3.4 Friction - SharpSchool

Etkina/Gentile/Van Heuvelen Process Physics 1/e, Chapter 3 3-28

acceleration and the values for their masses, you can use the expression you determined in the

previous Try It Yourself to determine the value of g.

Review Question 3.3

For problems involving objects moving upward or downward along inclined surfaces, we choose

the x -axis parallel to the surface, and the y -axis perpendicular to the surface. Why not use

horizontal and vertical axes?

3.4 Friction

In previous sections, we assumed that objects moved across absolutely smooth surfaces—

no friction. The vast majority of processes encountered in the real world involve some degree of

friction. In this section we will construct mathematical models of friction so that we can take

friction into account quantitatively.

Static friction

Consider some simple experiments that involve pulling a block with a spring scale, as

shown in Observational Experiment Table 3.2. The spring scale exerts an increasing force on the

block. Observe carefully what happens to the block.

Observational Experiment Table 3.2 Pulling a block with a spring scale

Observational experiments Analysis

A block is at rest on the horizontal surface of a desk.

A spring scale pulls lightly on the block that is at rest on a

horizontal surface; the block does not move.

The spring scale pulls harder on the block at rest on the

horizontal surface; the block still does not move.

ALG

3.3.8-

3.3.10


The spring scale pulls even harder on the block at rest on the

horizontal surface; the block finally starts moving.

Patterns

! In each of these experiments, the surface exerted a normal force on the block that balanced the

downward gravitational force exerted by Earth on the block.

! As the spring scale exerted an increasing force on the block to the right, the block remained stationary

(zero acceleration). The surface must have exerted an additional force - an increasing force on the block

toward the left.

! This resistive force must have a maximum value—once the spring scale exerted a strong enough force

on the block, the block started sliding.

From the above experiments we can infer that if you place an object (for example, a block)

on a surface of some other object (for example, a desk), and you exert a force on the block in an

attempt to get the block to slide, the surface will exert a force on the object in the opposite

direction so that the object does not begin to move. This is called a static friction force. It can vary

in strength, and has a maximum value. When the external force exerted on the object exceeds this

maximum resistive force, the block starts moving. This maximum resistive force that the surface

can exert on the block is called the maximum force of static friction.

Tip! A surface really exerts only one force on an object pressing against it. It is convenient to

break this force into two components: a perpendicular to the surface normal force S on ON

! and a

parallel to the surface static friction force s S on Of

!. (Fig. 3.15)

Figure 3.15 Force of a surface on an object


What determines the magnitude of the maximum static friction force?

To investigate this question, we use a similar experimental setup to that used in

Observational Experiment Table 3.2, except this time the characteristics of the block and the

surfaces are varied. For example, the smoothness of the surfaces can be varied. Suppose the block

is made of a fairly smooth plastic. We pull the block with a spring scale while the block rests on

three different surfaces: 1) a glass tabletop, 2) a wood floor, and 3) a rubber exercise mat. The

maximum static friction force is largest for the rubber mat, next largest for the wood floor, and

smallest for the glass tabletop. It appears that the maximum static friction force the surface exerts

on the block depends on the relative roughness of the contacting surfaces.

The contact area between the block and the surface can also be varied. Use a plastic block

that is shaped like a brick. Bricks have different lengths, widths, and heights. Consequently, the

faces of the block have three different areas. Use the spring scale to pull the block when resting on

each of these three different faces. Something unexpected happens: the maximum static friction

force exerted by the surface on the block is the same in all three cases. The maximum static

friction force does not depend on the area of contact.

The mass of the block can also be varied. Take a 1.0-kg, a 2.0-kg, and a 3.0-kg plastic

block and place them all on the same wood floor. Use a spring scale to pull each of them. The

result is that the maximum friction force is smallest for the 1.0-kg block, twice as large for the 2.0-

kg block, and three times as large for the 3.0-kg block. The maximum static friction force that the

floor exerts on the block is directly proportional to the mass of the block.

