A free-body diagram (F.B.D.) is the physical representation of an object (or part of an object) with all actions acting on it. In statics, this setup usually includes forces such as concentrated and distributed loads, support reactions, internal forces, and self weight; check out Chapter 13 for more on these force categories. These four categories of forces help depict all forces on a given object (although in non-statics problems, you may include other vector actions such as velocities, accelerations, distances, and many others). Consider an 8,000-pound elephant and an 8,000-pound commercial vehicle. Both have the same self weight (force due to gravitational effects on their masses), but the location of that weight tremendously affects the objects they're resting on in drastically different ways.
In this chapter, I show you some of the situations you need to be mindful of when tackling any statics analysis problem with an F.B.D. I start with a basic list of items you want to be sure to include on your free-body diagram. I also show you how you can extract additional free-body diagrams from within a given F.B.D.
The old saying that a picture is worth a thousand words may seem clichéd at first, but this simple statement accurately emphasizes the importance of a well-constructed graphical representation for statics analysis. In fact, without this pictorial description, you quickly discover that accurate engineering and physics solutions are next to impossible to achieve. In fact, even a simple photograph can sometimes serve as a basis for constructing a free-body diagram.
Tip
Think of an F.B.D. as being a snapshot of the support reactions and forces acting on an object at a particular instance.
To construct an F.B.D. in Cartesian components by starting with a basic picture, just follow these steps:
Sketch the complete object that you're going to be studying, such as a chair or a ladder.
In reality, the actual shape of the object really doesn't matter. However, knowing the exact locations (points of application) of every force and moment vector on that object is crucial. Actually drawing the object can give you a reference for measuring those locations.
Draw all external support reactions (physical movement restraints) as well as any springs or cables that are attached to the system.
Support reactions include pinned supports, ball-and-sockets, slider assemblies, and so on; I discuss them in more detail in Chapter 13, where I also show you how to draw basic two- and three-dimensional support reactions. Springs and cables are generally fairly easy to identify and can provide a quick confidence boost in the construction process; see Chapter 9 for more on these items.
Include complex supports, such as ramps and inclined supports.
Examples of these additional restraints include refrigerators on ramps and wedges under crates, along with any information about inclined support surfaces (usually expressed in degrees). Flip to Chapter 13 for more on inclined supports. In fact, if your object is resting on a ramp or incline, be sure to include the angle of the incline as a reference, even if you're focused on just the object.
Draw each external force acting on the systems, as described in Chapter 13.
Assume that all objects in the F.B.D. are rigid (meaning they aren't deformed by the force), even if the system has springs; springs aren't actually rigid, but you can consider a spring rigid at the exact instance that you draw the free-body diagram.
If your F.B.D. is of an isolated object that you're separating from a larger system or support, you also need to include contact forces, which are forces that arise when one object pushes on either another object or a support surface. However, you don't include contact forces on a free-body diagram when both objects (or supports) on either side of the contact surface are already included in the diagram.
When you're drawing an F.B.D., getting the forces onto the diagram is more important than worrying about their sign or the direction. As you discover in Chapter 16, until you start writing equations, you may not actually know the magnitude and sense of the vector for the unknown magnitudes at the time you're drawing the free-body diagram. If you don't know the magnitude (length) and sense (direction) of a vector when you draw it, simply guess a direction, include a label on the force, and apply it at its proper point of application along its appropriate line of action.
For example, say you want to apply the support reactions to a diagram of Figure 14-1a, which illustrates a man standing at the end of a diving board. Based on the sketch, you may be inclined to draw the support reactions as positive with respect to your Cartesian coordinate system as shown in Figure 14-1b, with all vertical loads acting upward and all horizontal loads acting to the left. And while being consistent is good practice for now, you may soon realize that it's not always correct. In fact, Figure 14-1c actually represents the correct F.B.D. of Figure 14-1a. Notice that the difference between Figure 14-1b and c is only in the direction of the vertical reaction Ay located at Support A — upward in b but downward in c.
Note
You always know the direction of a load resulting from self weight -- it's acting downward and applied at the center of mass, so be sure to go ahead and draw it in the correct direction.
In Part II, I emphasize the importance of being able to quickly create Cartesian vectors, which comes in handy when you realize that you often repeat the vector creation process many times on the same free-body diagram. To make your work easier later, get in the habit of including several additional pieces of information, including the following:
Coordinate axes: The coordinate axes are a useful reminder of any assumptions you've made about positive and negative directions. Including information regarding the positive directions of each of the x-, y-, and z-directions is especially useful because when you start writing the equations of equilibrium (which I introduce in Chapter 16), these directions can help you determine the sense of any unknown vectors.
Note
Although you typically take the Cartesian x-axis as being horizontal (and consequently the Cartesian y-axis as vertical), sometimes aligning your axis in some other direction is more convenient. You encounter this situation often when working with problems on ramps and inclined surfaces, or in problems that involve forces acting at some non- Cartesian orientation (or at some angle other than horizontal or vertical).
