When you think of friction, perhaps you think of the heat that's generated when you rub your hands together on a cold morning, or the scrapes and burns on your knees when you trip and fall over your dog on the way to the refrigerator in the middle of the night. Although these examples can be correctly called friction, in statics it has a slightly different meaning.
As you read through this chapter, you may realize that its contents are significantly different from the other chapters in Part VI of this text. In the other chapters of this part, I neglect the effects of friction on the problems in order to better explain the various techniques used to solve statics problems. However, in the real world, friction is an ever-present force that you must account for in all your calculations, so this chapter shows you how to do just that.
In this chapter, I describe the different types of friction and how to calculate their magnitudes, and then incorporate these values on a free-body diagram (F.B.D. — see Chapter 14). I also explore how the different objects in statics are affected by the various forms of friction.
In the old-time Western movies, a popular image shows a bartender sliding a beverage along the bar to a waiting customer at the other end. Somehow, magically, the bartender is able to hurl it down the bar and somehow make it stop exactly in front of his waiting patron. According to Newton's first law (which I discuss in Chapter 16), if the mug is moving down the bar, it continues in the same direction until an outside force acts on it. In this movie, that outside force is friction and it's what the bartender is counting on to save his beverages.
You can often blame friction for all sorts of funky behaviors. Stationary objects that refuse to move and tall objects that topple rather than slide are all victims of friction in one way or another.
Friction is caused by several factors that occur on the surface of all materials. One factor that affects friction is the combination of microscopic imperfections on the interfaces of all materials that rub and interlock with each other as one object slides past another. The rougher the surface, the more frictional resistance that can occur. Another factor that is present in friction is the adhesion (or stickiness) between materials. Adhesion is affected not only by the materials in contact but also the presence of lubricants (or lack thereof) on the contact surface.
Other factors that can affect frictional resistance include applied normal force, characteristics of the contact surface, and length/area of the contact surface.
Applied normal force: An applied normal force is the force acting perpendicular to the contact surface.
Characteristics of the contact surface: The frictional characteristics of a surface are measured by a numeric constant called the coefficient of friction (which is simply a ratio of force required to move an object to the contact forces between the friction surfaces).
Length/area of contact surface: The more contact length between two objects, the more interlocking and adhesion that can be developed.
Friction caused by two objects rubbing past each other is also known as dry friction (sometimes referred to as the Coulomb friction force) and is given by the inequality F ≤ μsN = FMAX, where F is the resisting friction force and FMAX is the force that must be overcome before an object can move. μs is the coefficient of static friction. N is the contact force (or the normal force) perpendicular to the interface between the two objects.
In mechanics, you actually run into two types of friction:
Because all the problems addressed in statics are typically not moving, in this book I focus exclusively on problems involving static friction. Table 24-1 shows approximate values for several common coefficients of static friction assuming a dry and smooth surface condition for each material.
Table 24.1. Common Approximate Coefficients of Static Friction
Material | Coefficient |
|---|---|
Steel on steel | 0.8 |
Glass on glass | 0.9 |
Teflon on Teflon | 0.05 |
Aluminum on steel | 0.6 |
Tip
You can greatly reduce the coefficient of static friction between surfaces by lubricating them with oil, grease, or water.
When you calculate the size of the static friction force FMAX (which happens to also be its magnitude) by using the coefficients of static friction and you determine that the force's location (or point of application) must be along the interface between the objects, you've fulfilled two of three requirements (see Chapter 4) for defining a force vector. But what about the sense (or direction) of the friction force vector?
Anytime you try to push a heavy object across the floor and it doesn't budge initially, you might move to another side and push from another direction. Regardless of which side of the object you push from, the fact that it doesn't initially move illustrates that the friction force is always fighting against you, or more specifically against the direction of intended motion, also called impending motion. This concept is the key to working with friction problems.
Impending motion refers to the direction that the object wants to move (or rotate) after it has overcome the resisting friction forces. Consider the block shown in Figure 24-1, which is subjected to a single horizontal load. From Newton's laws, you know that an object wants to move in the direction of a force applied to it.
After you've identified the direction of impending motion on an object, you're then ready to begin constructing the appropriate free-body diagrams and then you can start writing the equations of equilibrium. The following sections give you the lowdown on completing these tasks.
An F.B.D. of a problem including friction has all of the same information as a problem without friction. The major difference now is that you are also including the friction forces on the same diagram.
Consider the block shown in Figure 24-2, which is subjected to a horizontal force P applied to the edge of a 1,000-pound crate. The coefficient of static friction is given as µs = 0.3. The F.B.D. is shown.
