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🟦 3.1 Introduction to Free-Body Diagrams (FBDs)
🟦 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 the context. We call this the loading diagram.
In order to create a FBD, start with the loading diagram, and then free the body from the context. That means we have to remove supports, connections, elements, and members until you have isolated exactly what you want to study.
Here's the key concept: everything that is removed (cut, disassembled, not shown, etc.) is replaced with its effect (force and/or moment) on the body.
Inspect the flipbook very carefully. It contains 1 loading diagram (image 1) and 5 FBDs (images 2 through 6).
❏ Flipbook: a series of free-body diagrams
❏ Flipbook: a series of free-body diagrams
On slide 6, did you notice that the normal force was not drawn coincident with the center (centroid) of the monkey's mass? It is offset 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 precisely correct location -- not necessarily at the center (centroid) of the object.
In Physics, your professors would lump all of the mass of an object (for example, a calculator) into a single point.
Graphically they would draw the calculator as a solid dot, and then apply forces to that dot. In technical terms, they modeled the system as a particle and investigated particle equilibrium.
In Statics, only the simplest systems can be modeled as a particle. This is because we generally need to consider moment in our equilibrium studies. When you model a particle, you can only investigate translation.
This means that we have to draw the actual geometry of the object (or of the assembly of objects).
Your takeaway: do not draw Statics FBDs as a solid dot unless explicitly directed to model a particle.
❏ Physics FBDs vs. Statics FBDs
❏ Physics FBDs vs. Statics FBDs
FBDs are trickier than students think, and they are the foundation of the course.
You'll use them on almost every Statics problem you ever work.
❏ Sample FBD in a Statics course
❏ Sample FBD in a Statics course
🟦 3.2 The eight rules of FBDs
🟦 3.2 The eight rules of FBDs
Once you learn the eight rules for creating FBDs, you can apply them to any solid mechanics problem in this class and beyond.
❏ Rule 1. Construct the correct FBD.
❏ Rule 1. Construct the correct FBD.
Figure out what unknown the problem is asking for you to solve.
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 (N) 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
❏ 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 together
This is a FBD in static equilibrium. It represents the system of Blocks A and B together.
Since we didn't pull the blocks apart at c-c, the normal force on that surface cannot be drawn on this FBD. (If you were to draw it on this drawing, it would be a concept error.)
Conclusion: This FBD is not useful to us.
❏ FBD of Blocks A and B as a system
❏ FBD of Blocks A and B as a system
This FBD shows that F1 = F2 .
→ FBD of Block A alone
This is also a FBD in static equilibrium. We have isolated Block A. It 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.
Conclusion: This FBD will allow us to solve for N.
❏ FBD of Block A alone
❏ FBD of Block A alone
This FBD shows that N = F1 .
→ FBD of Block B alone
This FBD is again in static equilibrium, and 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).
Conclusion: This FBD will also allow us to solve for N.
❏ FBD of Block B alone
❏ FBD of Block B alone
This FBD shows that N = F2 .
❏ Rule 2. Personify the FBD.
❏ Rule 2. Personify the FBD.
We usually think about personification in literature courses. It's when an inanimate object is written in such a way that it "feels" human emotions. In mechanics, the FBD shows us what the body "feels" or experiences. The body is the protagonist of the story. The body in the FBD does not know what is happening in the system overall - it only "feels" (1) its own weight and (2) the forces transferred by adjacent bodies.
❏ Rule 3. Be ready for some ambiguity with respect to modeling self-weight.
❏ Rule 3. Be ready for some ambiguity with respect to modeling self-weight.
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 many Statics problems, it is common to neglect self-weight. In the example above, the self-weight of Blocks A and B has been intentionally excluded from the analysis, and that is OK.
❏ Rule 4. Think hard about Third Law Pairs.
❏ Rule 4. Think hard about Third Law Pairs.
You learned about N3L pairs in Lesson 01 - Force. 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. This was part of your Physics course, but there is no practical purpose for this type of FBD in Statics. Think of the contact forces as pushes or pulls: this will help you deduce which N3L pair force needs to be drawn on a given FBD.
❏ Rule 5. The line of action for each vector is critically important.
