Static and Kinetic Frictional Forces
When an object is in contact with a surface, there is a force acting on the object. The previous section discusses the component of this force that is perpendicular to the surface, which is called the normal force. When the object moves or attempts to move along the surface, there is also a component of the force that is parallel to the surface. This parallel force component is called the frictional force, or simply friction.

In many situations considerable engineering effort is expended trying to reduce friction. For example, oil is used to reduce the friction that causes wear and tear in the pistons and cylinder walls of an automobile engine. Sometimes, however, friction is absolutely necessary. Without friction, car tires could not provide the traction needed to move the car. In fact, the raised tread on a tire is designed to maintain friction. On a wet road, the spaces in the tread pattern (see Figure 4.18) provide channels for the water to collect and be diverted away. Thus, these channels largely prevent the water from coming between the tire surface and the road surface, where it would reduce friction and allow the tire to skid.
Courtesy Goodyear Tire & Rubber Co.
Figure zoom   Figure 4.18    This photo, shot from underneath a transparent surface, shows a tire rolling under wet conditions. The channels in the tire collect and divert water away from the regions where the tire contacts the surface, thus providing better traction.

Surfaces that appear to be highly polished can actually look quite rough when examined under a microscope. Such an examination reveals that two surfaces in contact touch only at relatively few spots, as Figure 4.19 illustrates. The microscopic area of contact for these spots is substantially less than the apparent macroscopic area of contact between the surfaces—perhaps thousands of times less. At these contact points the molecules of the different bodies are close enough together to exert strong attractive intermolecular forces on one another, leading to what are known as “cold welds.” Frictional forces are associated with these welded spots, but the exact details of how frictional forces arise are not well understood. However, some empirical relations have been developed that make it possible to account for the effects of friction.
Figure zoom   Figure 4.19    Even when two highly polished surfaces are in contact, they touch only at relatively few points.

Figure 4.20 helps to explain the main features of the type of friction known as static friction. The block in this drawing is initially at rest on a table, and as long as there is no attempt to move the block, there is no static frictional force. Then, a horizontal force is applied to the block by means of a rope. If is small, as in part a, experience tells us that the block still does not move. Why? It does not move because the static frictional force exactly cancels the effect of the applied force. The direction of is opposite to that of , and the magnitude of equals the magnitude of the applied force, fs = F. Increasing the applied force in Figure 4.20 by a small amount still does not cause the block to move. There is no movement because the static frictional force also increases by an amount that cancels out the increase in the applied force (see part b of the drawing). If the applied force continues to increase, however, there comes a point when the block finally “breaks away” and begins to slide. The force just before breakaway represents the maximum static frictional force that the table can exert on the block (see part c of the drawing). Any applied force that is greater than cannot be balanced by static friction, and the resulting net force accelerates the block to the right.
Figure zoom   Figure 4.20    Applying a small force to the block, as in parts a and b, produces no movement, because the static frictional force exactly balances the applied force. (c) The block just begins to move when the applied force is slightly greater than the maximum static frictional force .

Experimental evidence shows that, to a good degree of approximation, the maximum static frictional force between a pair of dry, unlubricated surfaces has two main characteristics. It is independent of the apparent macroscopic area of contact between the objects, provided that the surfaces are hard or nondeformable. For instance, in Figure 4.21 the maximum static frictional force that the surface of the table can exert on a block is the same, whether the block is resting on its largest or its smallest side. The other main characteristic of is that its magnitude is proportional to the magnitude of the normal force . As Section 4.8 points out, the magnitude of the normal force indicates how hard two surfaces are being pressed together. The harder they are pressed, the larger is , presumably because the number of “cold-welded,” microscopic contact points is increased. Equation 4.7 expresses the proportionality between and FN with the aid of a proportionality constant μs, which is called the coefficient of static friction.
Figure zoom   Figure 4.21    The maximum static frictional force would be the same, no matter which side of the block is in contact with the table.
Static Frictional Force
The magnitude fs of the static frictional force can have any value from zero up to a maximum value of , depending on the applied force. In other words, , where the symbol ≤ is read as “less than or equal to.” The equality holds only when fs attains its maximum value, which is

  (4.7)
In Equation 4.7, μs is the coefficient of static friction, and FN is the magnitude of the normal force.

