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6. Friction
A) Overview
This unit is devoted to the introduction of a single force, friction. Frictional forces are omnipresent in our world. Indeed, pretty much all science prior to Galileo, focused on what was directly observed, much of that being dominated by friction. Galileo, in his description of free fall, and Newton in his first law, took as fundamental the more idealized description of motions in the absence of friction. The effects of friction then can be âadded back inâ. We will learn in this unit exactly how to account for these effects of friction within Newtonâs framework.
In particular, we will adopt a simple model for frictional forces that is specified by two constants for each pair of surfaces that are in contact with each other. These constants are the coefficients of static and kinetic friction. We will then address specific examples to demonstrate how we can use these constants to account for frictional forces within Newtonâs framework.
B) Friction
In the last unit we introduced several new forces. The gravitational force is a fundamental force of nature that exists between any two objects of any size that have mass. In this unit, we will introduce a force, friction , that is more similar to the contact forces we discussed in the last unit.
In general the force between two surfaces that are in contact has components both perpendicular and parallel to the surfaces. The perpendicular component is the normal force we discussed last time. The parallel component is called the friction force.
The direction of the friction force is always such that it opposes any relative motion of the two surfaces. We distinguish between two kinds of friction forces. Kinetic friction refers to cases in which one surface moves relative to the other one, such as when a box slides across the floor, while static friction refers to cases in which the surfaces do not move relative to each other, such as when a person is pushing on a stationary heavy box.
The microscopic origins of these friction forces are complicated; they arise from the interactions of atoms on the surface of materials. We will not try to understand these interactions in this course. We will only be concerned with characterizing the friction forces on macroscopic objects using a simple model.
C) Kinetic Friction
When one object slides across another the frictional force between them is found to depend linearly on the perpendicular force between them. In other words, the frictional force is proportional to the normal force. The constant of proportionality, ÎźK, is called the coefficient of kinetic friction. This constant depends only on the properties of the two
f K = Îź K N
surfaces and not on the size or weight of the objects.
Weâll now do an example to see how this works. Suppose a box is given an initial shove after which it slides on a horizontal floor. If the coefficient of kinetic friction between the floor and the box is ÎźÎ, what is the acceleration of the box as it slows to a stop?
We will follow the problem solving procedure we developed last time. We first draw the free body diagram for the box as shown in Figure 6.1. The forces acting on the
box once it is in motion are the weight of the box, the normal force exerted by the ground on the box, and the kinetic friction force. The next step is to write down Newtonâs second law for both the x and y directions. â Îź K N = ma x N â mg = may = 0
The acceleration in the y direction is zero; therefore, the magnitude of the normal force is just equal to the weight of the box. Turning now to the x direction, we see that we can determine the acceleration since we now know the magnitude of the kinetic friction force. Namely, the acceleration in the x direction is just equal to the coefficient of kinetic friction times the acceleration due to gravity. The minus sign indicates that the direction of the acceleration is to the left. a (^) x =â Îź K g
Note that the acceleration does not depend upon the mass of the box! How does this come about? The reason is that the net force here is just the kinetic friction force which, by its definition, is proportional to the normal force. But the normal force here is equal to the weight of the box; therefore, we see that the kinetic friction force is proportional to
Figure 6. The free-body diagram for a box of mass m sliding across a
floor characterized by a coefficient fo sliding friction Îź K
between th box and the floor.
called the coefficient of static friction, Îź S, and depends only on the properties of the two
surfaces. This behavior of the static friction force that we have described is captured in an inequality; namely, that the static friction force is less than or equal to the coefficient of static friction times the normal force. Note the important difference between kinetic
f S ⤠Ο S N
and static friction: the kinetic force is always equal to Îź KN , while the static force is NOT
always equal to Îź S N. In fact, the static force is only equal to Îź S N just before the surfaces
began to move. In all other cases, the static force is less than this maximum force; just as it was for the normal force and for the tension force, the magnitude of the static frictional force must be determined from Newtonâs second law: The static frictional force is simply what is has to be to do what it does!
We can summarize what we have learned about friction so far by completing the plot we have created for static friction. As we increase T beyond the point at which the
box begins to move, we see that frictional force stays constant as T increases. The discontinuity at the point at which the box begins to move indicates that ÎźS > ÎźK.
E) Example: Box Sliding Down Ramp
Weâll now do some examples that illustrate the use of these frictional forces within the framework of Newtonâs laws. We will start with the calculation of the acceleration of a box as it slides down a rough ramp.
