As you move forward with your modeling of the world around you, which is a major step in the application of statics, one of the biggest obstacles you need to overcome is determining how to apply those pesky forces to your objects. Forces come from a wide range of sources and occur both internally and externally on the object, so you want to understand several basic rules for how these forces are applied.
In this chapter, I explain the differences between internal and external forces, explain more about where these forces come from, and show you several common scenarios in which you encounter them. I also show you some simple steps for working with forces in ropes, cables, and springs, as well as how to handle gravitational forces on objects (or self weight). Finally, I introduce the principle of transmissibility, which allows you to move forces to different locations on an object and proves very useful in statics. This chapter serves as a stepping-stone into crafting the pictures you use when applying statics to the world around you (which I cover in Part IV).
You can typically separate forces into two basic categories: internal and external.
External forces: External forces are created by an action external to the object itself. Examples of external forces include the force created by a ball striking a wall or the weight of snow resting on the roof of your house. External forces are often the easiest to calculate because they're usually caused by a visible phenomenon and can be measured.
Internal forces: Internal forces are developed inside the object as a response to the applied external loads (or external forces) and system restraints. Tension in a cable, compression in a column, and torque in a drive shaft are all examples of internal forces. One of the major difficulties in working with internal forces is determining the type and direction of each; I show you some of the techniques for identifying internal forces beginning in Chapter 20.
Because internal and external forces are both force vectors, you need to include all the usual information such as magnitude (the numerical value of the vector), sense (direction of the vector's action), and point of application (physical location where the vector is acting), as I discuss in Chapter 4. In addition to distinguishing between internal and external forces, you have to make an assumption about the type of system you're dealing with. Rigid bodies are objects that don't experience deformation when the vector acts upon them. If internal forces act on a deformable body, however, the object undergoes a physical change or deformation, which can alter both your internal and external load calculations or change any relevant geometric dimensions.
Note
In reality, all objects are deformable, but for the simple purpose of understanding the basic fundamentals of statics analysis, here I assume my objects are always rigid bodies.
A concentrated force is a force that acts at a point, or more realistically, on a very small area. Drawing an external concentrated load requires a single arrowhead, as shown in Figure 9-1. Remember that you usually place the vector's tail at the point of application on the object.
In Figure 9-1a, a man is pulling on a rope with a tension of 100 Newton; this force causes the rope to tighten. In this case, you place the single-headed arrow such that the tail of the force is applied where the man's hands attach to the rope.
Conversely, in Figure 9-1b, a 250-pound man is sitting on a chair. Because the force of the man is pushing down on the chair, the single-headed arrow's head is applied at the point of application.
In reality, your calculations produce the same equations and solutions whether the head or the tail of the vector is applied at the point of application as long as the sense, magnitude, and line of action are all the same.
The following sections introduce several types of concentrated forces.
In statics, you idealize most contact forces as being concentrated forces. Normal contact forces are forces that develop from one object pushing on another. They can be caused by self weight, applied external load, or any other force. To illustrate this idea, consider the block sitting on the ground in Figure 9-2a.
To see the contact forces, you need to separate the block from the ground in the picture. In Figure 9-2b, the force from the block pushes down on the ground. Conversely, if you examine the block by itself (as in Figure 9-2c), you see that in order to keep the block from falling through the ground, a second contact force is required to hold the block in its position. This reaction force from the ground onto the block is balanced by the contact force from the block onto the ground.
Contact forces for both the ground and the block are always perpendicular to the contact surface. For this reason, you sometimes see contact forces referred to as normal forces because they're normal or perpendicular to the surface.
Note
The point of application of contact forces is always somewhere along the contact surface, but it's exact location depends on the type of problem — you want to pay special attention to this detail when you encounter friction problems (such as the ones I cover in Chapter 24) and rotation problems (such as those in Chapter 12).
Friction is a type of external force that develops from resistance as one object tries to slide past another. From experience, you know that if you go up to an object such as a refrigerator and push on it with a very small force, it doesn't move. However, if you continue to push and eventually push hard enough, the refrigerator starts to slide. It's those hidden friction forces that initially prevent the refrigerator from sliding. For a visual of this concept, consider the block in Figure 9-3a, which is subjected to a force that causes it to begin sliding along the ground.