It should be emphasized that Equation 4.7 relates only the magnitudes of and , not the vectors themselves. This equation does not imply that the directions of the vectors are the same. In fact, is parallel to the surface, while is perpendicular to it.

The coefficient of static friction, being the ratio of the magnitudes of two forces , has no units. Also, it depends on the type of material from which each surface is made (steel on wood, rubber on concrete, etc.), the condition of the surfaces (polished, rough, etc.), and other variables such as temperature. Table 4.2 gives some typical values of μs for various surfaces. Example 9 illustrates the use of Equation 4.7 for determining the maximum static frictional force.
Table 4.2    

Approximate Values of the Coefficients of Friction for Various Surfaces1
Materials
Coefficient of Static Friction, μs
Coefficient of Kinetic Friction, μk
Glass on glass (dry)
0.94
0.4 
Ice on ice (clean, 0 °C)
0.1 
0.02
Rubber on dry concrete
1.0 
0.8 
Rubber on wet concrete
0.7 
0.5 
Steel on ice
0.1 
0.05
Steel on steel (dry hard steel)
0.78
0.42
Teflon on Teflon
0.04
0.04
Wood on wood
0.35
0.3 
Analyzing Multiple-Concept Problems

Example   9   The Force Needed To Start a Skier Moving

A skier is standing motionless on a horizontal patch of snow. She is holding onto a horizontal tow rope, which is about to pull her forward (see Figure 4.22a). The skier's mass is 59 kg, and the coefficient of static friction between the skis and snow is 0.14. What is the magnitude of the maximum force that the tow rope can apply to the skier without causing her to move?
Figure zoom   Figure 4.22   

(a)  
Two horizontal forces act on the skier in the horizontal direction just before she begins to move.
(b)  
Two vertical forces act on the skier.


Reasoning   When the rope applies a relatively small force, the skier does not accelerate. The reason is that the static frictional force opposes the applied force and the two forces have the same magnitude. We can apply Newton's second law in the horizontal direction to this situation. In order for the rope to pull the skier forward, it must exert a force large enough to overcome the maximum static frictional force acting on the skis. The magnitude of the maximum static frictional force depends on the coefficient of static friction (which is known) and on the magnitude of the normal force. We can determine the magnitude of the normal force by using Newton's second law, along with the fact that the skier does not accelerate in the vertical direction.
Knowns and Unknowns   The data for this problem are as follows:
Description
Symbol
Value
Mass of skier
m
59 kg
Coefficient of static friction
μs
0.14
Unknown Variable
    
Magnitude of maximum horizontal force that tow rope can apply
F
?

Modeling the Problem
Newton's Second Law (Horizontal Direction) Figure 4.22a shows the two horizontal forces that act on the skier just before she begins to move: the force applied by the tow rope and the maximum static frictional force . Since the skier is standing motionless, she is not accelerating in the horizontal or x direction, so ax = 0 m/s2. Applying Newton's second law (Equation 4.2a) to this situation, we have
Since the net force ΣFx in the x direction is , Newton's second law can be written as . Thus,
We do not know , but its value will be determined in Steps 2 and 3.
The Maximum Static Frictional Force The magnitude of the maximum static frictional force is related to the coefficient of static friction μs and the magnitude FN of the normal force by Equation 4.7:

  (4.7)
We now substitute this result into Equation 1, as indicated in the right column. The coefficient of static friction is known, but FN is not. An expression for FN will be obtained in the next step.
Newton's Second Law (Vertical Direction) We can find the magnitude FN of the normal force by noting that the skier does not accelerate in the vertical or y direction, so ay = 0 m/s2. Figure 4.22b shows the two vertical forces that act on the skier: the normal force and her weight . Applying Newton's second law (Equation 4.2b) to the vertical direction gives
The net force in the y direction is ΣFy = +FN - mg, so Newton's second law becomes +FN - mg = 0. Thus,
We now substitute this result into Equation 4.7, as shown at the right.