Figure 6.5 shows a free-body diagram of all of the forces acting on the box. In the absence of friction between the ramp and the box, there are two forces acting on the box: its weight ( mg ), which points downward, and the normal force N that the ramp exerts on the box. The direction of this normal force is perpendicular to the ramp. Right away we can see that there is a net force directed down the plane which gives rise to the acceleration down the plane. If there is friction between the box and the ramp, there will
Figure 6. A plot of the magnitude of the fritional force as a function of the tension of the rope used in the example in Fig 6.2 The foprce is discontinuous at the point where the box begins to move.
be an additional force that opposes this motion. Therefore, this frictional force is directed
up the plane and has a magnitude equal to Îź KN , when the box is sliding.
We now need to choose a coordinate system. As we mentioned last time, choosing one axis to be parallel to the acceleration often simplifies the calculation. Therefore, we will choose our x-axis to point down the ramp and the y-axis to be perpendicular to the ramp.
Finally we write down Newtonâs second law for both the x and y directions. To write down the y equation, we need to find the y-component of the weight. Since the normal force is perpendicular to the ramp and the weight is perpendicular to the
horizontal, we see that the angle between these two forces must also be equal to θ, the
angle the ramp makes with the horizontal. Therefore, we can use trigonometry to
â Fy = N â mg cos^ θ= may =^0
determine that the y -component of the weight is just equal to the weight times the cosine of the angle the ramp makes with the horizontal. Writing down Newtonâs second law for the y direction, we obtain the magnitude of the normal force. This result holds for both the friction and frictionless cases.
N = mg cos θ
We can now obtain a value for the magnitude of the frictional force from our result for the magnitude of the normal force. The x -component of the weight is equal to the product of the weight and the sine of the angle the ramp makes with the horizontal. We can now write down Newtonâs second law in the x -direction in terms of this frictional
â Fx^ = mg sinθâ^ Îź KN = ma x
force and the component of the weight down the ramp. We can substitute our result for the normal force into this equation and then solve for the acceleration. We obtain the
Figure 6. A free-body diagram for a box of mass m sliding down a ramp. The friction between the ramp and the box is specified by the coefficient of
kinetic frcition, Îź K.
maximum value ( Îź S N ). Substituting this value for the frictional force back into our
f max =Ο S N =Ο Smg cos θ max
general equation, we obtain our result for the maximum angle the ramp can make with the horizontal to prevent the box from sliding down the ramp.
Ο S mg cosθmax = mg sin θ max
tanθ max= Ο S
G) Example: Car Rounding a Corner
Weâll close this unit by doing one more example involving friction, namely, that of a car of mass M rounding a circular turn of radius R. If the coefficient of static friction
between the tires and the road is Îź S , how fast can the car go around the turn without
skidding off the road?
Perhaps your first question here is why in the world are we giving you the static coefficient of friction when the car is clearly moving? The answer to this question is that the tires are rolling: the surfaces of the tires are not sliding relative to the surface of the road, since if they were, the car would already be skidding! During normal driving it is the static friction between the tires and the road that makes a car speed up, slow down, and turn corners!
We start, as always, by drawing a free body diagram for the car as shown in Figure 6.7. From this diagram, we see it is just the frictional force f that is responsible
for the centripetal acceleration of the car. Next, we choose the x -axis to point toward the center of the circle to align it with the direction of the acceleration. We choose the y -axis to be vertically up.
Figure 6. The free-body diagram for a car of mass m moving at constant speed v in a circle of radius R ..
Writing Newtonâs second law for the y -direction, we see that the magnitude of the â Fy = N â mg = may =^0
normal force is just equal to the weight of the car. Writing Newtonâs second law for the x -direction, we see that the frictional force must be equal to the mass of the car times the
â Fx^ = f = ma x
centripetal acceleration. We know the centripetal acceleration is equal to the square of the speed divided by the radius of the turn. Therefore as the carâs speed increases, the
R
v ax
2
frictional force must also increase. There is a limit though, as to how much this frictional
force can increase. This limit is determined by the coefficient of static friction, Îź S. This
maximum frictional force then produces the maximum centripetal acceleration which determines the maximum velocity. Therefore, we obtain the maximum speed that the car
f max =Îź SN = Îź S mg
R
v a
2 max max =
R gR m
f
v max= a max R = max = Îź S
can make the turn without skidding is proportional to the square root of the product of the turning radius and the coefficient of static friction. This result makes sense: the carâs maximum speed should increase if either the friction increases or the turning radius increases.