Just like for normal contact forces, you can separate the block from the ground and expose the internal forces. The sense of the friction force on the object that wants to move goes against the direction of the motion of the object (as shown in Figure 9-3b). The friction force on the opposite side of the contact surface (on the ground in Figure 9-3c) is generally in the same direction as the motion of the moving object. However, both of these force vectors are applied parallel to the contact surface.
If you don't quite fully understand friction at this point, don't worry! I explain it in a whole lot more detail in Chapter 24.
A few concentrated loads that you want to remember are moments and support reactions. Both of these concepts play significant roles in your application of statics.
Moments: Moments are physical effects that cause an object to want to rotate. In Chapter 12, I show you how to draw and calculate a moment.
Support reactions: Support reactions are external conditions that restrain an object from moving and/or rotating in a specified direction. In Chapter 13, I show you how to draw and represent support reactions.
Other unique behaviors can result in concentrated loads, but these cases are special. I address those as I come across them throughout the book.
Internal loads are created through the application of external forces and effects and are always present on rigid bodies. You only see them when an object is physically cut. You can classify internal forces into three major categories: internal axial loads, internal shear forces, and internal moments.
Internal axial loads: Internal axial loads are the simplest of the internal forces and are present in many objects, such as ropes and cables, simple columns, and springs. Internal axial loads are concentrated forces that act parallel to the longitudinal axis of the object. The senses of internal axial loads are always referred to as either tension (getting longer) or compression (getting shorter).
Internal shear loads: Internal shear loads are concentrated forces that act perpendicular to the longitudinal axis of the object.
Internal moments: Internal moments are actions that cause the object to rotate around a point or an axis and always appear when internal shear loads are developed.
Don't worry too much about shear and moments at this point; I discuss those in much more detail in Chapter 20. In this section, I focus on internal axial loads.
Figure 9-4a shows a bar subjected to tensile forces (P) applied on each end. This bar is perfectly balanced by these two forces acting in equal but opposite directions. If you cut the bar into two pieces and look at the lower portion (shown Figure 9-4b), you see that a new force, (PINT) is developed inside the bar in order to help hold that portion of the bar in its original position. In order to hold the bar in place, PINT must equal the applied load P. Likewise, if you look at the upper portion of the bar (shown in Figure 9-4c) by itself, another internal force is created to keep this part balanced as well. This internal force PINT is also equal to the applied load P.
If you examine the direction of the same force PINT on the upper and lower portions, you see that each of the internal forces are of the same magnitude P but have different directions depending on which side of the cut line you're working with (see Figure 9-4d). In Chapter 13, I show you how to better represent these internal forces, and I show you how to start writing equations and working with them beginning in Part VI.
Forces in ropes and cables are always in an axial direction, along the line of the rope or cable, and are always in tension. These forces act along the line of action of the rope or cable. That is, the position vector of the rope is in the same direction as the unit vector of the force in the cable. For these types of objects, you often make use of dimensional data and coordinates.
In Figure 9-5, a force of magnitude 150 pounds is applied to the end of the rope. Because you know the location on the wall is a height of 30 feet, the Cartesian coordinates of the knot on the wall are (0,30,0). (Flip to Chapter 5 for more on Cartesian coordinates.) The end of the rope is located at coordinates of (10,10,10).
Tip
In Chapter 5, I also mention that you can use a unit vector to define the direction of a line of action. In the case of these rope problems, you always know the line of action because the orientation of the rope itself is on that same line of action.
To write a force vector for the force on the rope, follow these steps:
Create a position vector between the ends of the rope.
Define the end of the rope at the wall as Point 1 and the free end of the rope as Point 2. You can calculate the position vector that defines the rope from Point 1 to Point 2 from the basic position vector equation:
Point 2 is the end point of the position vector (the head) and Point 1 is the start point of the position vector (the tail); substitute the coordinate values from this example to produce a position vector for the ends of the rope:
Calculate the magnitude of the position vector from Step 1.
To calculate the magnitude of a three-dimensional vector, you utilize the Pythagorean theorem, as I discuss in Chapter 5.
Use the position vector from Step 1 and the magnitude from Step 2 to calculate the unit vector that describes the line of action of the rope.
After you have the position vector and the magnitude (or length of the rope), you can create a unit vector by dividing the vector itself by its magnitude (see Chapter 5).