By looking at the applied force, logic tells you that the impending motion is in the direction of the applied load, or to the right. Because the object isn't currently moving (after all, the motion is impending, right?), you can apply the equations of static equilibrium:
The first equation gives you the value of the normal force on the contact interface, N = 1,000 pounds. You use this normal reaction force to determine the friction limit FMAX in the following section. You want to apply the other equilibrium equation (in the x-direction) to find the actual friction force acting on the interface due to the applied force P:
This equilibrium equation tell you that if the applied load is 500 pounds, the resisting friction force must also be 500 pounds to ensure equilibrium.
To determine if the object moves due to the 500 pound load, you need to examine whether that load is more than the friction limit FMAX:
Note
If the friction force F at the interface is less than the friction limit FMAX (which is the limiting force before motion occurs), the object doesn't move. If the friction force exceeds the friction limit, there is enough force on the object to overcome friction and cause it to start moving in the direction of impending motion.
Plugging the numbers for Figure 24-2 earlier in the chapter into the formula, you get FMAX = (0.3)(1,000 lb) = 300 lb.
This result indicates that in order for the object to slide, the friction force F must be greater than the friction limit FMAX = 300 pounds. Thus:
Thus, in this example, you've determined that you have sufficient applied force to overcome the friction limit, and the crate therefore moves.
Sometimes, contact points on an object can move in different directions. An example of this situation is shown in Figure 24-3a, which presents a 150-Newton plank 4 meters long resting on the ground at Point A and against the wall at Point B. The coefficient of static friction at Points A and B is µ = 0.3.
To draw the complete F.B.D. for this plank, you must account for friction and normal forces at every contact point.
The end of the plank at Point A wants to move to the left, and the top of the plank at Point B wants to slide down the wall. Thus, FA, the friction at Point A, must be to the right to oppose the impending motion at Point A (see Figure 24-3b).
At the same time, Point B wants to move downward due to the weight of the plank, and FB, the friction force at Point B, must be acting upward.
In addition to the friction forces, a normal force NA occurs at Point A, and another, NB, occurs at Point B.
In total, you have four unknowns acting on the F.B.D. of this plank. As it's currently drawn, this F.B.D. (see Figure 24-3c) is statically indeterminate to the first degree, meaning that there's one more unknown than there are available equilibrium equations.
Note
For a two-dimensional equilibrium problem, you have only three equilibrium equations to work with. Fortunately, by taking advantage of the relationship between the friction limit and the normal contact force, you can both simplify the F.B.D. and provide extra equations in addition to equilibrium equations. I"show you how to do this in the following section.
At every point where a friction force is applied, you always find a normal force applied at the same location. And for an object to move, the friction force must be greater than the friction limit: F > FMAX = µsN.
Note
The friction limit is also a function of the normal force at that same location.
Together, the friction limit and the normal force form rectangular components of a single resultant. You can then calculate the magnitude of the resultant of these two forces as
which means that you can compute the resultant by knowing only the normal force and the coefficient of static friction (which is already a known quantity), as shown in Figure 24-4a. Employing the head-to-tail technique I describe in Chapter 6, you can calculate the magnitude R and the direction of the friction angle, φ, of a resultant of those two forces as shown in Figure 24-4b. All you have to do is apply a little bit of basic trigonometry.
Thus, the angle that the friction resultant makes with respect to the normal force is also a function of the static friction coefficient, or is a constant, because the normal forces in the calculation cancel each other.
By creating a new single resultant for the friction limit and the normal force, you can eliminate having to work with two unknown forces. Instead, you can replace this system with a single unknown (in terms of the normal force) acting at a known angle, which you can compute from the coefficient of static friction.
For the earlier plank example of Figure 24-3, you can compute the friction angle as
The combined resultant at each contact point (because µs is the same at both points) is given as the following:
Thus, you can replace the friction and normal pairs at both Point A and Point B with a single resultant force at a new orientation. I've gone ahead and done this for you in Figure 24-5.
By replacing the normal and friction components with a single resultant R at a new angle φ at each contact point, you're now left with two unknowns, NA and NB, on the free-body diagram. Because you have three equations to work with, you're now able to solve for these components.
You can then solve these two equations simultaneously to give you the normal forces NA and NB that are required to keep the 150-Newton plank in equilibrium. The results: NA = 138.2 Newton, and NB = 41.4 Newton.
To determine whether the plank starts to slide, you then must calculate the friction limit for each contact point (flip to "Finding the friction limit FMAX" earlier in the chapter for more on this calculation).
From this F.B.D., you can also calculate the actual friction forces atPoints A and B by calculating the components of the resultant force and its orientation angle φ:
Because you know the actual friction force at each point, as well as the friction limit, you can start to draw some conclusions:
Because FA = 41.3 N ≤ FA MAX = 41.5 N, Point A can't move.