❏ Rule 5. The line of action for each vector is critically important.
Remember from Lesson 02 - Moment that moment summation equations require correct calculation of the moment arm. Incorrect lines of action for vectors will yield incorrect moment summation equations.
❏ Rule 6. The point of application for each vector is important (in the early stages of this course).
❏ Rule 6. The point of application for each vector is important (in the early stages of this course).
In general, in the beginning of Statics, we will draw pushes and pulls properly. In Lesson 05 - Loads and Static Equivalency, we'll focus less on the point of application so that we can manipulate the vectors mathematically.
❏ Rule 7. Loading diagrams are not FBDs.
❏ Rule 7. Loading diagrams are not FBDs.
They're certainly related, but also quite different. Loading diagrams show the context; they are the problem statement. You can't run any equations on a loading diagram. The FBD frees the body from the context.
❏ Rule 8. Your calculations in Statics must be based on a FBD.
❏ Rule 8. Your calculations in Statics must be based on a FBD.
In Statics, you cannot write any equations until you have drawn the FBD! Different FBDs result in different equations. That means that equations (even if they are computationally correct) are not comprehensible to other engineers unless they are based on a drawn FBD.
🟦 3.3 Visual communication and drawing
🟦 3.3 Visual communication and drawing
❏ Drawing is a mode of thinking
❏ Drawing is a mode of thinking
Statics is certainly based in principles from Physics (mostly Newton's Third Law), but its strongest prerequisite connection is geometry.
If you think back to your geometry class, you'll (hopefully) remember constructing a lot of drawings.
Trust me. 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 develop good problem-solving skills.
In Statics, most professors value a correct FBD more highly than a correct numerical answer.
❏ A loading diagram
❏ A loading diagram
We will solve this problem later this semester. You will have to draw FBDs that correspond.This is not a FBD. It's a loading diagram.
❏ Tips for technical drawings
❏ Tips for technical drawings
Overcome any perceptions you may have that you "can't draw" or that "engineers are good at math, not drawing." A positive attitude is immensely helpful. Take advantage of this opportunity to hone your graphical communication skills.
Draw the geometry to scale. Graph paper is ideal. If you don't have graph paper, make an effort to draw lengths and angles reasonably accurately.
Draw the FBDs about 2 or 3 times larger than what you think is necessary. That will make it easier to legibly add length dimensions and angular dimensions as you solve problems.
❏ Another loading diagram
❏ Another loading diagram
We will also solve this problem later this semester. You will have to draw FBDs that correspond.This is not a FBD. It's a loading diagram.
🟦 3.4 The Equations of Equilibrium (E.o.E)
🟦 3.4 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 focus on 3D problems in the latter part of the course.
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. Or - the net x-direction force is zero.
(2) The sum of the forces in the y-direction must equal zero. Or - the net y-direction force is zero.
(3) The sum of the moments about any z-axis must equal zero. Or - the net moment is zero.
These three conditions define the state of static equilibrium.
❏ The Equations of Equilibrium for 2D and 3D models
❏ The Equations of Equilibrium for 2D and 3D models
❏ Key points (for a 2D problem):
❏ 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.
a net force = translational motion
a net moment = rotational motion
a net force and a net 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, which will be the focus of Lesson 04 - Connections and Reactions.
🟦 3.5 Example problems for evaluating static equilibrium
🟦 3.5 Example problems for evaluating static equilibrium
❏ Scenario #1
❏ Scenario #1
Let's explore the idea of static equilibrium with an example. Let's take a small block of wood as an example.
❏ FBD of Scenario #1
❏ FBD of Scenario #1
Determine whether or not the body translates by checking for a net force in the x-direction and/or y-direction:
We also need to determine whether or not the body rotates. If a body is in static equilibrium, then the sum of the moments about any point equals zero.
Point E looks appealing: the four vectors are coincident with E and therefore do not cause a moment about E. Since the sum of the moments about E is equal to zero, the body does not 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 that the body does not rotate.
Remember: if the body is in static equilibrium, then you can sum moments about any point (axis) and calculate a net moment of zero.
❏ Scenario #2
❏ Scenario #2
In this scenario, the horizontal 4N force has a line of action that coincides the block's center (C).
