zero

Since the forces coincide at the same point, a moment summation equation will not help you solve any unknowns.

❏ Tensile cables in Rome, Italy

Largo Argentina, June 2023, photo by S. Reynolds

Here is an example of a concurrent force problem.

A connection ring is pulled by three cables (1, 2, and 3). The system is in static equilibrium as shown.

You intend to tension cable 3 to 100 kips (kilopounds) of force.

What is the force in cable 1 and cable 2 in this scenario?

Our strategy is to count the number of unknowns in the FBD and make sure we have enough equations to solve the system. The solution is copied below. You should have worked this type of problem in Physics, so this should be a review.

Notational note: It would also be acceptable to use the symbol T for the tensions in the three cables. F is a generic symbol for force. T is more specialized.

❏ Three cables pull on a connection ring

🟦 3.9 Example: using moment equilibrium to solve for unknowns

In this flipbook, you'll see an example of how we can use the moment equilibrium equation to solve for unknowns.

Remember that the moment equilibrium equation can be applied with respect to rotation about any axis. Be sure to apply the sign conventions we learned:

a moment (force times distance) that tends to rotate the body counterclockwise is considered positive
a moment (force times distance) that tends to rotate the body clockwise is considered negative

❏ Flipbook: two springs arrest the tendency to rotate

🟦 3.10 Pulley sheaves, pins, and cables

Explore the interactive visualization of a two pulley system in static equilibrium.

For this problem, we will assume that the weight (W=mg) of the two green boxes is significant and should be included in the problem. (They both weigh the same amount.)

All of the other bodies have negligible weights, so we will neglect those self-weights from our model.

Fall 2024 students: in-class, we will go through terminology for pulley sheaves, pins, and cables.

❏ Interactive: a two-pulley system

🟦 3.11 Qualitative equilibrium

Fall 2024 students: we will go through this concept in class together, OK?

➜ Practice Problems

Part I.

Please work questions 1, 2, 3, 4, 5, and 6 from the page linked below. Make sure that each FBD is constructed properly. Additionally, practice using the correct symbols for vectors (see Section 1.12 for a refresher). Engineers rarely use the generic symbol F for force. We prefer more specialized symbols (T for tension, N for normal compressive force, V for shear, etc.). Fs for a spring force is fine. And Ffr for a friction force is OK too.

http://mechanicsmap.psu.edu/websites/2_equilibrium_concurrent/2-5_equilibrium_analysis_concurrent/equilibriumconcurrent.html

Part II.

Also, please work problem 3 from this page:

http://mechanicsmap.psu.edu/websites/3_equilibrium_rigid_body/3-6_equilibrium_analysis_rigid_body/equilibrium_analysis_rigid_body.html

Part III.

Tinker with this Geogebra interactive. (Note: I did not create this one.)

Model c=4, d=5, and e=6.

Place c at x=0 and e at x=10. Keep support A at 6.

What is the location of d that creates static equilibrium? Calculate it and then use the model to check your work.

https://www.geogebra.org/m/HxQ6MKCd

Part IV.

In your own words ...