Because FB = 41.2 N >FB MAX = 12.4 N, Point B is able to move.
In order for the plank to move, both Point A and Point B must be able to move. After all, unless the plank stretches, one point can't move without the other also moving. In the Figure 24-3 example, based on the calculations, Point A can't move because its friction force isn't sufficient to overcome the friction limit (though it's actually very close). Even though Point B can move, you must remember that both points must move for the entire plank to move. Your final conclusion would thus be that the plank is unable to slide.
Say you're back in your kitchen, trying to move that refrigerator yet again. First you push horizontally at a low point on the refrigerator (as shown in Figure 24-6a), and it begins to slide. Great! You push for a while and then take a breather. When you return to the task, you decide to push on the upper corner. But instead of sliding, what happens? Chances are, if it's a full-height refrigerator and it's empty, the refrigerator starts to lean and then fall over without ever sliding. This phenomenon of falling over before sliding is called tipping.
To quantify tipping, you need to locate the tipping point, which is usually at a corner or edge along the contact surface of the object. When tipping occurs, the contact surface disappears, with the tipping point remaining as the last point in physical contact with the original interface (see Figure 24-6b).
To begin analyzing tipping problems, you first need to construct an F.B.D. (surprise, surprise!). Referring to the free-body diagram in Figure 24-6c, you see that all of the old familiar F.B.D. information has been included: external forces, self weight, and support reactions (or contact forces and friction forces in this case). See Chapter 13 for more on the basic parts of an F.B.D.
However, a new parameter — location of the normal force x — has made an appearance. In most F.B.D.s, you assume the normal force is acting in line with the center of mass of the object and don't worry about the location of the normal. With tipping and friction problems, the location of the normal force matters. To start the analysis, you first apply the translational equilibrium equations from earlier in the chapter to determine the normal force N and the friction force F:
Note
The applied force P and friction force F have the same magnitude but opposite directions, and they're separated by a distance y. Together, these two forces cause a moment (or a rotational effect that I discuss in Chapter 12) on the object, which causes the object to want to rotate in a clockwise direction about the tipping point. To maintain equilibrium, the normal force N shifts away from the center of the object in response, to create a balancing couple with the weight. The normal force N moves toward the tipping point (which causes x to decrease). If the force becomes too great, the distance x may actually become a negative value. When this situation happens, the normal must be located outside the boundaries of the object to balance the overturning moment from the applied load P, which is physically impossible. Your fridge is no longer stable and tips over.
To continue your analysis of tipping, you first establish your Cartesian coordinate system (see Chapter 5) by placing the origin at the tipping point location. Then you write the rotational equilibrium equation for this F.B.D. about the tipping point (because tipping is a rotational behavior). For Figure 24-6, the equation is
The terms of this moment equation indicate that each term is set by a physical parameter of the object or a known applied point load and doesn't vary, except for the position of the normal force with respect to the tipping point x. As the applied load P increases, the x parameter for the normal force N changes in order to maintain equilibrium. This position is what helps you determine whether an object will tip.
If you apply a force, say P = 100 pounds, at the base of the object (or y = 0), tipping doesn't occur. Say the fridge in Figure 24-6 has a base of 12 inches, is 60 inches high, and weighs 150 pounds. Substituting these values into the moment equation, you can then solve for x:
Because x is positive, you know your assumption of the location of the normal force was correct. Now watch what happens if you move the same load to the top of the refrigerator, at y = h = 60 inches.
Thus, by simply moving the force, you've also moved the location of the normal force; in this case, you've moved it to the right of the tipping point, or outside the physical boundary of the object, so the fridge tips over.
Warning
Note that the sign of x doesn't actually correspond to a Cartesian coordinate value. In this particular example, a negative x actually corresponds to a positive Cartesian position, so be careful!
As the preceding section indicates, if you push horizontally at a low point on a refrigerator, it's more likely to slide. And if you push with the same force at the top of the refrigerator, it's more likely to tip. But what if you place your refrigerator on a sloped floor and start to increase the angle of the ramp θ without applying a force at all. Which friction phenomenon occurs first?
Consider the 300-pound refrigerator acting on the ramp, as shown in Figure 24-7a.
To start your analysis, you must first write the equilibrium equations for the F.B.D. If you establish a coordinate system that is inclined with the ramp, with the origin at the presumed tipping point, you can create a simpler equation than if you had used regular horizontal and vertical components. Here's what the equations look like for Figure 24-7:
Tip
Remember, these equilibrium equations are unique to each problem you solve. Just draw your F.B.D. (such as the one in the F.B.D. portion of Figure 24-7b) and then write the equilibrium equations based on your Cartesian axes.