Solution   Algebraically combining the results of the three steps, we have
The magnitude F of the maximum force is
If the force exerted by the tow rope exceeds this value, the skier will begin to accelerate forward.

Related Homework: Problems 44, 108, 118

The physics of rock climbing.
Static friction is often essential, as it is to the rock climber in Figure 4.23, for instance. She presses outward against the walls of the rock formation with her hands and feet to create sufficiently large normal forces, so that the static frictional forces can support her weight.
© Corey Rich/Aurora Photos
Figure zoom   Figure 4.23    In maneuvering her way up El Matador at Devil's Tower National Monument in Wyoming, this rock climber uses the static frictional forces between her hands and feet and the vertical rock walls to support her weight.

Once two surfaces begin sliding over one another, the static frictional force is no longer of any concern. Instead, a type of friction known as kinetic* friction comes into play. The kinetic frictional force opposes the relative sliding motion. If you have ever pushed an object across a floor, you may have noticed that it takes less force to keep the object sliding than it takes to get it going in the first place. In other words, the kinetic frictional force is usually less than the static frictional force.

Experimental evidence indicates that the kinetic frictional force has three main characteristics, to a good degree of approximation. It is independent of the apparent area of contact between the surfaces (see Figure 4.21). It is independent of the speed of the sliding motion, if the speed is small. And lastly, the magnitude of the kinetic frictional force is proportional to the magnitude of the normal force. Equation 4.8 expresses this proportionality with the aid of a proportionality constant μk, which is called the coefficient of kinetic friction.
Kinetic Frictional Force
The magnitude fk of the kinetic frictional force is given by

  (4.8)
In Equation 4.8, μk is the coefficient of kinetic friction, and FN is the magnitude of the normal force.

Equation 4.8, like Equation 4.7, is a relationship between only the magnitudes of the frictional and normal forces. The directions of these forces are perpendicular to each other. Moreover, like the coefficient of static friction, the coefficient of kinetic friction is a number without units and depends on the type and condition of the two surfaces that are in contact. As indicated in Table 4.2, values for μk are typically less than those for μs, reflecting the fact that kinetic friction is generally less than static friction. The next example illustrates the effect of kinetic friction.
Analyzing Multiple-Concept Problems

Example   10   Sled Riding

A sled and its rider are moving at a speed of 4.0 m/s along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?
Figure zoom   Figure 4.24   

(a)  
The moving sled decelerates because of the kinetic frictional force.
(b)  
Three forces act on the moving sled, the weight of the sled and its rider, the normal force , and the kinetic frictional force . The free-body diagram for the sled shows these forces.


Reasoning   As the sled slows down, its velocity is decreasing. As our discussions in Chapters 2 and 3 indicate, the changing velocity is described by an acceleration (which in this case is a deceleration since the sled is slowing down). Assuming that the acceleration is constant, we can use one of the equations of kinematics from Chapter 3 to relate the displacement to the initial and final velocities and to the acceleration. The acceleration of the sled is not given directly. However, we can determine it by using Newton's second law of motion, which relates the acceleration to the net force (which is the kinetic frictional force in this case) acting on the sled and its mass.
Knowns and Unknowns   The data for this problem are listed in the table:
Description
Symbol
Value
Comment
Explicit Data
      
Initial velocity
v0x
+4.0 m/s
Positive, because the velocity points in the +x direction. See drawing.
Coefficient of kinetic friction
μk
0.050
  
Implicit Data
      
Final velocity
vx
0 m/s
The sled comes to a stop.
Unknown Variable
      
Displacement
x
?
  