❏ FBD of Scenario #2
❏ FBD of Scenario #2
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.
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. The net moment on the block is zero, so there is no rotation.
❏ Scenario #3
❏ 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.
❏ FBD of Scenario #3
❏ FBD of Scenario #3
The net force in the x-direction means that the block is in rightward translational motion.
Sum moments about any point to determine whether or not the block rotates. For convenience, we choose to we sum moments about the block's center (C):
We conclude that this block is both translating rightwards and rotating clockwise.
🟦 3.6 Example problem: how to solve for an unknown
🟦 3.6 Example problem: how to solve for an unknown
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 quickly determine by inspection that a 1# leftward force is needed).
This example is not here for the content. It's here to highlight the problem-solving process.
❏ A ruler lays flatwise on a table and forces are applied
❏ 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.
The purpose of this teaching tool is to specify the line of action of an unknown vector while discussing the direction or sense with students.
For instance, I might say: "can we deduce whether FC is leftwards or rightwards?"
❏ A double-headed arrow is a mental placeholder only (never draw it)
❏ A double-headed arrow is a mental placeholder only (never draw it)
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)
❏ 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 use this approach throughout the entire course.
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.
Important: Please adopt and emulate this type of reasoning throughout the course when solving unknowns.
❏ Closely emulate this calculation example
❏ Closely emulate this calculation example
🟦 3.7 Example: a collinear force system (a stack of books)
🟦 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 (their centers line up).
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
❏ Flipbook: A stack of books is supported by a table
Concept question #1: did the FBDs in the flipbook follow THE EIGHT RULES FOR CONSTRUCTING FBDs?
Yes. Remember: you can only draw a vector on a FBD when the body causing that force is not depicted.
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 collinear problems 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)
🟦 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 pretension them.
At each point where multiple cables connect, we can perform a concurrent force analysis.
We can use two E.o.E. to solve concurrent force problems:
the sum of forces in the x-direction equals zero
the 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
❏ 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?
Our strategy is to count the number of unknowns in the FBD and make sure we have enough equations of equilibrium 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 be better to use the symbol T for the tensions in the three cables. F is a generic symbol for force. T is more specialized. I need to update the image accordingly.
❏ Three cables pull on a connection ring
❏ Three cables pull on a connection ring
Note: while most people would solve this problem using the approach above, there is another way to do it. It's a graphical method. I plan on including that method in this textbook in a future update. For now, I'm placing a small image here to remind me to add this content later. Current students can ignore this small image.
🟦 3.9 Example: using moment equilibrium to solve for unknowns
🟦 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 properly:
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
❏ Flipbook: two springs arrest the tendency to rotate
🟦 3.10 Pulley vocabulary: sheaves, pins, and cables
🟦 3.10 Pulley vocabulary: 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.
Here is some useful vocabulary for pulley systems.
Load is applied into the system through the cable. The cable is taut, and if we neglect friction, simply changes the direction of the force P.
The cable bears down on the pulley sheave (or wheel). The sheave (pronounced shiv) rotates freely about the shaft of the nut-and-bolt connection. This connection is called a pin, and you'll learn more about pin connections in Lesson 04.
The pin bears on plates, and the entire pulley assembly is connected to a plane of fixity (likely reinforced concrete) by anchor bolts.
❏ Parts of a pulley
❏ Parts of a pulley
🟦 3.11 Qualitative equilibrium
🟦 3.11 Qualitative equilibrium
This section will be added soon. Current students: I'll go over this in class.
➜ Practice Problems
➜ Practice Problems
Scrambled answers (magnitudes, no units):
3.31 4.17 5.5 6 8.66 10 13 15 30 144 280 380 1333 1333
Problem 1.
Four books are stacked on a table.
Book 1 weighs 1#, book 2 weighs 2#, book 3 weighs 3#, and book 4 weighs 4#. Each book has a centroid (center of mass).
The contact points are labeled A, B, C, and D.
First, construct FBDs of each book. The FBDs should contain symbolic forces (e.g. N_A).
Then, use the E.o.E. to solve for N_B and N_D.
For a sign convention, do not use any negative signs. Write the magnitude with a (T) to indicate tension (a pull force) and (C) to indicate compression (a push force).