Next, you list all the possible friction phenomenon that can occur. For this simple problem, you have two possibilities, which I discuss in the following sections.
Tip
Without knowing which occurs first, you must make a guess. Choose one of the two cases and then verify that the assumptions behind that case are correct. Friction requires a lot of guess-and-check type of calculations.
For sliding to occur first, you need to prove that the friction force at the bottom of the refrigerator exceeds the friction limit µsN. For sliding to occur before tipping happens, the normal force is still located somewhere along the contact surface and the x dimension is positive. If sliding occurs, then
Thus, at a ramp angle of 16.7 degrees, the refrigerator starts to slide. But now you must confirm your assumption about the location of the normal force x by checking the moment equation:
The positive sign indicates that the normal force is assumed to be acting on the correct side of the tipping point, and more importantly, is still in contact with the interface.
For tipping to occur first, you assume that the location of the normal is at the tipping point (or x = 0) and then compute the corresponding friction force. Finally, to verify, you check that the friction force F is less than the friction limit FMAX.
To check tipping for Figure 24-7, you start with the moment equation and assume x = 0 (that the normal is located at the tipping point).
If you compare this angle with the result from the preceding section, you see that you need a steeper angle to cause tipping than you do to cause sliding.
Tip
If you haven't already calculated the angle for sliding to occur (perhaps you chose to check tipping before sliding), you can check it now by determining the friction force F and comparing it to FMAX.
Thus, F > FMAX at the angle required to cause tipping, which implies that sliding would have already occurred. The calculations in the preceding section verify this result.
You encounter a wide variety of other types of friction problems in statics, but the most frequently encountered problems typically involve wedges or belt/pulley friction.
Wedges are small mechanical devices that transmit (and usually increase) an applied force in another direction. Usually, wedges are very long in comparison to their thickness, which often eliminates tipping as a concern. (Flip to "Timber! Exploring Tipping" earlier in the chapter for more on this concept.) Consider Figure 24-8, which shows a force P being applied to Block A. As force P increases, Block A moves to the right. Because of the sloped interface between Blocks A and B, Block B moves upward as Block A moves to the right.
To analyze problems involving wedges, you need to draw several F.B.D.s; you don't have to draw them in the order I list here, but make sure you draw them all:
Draw one F.B.D. for each individual wedge.
Normally, you'd draw the combined system F.B.D. in Step 2 first, but in this example, looking at the separate diagrams first is helpful. You can actually determine the directions of the friction forces by looking at the individual behaviors of each block. Make sure to include them in the same directions on the combined system in Step 2.
Starting with Block A, you know that when the force is applied, the impending motion of Block A is to move to the right. (I discuss impending motion earlier in the chapter.) Thus, the friction on both surfaces of the wedge must oppose the motion of direction. The friction force FA is to the left and the interface friction force FAB is upward and to the left. You also must include both the normal force from the ground (NA) and the normal force from Block B sitting on top (NAB). See Figure 24-9.
Note
Normal forces always act perpendicular to their contact surfaces.
Similarly, you know that Block B wants to move upward, so the friction forces on it must oppose that impending motion. Thus, the friction at the wall (FB), is acting downward, and the interface friction (FAB) is acting down and to the right. (Notice that the force FAB on Block B is in the opposite direction of the same force on Block A, which ensures equilibrium.)
Draw an F.B.D. of the combined systems.
You also need to draw a combined diagram of Blocks A and B. Remember to include the normal and friction forces from the contact surfaces with the wall and floor on your F.B.D.; you can get those from the individual block diagrams in Step 1. See Figure 24-10.
Write equilibrium equations for each individual block and the combined F.B.D.
You sum forces in the x-and y-directions for each of the three diagrams in Steps 1 and 2 to develop relationships between unknown forces and the applied load P.
Check the friction limits for each friction force.
Remember, in order for Block A to move, the friction at the floor and at the interface with Block B must be larger than the friction limits at each location. If Block A can move, you must then check that Block B can also move. If Block B doesn't move, Block A can't move.
Another type of friction problem that you encounter involves pulleys and belt/cable friction. This friction is caused by the motion of a belt or cable relative to the surface of a pulley or drum assembly (see Figure 24-11). The direction of movement of the cable determines the direction of the friction force.
However, you normally express the force on one side of the pulley in terms of the force on the other when examining belt friction problems.
Note
For frictionless cases, the tension on both sides of a pulley or drum is assumed to be equal. If friction is present, that's no longer the case. The tension of the belts on either side of a pulley subjected to belt friction is given by the following equation:
where β is the angle of wrap of the belt around the pulley, expressed in radians. If the belt wraps 180 degrees, the angle of wrap is π or 3.14 radians. The larger the angle of wrap, the more force is necessary to overcome friction.