Origin: The origin is the location where the Cartesian axes for your system intersect; it serves as a handy reference location and customarily has the coordinates (0,0,0).
Labels for points of interest: You also should label some key points of interest on your diagram such as the following:
All support and internal hinge locations (see Chapter 13)
All locations of applied external concentrated loads (see Chapter 13)
All locations where forces' lines of action cross the object or other lines of action (see Chapter 4)
All resultant force points of application (see Chapter 7)
All locations at the start and end of the free-body diagram (see Chapter 10)
All changes in the geometry of the object
All centers of gravity and centers of mass (see Chapter 11)
Force vector components: You can also use dotted arrows to indicate perpendicular components and their senses with respect to your assumed Cartesian axis direction. Calculate the components of the force vectors and sketch them as individual forces on the free-body diagram.
Dimensions and angles of supports and forces: You also want to include dimensions that relate all of the locations discussed in the preceding bullets. Be sure to also include information such as angles or proportion triangles that may help describe the orientation of any lines of action for forces on your free-body diagram. (Chapter 5 gives you the skinny on proportion triangles.) You'll also need to include the angles of any inclined supports (see Chapter 13).
Tip
If you provide adequate linear dimensions, angular dimensions may not be necessary. A little trigonometry can be used to compute angles on free-body diagrams.
Isolation boxes let you take larger objects and zoom in on specific features. An isolation box tells you when an internal force needs to be included on a free-body diagram — whenever an isolation box crosses a physical object, you must include an internal force on the object.
Several features that you may want to explore in further detail on a given free-body diagram include support reactions and internal forces in members. The following sections show you how you can use an isolation box to extract smaller (and sometimes more manageable) pieces of a free-body diagram while preserving the overall behavior of both the larger system and the isolated portion.
Depending how you cut the picture when you draw a free-body diagram, you can greatly reduce or increase the complexity of the system. For example, Figure 14-2a shows a system of two cables oriented at different directions connected to a ring that's suspending a 100-kilogram box.
According to the procedure outlined in "Getting Your F.B.D. Started" earlier in this chapter, the first diagram you should draw is the free-body diagram of the entire system, which I show in Figure 14-2b. The system has two support reactions each at location A and location B generated by the pinned supports. The only load applied to the system is the self weight of the box, which has a force of 981 Newton (100 kilograms · 9.81 meters per second squared = 981 Newton).
A total of five forces are acting on the system in Figure 14-2a, four of which are unknown at this time. You won't actually calculate the unknown forces at this time (I explain how to do that beginning in Chapter 16), but I do show you the free-body diagrams you need to create as part of this process.
In this system, suppose you're interested in calculating the internal forces in each of the rope sections. In order to see these internal forces, you need to create isolation boxes that cut these objects and expose these forces. Because all the connecting objects are cables, the internal forces are rather simple and include only axial forces. The sense of the internal cable forces is determined from the knowledge that cables are only capable of transmitting tension (or pulling on the free-body diagram). I discuss more about cable requirements in Chapter 22.
The main free-body diagram of interest in Figure 14-2a is at the ring at location C, where all three cables meet. To draw the free-body diagram of just the ring at C, you use an isolation box around the ring.
To construct an isolation box, stick to the following basic steps:
Identify the object or detail of interest.
Decide what part of a structure you need to examine in further detail. In this example, you're interested in what's happening at the ring at location C.
Construct a closed polygon around the object or detail of interest.
Draw a box, circle, or some closed polygon shape — usually with dashed lines — to differentiate between the isolation box and the physical object itself (refer to Figure 14-2a). Make sure that the isolation box is a closed polygon. Labeling the box is handy if you're going to be using multiple isolation boxes.
Extract the contents of the isolation box.
Copy everything inside the isolation box to a new picture. Include external forces, supports, self weight, and all relevant dimensions and Cartesian coordinate system data.
Include all internal forces revealed as a result of the cutting process.
The chapters of Part VI deal with structures with different types of internal forces. Depending on the type of system you are working with, different internal forces may appear when you cut an object. For this example, you're dealing with a system of axially loaded cables, which I introduce in Chapter 9.
Look at the free-body diagram again. At any location where the isolation box crosses a physical object, internal forces are revealed and must be included. In this example, the isolation box for the ring at location C cuts three different cables, so you must apply a concentrated axial force at each of these locations.
The isolation box for the ring at location C cuts each of the three cables, exposing their internal forces on the free-body diagram. Figure 14-3 illustrates what a properly constructed free-body diagram for the ring at C looks like. Notice how even though you don't know the magnitude of the internal force, you do know the sense of the force because cable forces always act in line with the cable and they're always in tension (or pulling on the object).