The following quick check verifies that this result is indeed a unit vector:
Note
u12 as created from the dimensional data of ropes and cables always defines the same line of action as the line of action of the force, or
Create the force vector by plugging the information gathered into the force vector formula.
Here's what your example looks like in the formula:
or
which defines the force acting on the end of the rope at (10,10,10) as a Cartesian vector.
This vector notation also includes additional useful information. For example, based on your final calculation, you now know that the rope is experiencing a force of +61.2 pounds parallel to the positive x-direction, −122.4 pounds parallel to the positive y-direction (or +122.4 pounds in the negative y-direction), and +61.2 pounds parallel to the positive z-direction. These are the Cartesian components of the force vector F (which I discuss in more detail in Chapter 8).
Verify that your force vector is correct by verifying its magnitude.
To verify the magnitude, you just use the Pythagorean theorem a second time and plug in the component values for the x-, y-, and z-components.
Notice that the exact value of the magnitude is only approximately equal to 150 pounds because the scalar force components were only taken to one decimal place.
Another type of axial force that you encounter is a mechanical spring object. When you think of a spring, you probably picture that wonderful toy you owned as a child and the countless hours you spent flopping it down your stairs (or trying to straighten it out). In statics, springs are a bit different. They may resemble toy springs, but they actually behave quite differently. Figure 9-6a illustrates a common depiction of a spring object.
Several features are very important when discussing springs, including line of action, stretch, and spring constants:
Line of action: As with the depiction of force vectors (see Chapter 4), the line of action of a spring refers to the direction of the longitudinal axis. The two points that connect the spring to other objects or supports always occur along the line of action of the spring. As a result, the internal axial force of the spring is always along this line.
Stretch: The stretch refers to how much the spring is displaced from its original unstretched length. Stretching a spring to make it longer is tension or elongation (shown in Figure 9-6b), whereas stretching it to make it shorter is called compression (Figure 9-6c). The more you stretch a spring, the more force is developed in the spring.
Spring constants: The spring constant is a measure of how much force you need to compress or elongate a spring from its unstretched position. Factors that affect spring constants include the cross-sectional area of the spring, the unstretched length of the spring, and the material the spring is made of.
The following sections look at stretch and spring constants; for more on line of action, check out Chapter 4.
The force in a spring is directly related to the amount and direction of the stretch (or deformation) of the spring. A negative stretch indicates a compressive force; a positive stretch denotes an elongating force. You can compute the stretch (or spring deformation) from
Spring constants define how stiff a spring actually is and are a measure of how much force is necessary to stretch (or compress) a spring a distance of one unit. In SI units, the unit of the spring constant is usually Newton per meter (N/m), and in U.S. customary units, it's pounds per inch (lb/in) or pounds per foot (lb/ft). For example, Figure 9-7 depicts a spring with an unstretched length of 6.5 inches and a spring constant of 3,000 pounds per inch. A force applied to the spring compresses the spring to a final length of 5.25 inches.
The equation that represents the magnitude of the internal force in a spring is given by
where k represents the spring constant parameter and δx represents the spring's deformation from its unstretched state. You can compute the force in the example spring by plugging in the numbers:
The negative sign on the force indicates that the force on Figure 9-7's spring is a compressive force. This finding supports the idea that a shorter spring is a compressed spring.
Note
If you want to determine the direction of the force, you can create a unit vector for the line of action as I demonstrate in the "Forces in ropes and cables" section earlier in this chapter.
For many people, self weight is an ominous concept, but don't worry — you're not climbing on any scales here. In statics, the self weight of an object is a measure of the force created by gravity's effects on the object's mass.
Gravity: Gravity is a force of attraction between masses or particles. On Earth, at sea level, the average gravitational acceleration constant is taken as 9.81 meters per second squared (m/s2, SI units) or 32.2 feet per second squared (ft/s2, U.S. customary units). These values actually vary slightly by location because the Earth isn't a perfectly round sphere. However, this variation is very minor, so I use these average values throughout the book.
Mass: Mass is a measure of the number of atoms in an object in conjunction with the density of each of those atoms; the mass of an object is a fundamental property of that object. Mass is measured in kilograms (kg) in SI units and in slugs (lb-s2/ft) in U.S. customary units.