Note that I'm building this site using a web editor that doesn't have a easy way to do subscripts. Please interpret N_B as N with a subscript as B.
Problem 2.
Three bodies (you can think of them as a pile of books, if you like) are supported by two springs.
Body 1 weighs 1#, body 2 weighs 2#, and body 3 weighs 3#.
Spring A (S_A) and Spring B (S_B) are superglued to the books at Points A and B.
If the force in spring A is 0.5# of tension, solve for the force in spring B.
Be sure to start with a useful FBD, because you can only run an E.o.E. after constructing the FBD.
For a sign convention, do not use any negative signs. Write the magnitude with a (T) to indicate tension (a pull force) and (C) to indicate compression (a push force).
Problem 3.
Composite body (made of shapes 1, 2, and 3, perhaps a pile of books) weighs 6#.
Construct a FBD and solve for the unknown spring force (S_C).
The structure is symmetric.
For a sign convention, do not use any negative signs. Write the magnitude with a (T) to indicate tension (a pull force) and (C) to indicate compression (a push force).
Problem 4.
A sphere (weighing 350 N) is supported by a horizontal strap and (frictionless) surface d-e.
Construct a FBD by removing surface d-e and cutting through the strap at f-f. Express the unknowns symbolically.
Then, solve for the normal force at A (N_A) numerically.
For a sign convention, do not use any negative signs. Write the magnitude with a (T) to indicate tension (a pull force) and (C) to indicate compression (a push force).
Problem 5.
A sphere (W = 350 Newtons) is supported by (frictionless) surfaces f-g and g-h.
Construct a FBD of the sphere, using symbols.
Then solve for N_B and N_D numerically.
For a sign convention, do not use any negative signs. Write the magnitude with a (T) to indicate tension (a pull force) and (C) to indicate compression (a push force).
Problem 6.
Three solid cylinders are supported by symmetric (frictionless) surfaces a-a. Each cylinder weighs 10#. The structure is symmetric and the dashed gray lines are 30 degree increments in polar coordinates.
Because we're neglecting friction, the system is ONLY in static equilibrium if symmetric surfaces a-a are inclined at a specific angle.
Solve for that angle of inclination (alpha). Report your answer as a magnitude, in degrees.
Problem 7.
The same three cylinders are now placed inside a hollow box as shown.
Now you want to solve for the normal force at M and Q.
Again, think through different FBD options.
Construct the FBD and then report N_M and N_Q numerically.
For a sign convention, do not use any negative signs. Write the magnitude with a (T) to indicate tension (a pull force) and (C) to indicate compression (a push force).
Problem 8.
A box of negligible weight is supported by two springs, A and B. Two horizontal forces are applied as shown.
Construct a FBD of the box.
Then solve for the force transfer at A and B (S_A and S_B).
For a sign convention, do not use any negative signs. Write the magnitude with a (T) to indicate tension (a pull force) and (C) to indicate compression (a push force).
Since the problem statement says "negligible weight," we will omit self-weight from this analysis. This would be unusual in a Physics course, but it's a common assumption in Engineering problem-solving. If it bothers you, you can write W ≈ 0.
Problem 9.
A 5N hammer is suspended from Node B. The hand winch and electric winch are used to wrap cable length around the two drums until segment AB is horizontal, and segment BC has a rise of 5 over a run of 12.
First, construct a FBD of the metal connection ring at B.
Then, solve for the force in cable segment BC (T_BC).
For a sign convention, do not use any negative signs. Write the magnitude with a (T) to indicate tension (a pull force) and (C) to indicate compression (a push force).
It's assumed that the metal connection ring has a negligible weight compared to the other forces in the system, so we neglect it from our analysis.
Problem 10.
The same hammer is in a new configuration. Someone used the hand winch and electric winch to unwrap cable from the drum to create the geometry shown here.
Compared to the last problem, do you think that there will be more or less force in cable segment BC? Make an assumption to build your engineering intuition.
You know the drill! Construct a FBD and then solve for the force in segment BC (T_BC).
For a sign convention, do not use any negative signs. Write the magnitude with a (T) to indicate tension (a pull force) and (C) to indicate compression (a push force).
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