Another object of interest in this example may be the box itself. Because you already know the mass of the box, you know that there's a self weight force present on this system. Sketching an isolation polygon around the box requires that you also cut cable CD. As before, this move results in an unknown internal cable force pulling on the box at D as shown in Figure 14-4.
Take a look at the force vector FCD in Figures 14-3 and 14-4. Notice that the internal force in the cable CD is oriented in one direction in 14-3 and the opposite in 14-4. Both represent tension in the cable because they're pulling on their respective free-body diagrams. However, the sense of the force changes depending on which isolation box you're working with. In this case, that's perfectly acceptable. Just make sure that internal forces on opposite sides of a cut line (or the location where an isolation box crosses the physical object) are equal and opposite in magnitude, sense, and direction.
You can also use an isolation box to capture unknown support reactions and relate them to internal forces. On this diagram, you include the two unknown support reactions Ax and Ay and the unknown revealed cable force FAC with a 30-degree dimension indicating the orientation of the cable (and as a result, its internal force) as shown in Figure 14-5.
The art of using isolation boxes takes some trial and error to get used to, but after you get the basics down, they provide the framework for quickly moving through a system with calculation and equation techniques that I discuss in later chapters.
One of the potential problems of isolation boxes is the increase in the number of unknown reactions or internal forces that you must include on the free-body diagram. For example, consider the simple truss structure shown in Figure 14-6a, which already indicates the support reactions. If you use an isolation box to cut only members CD, CG, and HD as shown in Figure 14-6b, notice how the complexity of the free-body diagram greatly increases. You've cut each member at two different places, exposing six additional internal forces as shown in Figure 14-6b — one on each side of each member within the isolation box.
The free-body diagram of the remaining portion of the structure also has these same six revealed internal forces (as shown in Figure 14-6c). However, each pair of forces is acting along the same line of action, with one of the forces acting in one direction and the other acting in the opposite direction for each member. Vector addition tells you that the net effect on that member is zero (for example, FCD – FCD = 0).
Your isolation box hasn't really yielded a whole lot of new information. What you want to do is cut the objects in your free-body diagram in a way such that only one of any unknown internal force is present on any free-body diagram. The simplest way to accomplish this feat is to require cutting all the way through the object or system. Figure 14-7 illustrates a better way of cutting this structure, resulting in only one of each unknown internal force on the system.
In this example, notice that you only have four unknown forces to deal with, whereas before you had six — all of which cancelled each other because they were acting in opposite directions with the same magnitude along the same line of action.
By cutting the system entirely into two pieces with your isolation box, you actually produce two free-body diagrams with the same unknown internal member forces. This result is useful because when you start writing equilibrium equations (which I show in Chapter 16), you can use either of the free-body diagrams. Sometimes, one free-body diagram is significantly simpler, meaning it has fewer loads, fewer reactions, or more-convenient geometry.
When multiple objects interact with each other, forces inherently exist between the objects. The chair you're sitting on applies a force to the carpeted floor below (if you don't have carpet, pretend you do); you can see the effect when you move the chair and see the impression of the chair legs that remains behind. Likewise, the carpeted floor itself exerts a force on the legs of the chair, preventing it from falling through the floor into your basement.
In systems with multiple objects connected, such as with a clamp or pair of pliers, looking at each of the individual pieces is useful. When exploding an object (separating connected pieces), your isolation box typically has to cut through a pin, bolt, or some other sort of connecting element. Cutting through pins or connections is one of the tricks I explore more in detail in Chapter 21.
In problems where one object is resting on top of another, such as the block resting on a ramp as shown in Figure 14-8a, the forces interacting between the two objects are what tell the story.
The only forces acting on the system as drawn are the weight of the block itself and the external applied load P, which is 200 Newton. So if the block is sitting on an incline and an applied load is pushing the block down the incline, what's holding the block in place? The answer is friction, which is an invisible force that exists between two objects as they attempt to move past each other. (Don't worry; I discuss more about friction in Chapter 24).
When you draw two isolation boxes, the internal balancing forces actually show up. Figure 14-8b shows the free-body diagram of just the block. In this picture, if you draw only the 200-Newton applied force and the 150-Newton self weight, you see that the block would clearly want to move downward and to the left as drawn.
However, this scenario can't physically happen. Clearly, the ramp has to be pushing back on the block; otherwise, the block would fall through the ramp, which is physically impossible. Think of the support reaction of the block as a quasi-roller support (sort of like the roller supports I discuss in Chapter 13). To counter the downward forces, a vertical supporting force also has to be present to maintain the block in its original position. A portion of this resistance is due to the normal force N.
To prevent the block from sliding to the left, a second force has to be introduced that has a component pushing to the right. The only way for this situation to occur is for a force vector to be developed that's acting up the ramp. To balance these two forces that are acting on the block at the interface of the block and ramp, a second set of forces equal in magnitude but opposite in sense have to be acting on the ramp as shown in Figure 14-8c.