Modeling the Problem
Displacement To obtain the displacement x of the sled we will use Equation 3.6a from the equations of kinematics:
Solving for the displacement x gives the result shown at the right. This equation is useful because two of the variables, v0x and vx, are known and the acceleration ax can be found by applying Newton's second law to the accelerating sled (see Step 2).
Newton's Second Law Newton's second law, as given in Equation 4.2a, states that the acceleration ax is equal to the net force ΣFx divided by the mass m:
The free-body diagram in Figure 4.24b shows that the only force acting on the sled in the horizontal or x direction is the kinetic frictional force . We can write this force as -fk, where fk is the magnitude of the force and the minus sign indicates that it points in the -x direction. Since the net force is ΣFx = -fk, Equation 4.2a becomes
This result can now be substituted into Equation 1, as shown at the right.
Kinetic Frictional Force We do not know the magnitude fk of the kinetic frictional force, but we do know the coefficient of kinetic friction μk. According to Equation 4.8, the two are related by

  (4.8)
where FN is the magnitude of the normal force. This relation can be substituted into Equation 2, as shown at the right. An expression for FN will be obtained in the next step.
Normal Force The magnitude FN of the normal force can be found by noting that the sled does not accelerate in the vertical or y direction (ay = 0 m/s2). Thus, Newton's second law, as given in Equation 4.2b becomes
There are two forces acting on the sled in the y direction, the normal force and its weight [see part (b) of the drawing]. Therefore, the net force in the y direction is
where W = mg (Equation 4.5). Thus, Newton's second law becomes
This result for FN can be substituted into Equation 4.8, as shown at the right.

Solution   Algebraically combining the results of each step, we find that
Note that the mass m of the sled and rider is algebraically eliminated from the final result. Thus, the displacement of the sled is

Related Homework: Problems 48, 49, 85

The physics of walking.
Static friction opposes the impending relative motion between two objects, while kinetic friction opposes the relative sliding motion that actually does occur. In either case, relative motion is opposed. However, this opposition to relative motion does not mean that friction prevents or works against the motion of all objects. For instance, the foot of a person walking exerts a force on the earth, and the earth exerts a reaction force on the foot. This reaction force is a static frictional force, and it opposes the impending backward motion of the foot, propelling the person forward in the process. Kinetic friction can also cause an object to move, all the while opposing relative motion, as it does in Example 10. In this example the kinetic frictional force acts on the sled and opposes the relative motion of the sled and the earth. Newton's third law indicates, however, that since the earth exerts the kinetic frictional force on the sled, the sled must exert a reaction force on the earth. In response, the earth accelerates, but because of the earth's huge mass, the motion is too slight to be noticed.

Check Your Understanding
(The answer is given at the end of the book.)
14.  
 
Suppose that the coefficients of static and kinetic friction have values such that μs = 1.4 μk for a crate in contact with a cement floor. Which one of the following statements is true?

(a)  
The magnitude of the static frictional force is always 1.4 times the magnitude of the kinetic frictional force.

(b)  
The magnitude of the kinetic frictional force is always 1.4 times the magnitude of the static frictional force.

(c)  
The magnitude of the maximum static frictional force is 1.4 times the magnitude of the kinetic frictional force.
15.  
 
A person has a choice of either pushing or pulling a sled at a constant velocity, as the drawing illustrates. Friction is present. If the angle θ is the same in both cases, does it require less force to push or to pull the sled?
16.  
 
A box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling force can point horizontally, or it can point above the horizontal at an angle θ. When the pulling force points horizontally, the kinetic frictional force acting on the box is twice as large as when the pulling force points at the angle θ. Find θ.
17.  
 
A box rests on the floor of an elevator. Because of static friction, a force is required to start the box sliding across the floor when the elevator is (a) stationary, (b) accelerating upward, and (c) accelerating downward. Rank the forces required in these three situations in ascending order—that is, smallest first.




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