Self weight is usually either treated as a single concentrated value, which I discuss in this section, or spread over a continuous region, which I discuss in Chapter 10. Regardless, the basic formula needed to calculate the self weight, W of an object is given by
where m represents the mass of the object and g represents the acceleration due to gravity. In this form, the weight W of an object is a vector force in the direction of the gravitational acceleration g. On Earth, you always assume that gravity is acting locally downwards. (Of course, if you're standing on your head, your sense of up and down may be a bit skewed.) Gravity always acts towards the center of the Earth. I discuss more about the location of application of this force in Chapter 11.
When working with self weight, you may sometimes encounter problems that don't directly state the mass or the weight of an object. However, you may be able to use other terms, such as the following, to calculate those figures:
Density: The density of an object is a measure of the amount of mass of an object contained within a certain volume of that object. For example, dropping a bowling ball on your foot is a lot different than dropping an air-filled balloon of the same size on your foot, and that's because of the effects of density. Temperature is a major factor in the density of an object. The density of an object is measured in units of kilograms per cubic meter (kg/m3) for SI units, in lb-s2/(ft-ft3) (I won't bore you with the spelled-out version of that one) in U.S. customary units. Density is usually expressed by the Greek variable ρ, or rho.
Specific gravity: Specific gravity (sg) is the density of the material relative to the density of water, which has a specific gravity of 1.0. One useful feature of specific gravity is that the value is unitless and remains the same regardless of which measurement system you use. So a specific gravity of 1.0 in the metric system is a specific gravity of 1.0 in the U.S. customary system.
Note
For example, carbon steel has a specific gravity of approximately 7.8, meaning that it's nearly eight times as heavy as the same volume of water. Ice, on the other hand, has a specific gravity of approximately 0.92 (which means that it's lighter than water, and one of the reasons that it floats!).
Note
If the problem gives you the weight of the object, you must include it in your calculations. If the mass, density, or specific gravity is provided, you can calculate the weight from the formulas presented in this section. In most structural problems, the applied forces on an object are often much, much greater in magnitude than the actual self weight of the object. So, if you choose to neglect the self weight when the applied loads are large, you'll probably be okay. However, keep in mind that if you ignore the self weight of an object weighing 200 pounds when the applied loads are only 5 pounds, your results will end up being highly inaccurate.
Specific weight (γ) is the weight per unit volume and is a relationship that relates density (mass per unit volume) of a material with gravitational effects. The formula is:
where γ is the specific weight, measured in Newton per cubic meter or pounds per cubic foot in SI and U.S. customary units, respectively.
Tip
Mass is generally distributed throughout each particle of an object. However, instead of calculating each weight for each particle (and you may have a lot of particles) in an object, you can simplify your work with mass and weight by making an assumption about the location at which the weight or mass is acting. Compute the grand total of all the particle masses and then express this value as a single value, or lumped mass. You can compute the lumped mass of an object, m, from the following relationship:
where ρ is the density of the object (as defined earlier in this chapter), and V"represents the volume of the object measured in cubic meters or cubic feet in SI and U.S. customary units, respectively. So, if you know the density of a cube of water is approximately 1,000 kilograms per cubic meter at 60 degrees Fahrenheit and that the cube has a volume of 0.35 cubic meters, you can calculate the mass of that water cube as
This method is valid for prismatic objects, or simple objects having constant dimensions in each direction.
The principle of transmissibility implies that a force vector acting on a rigid body results in the same behavior regardless of the point of application of the force vector, as long as the force vector is applied along the same line of action. It's a concept that's very important within vector mechanics. In fact, it forms the basis for one of the three major assumptions that Isaac Newton proposed with respect to rigid body mechanics — it doesn't affect the calculations for equations of equilibrium (covered in Chapter 16) because ultimately, the direction, point of application, and magnitude of the applied vector are still the same. Figure 9-8 shows a graphic representation of the principle of transmissibility for rigid bodies.
Notice that for the rigid body shown, a force P1 acting to the right and another force P2 acting to the left result in the same net behavior as long as the forces maintain their original magnitude and sense, and act on the object along the same line of action. That is, the point of application of forces can occur anywhere along the same line of action on the object without changing the resulting behavior of the object.