Or is it? This is a hypothesis based on the above observations. The next step is to test the

hypothesis. Remember, in a testing experiment the goal is to disprove the hypothesis. So, try to

construct a situation in which the maximum static friction force exerted by the surface on the block

changes, but without changing the mass of the block or the roughness of the surfaces. Use a spring

scale to pull the block, but simultaneously push down on the block exerting a series of specific

downward forces on it. For each of these downward forces, we use the spring scale to determine

the maximum static friction force the surface exerts on the block (see Fig. 3.16). The results of

these experiments are reported in Table 3.3. We used a 1-kg block with a 0.1-m2 surface area.

Figure 3.16 Effect of normal force on friction force


Table 3.3 Maximum static friction force when a block presses harder against a surface.

Mass of

the block

Extra downward

force exerted on

the 1-kg block

Normal force

exerted by the

surface on the block

Maximum

static friction

force

Ratio of maximum static

friction force and normal

force

1.0 kg 0.0 N 9.8 N 3.0 N 0.31

1.0 kg 5.0 N 14.8 N 4.5 N 0.30

1.0 kg 10.0 N 19.8 N 6.1 N 0.31

1.0 kg 20.0 N 29.8 N 9.1 N 0.31

The block mass is the same for all the experiments in Table 3,3 but the maximum static

friction force varied. So, we disproved the hypothesis that the maximum static friction force

depends on the mass of the block. However, the ratio of the maximum static friction force and the

normal force is the same for all of the measurements (the last column in the table). It appears that

the maximum static friction force is directly proportional to the magnitude of the normal force:

S on O max S on Osf N2

For the experiments in Table 3.3, the proportionality constant is 0.31.

The finding S on O max S on Osf N2 makes sense if you we look back at Fig. 3.15 – the normal force

and the friction force are two perpendicular components of the same force – the force that a

surface exerts on an object!

If we repeat these experiments only using a different type of block on a different surface,

the results are similar. The ratio of the maximum friction force and the normal force are the same

for all of the different values of the normal forces. However, the proportionality constant is

different for different surfaces; the proportionality depends on the types of contacting surfaces.

The proportionality constant is greater for two rough surfaces contacting each other and less for

smoother surfaces.

This ratio maxs

s

f

N3 " !is called the coefficient of static friction

s3 for a particular pair of

surfaces. It’s a measure of the relative difficulty of sliding two surfaces across each other. The

easier to slide one surface on the other, the smaller the value of s

3 . You have a physical

experience of s

3 when you try to walk on ice and on dry pavement. The value of s

3 is somewhat

greater for the pavement than for the slippery ice.

From the definition of the coefficient of static friction s

3 , we see that it has no units as it

is the ratio of two forces (actually the ratio of two components of the same force). It usually has

values between 0 and 1, but can be greater than 1. Its value is about 0.8 for rubber car tires on a

dry highway surface, and is very small for bones in healthy body joints separated by cartilage and

synovial fluid. Some values for different surfaces are listed in Table 3.4.


Static friction force When two objects are in contact and we try to pull one across the other,

they exert a static friction force on each other. This force is parallel to the surfaces of two

objects and opposes the tendency of one object to move across the other. The static friction

force provides flexible resistance (as much as is needed) to prevent motion—up to a

maximum value. This maximum static friction force depends on the roughness of the two

surfaces (on the coefficient of static friction s

3 between the surfaces), and on the magnitude

of the normal force N exerted by one surface on the other. The magnitude of the static

friction force is always less than or equal to the product of these two quantities:

0s s

f N34 4 !! "3.10)

Keep in mind the assumptions used when constructing this model of friction. We used relatively

light objects resting on relatively firm surfaces. The objects never caused the surfaces to deform

significantly (for example, it was not a car tire sinking into mud). Equation (3.10) is only

reasonable in situations where this condition holds.

An amazing example of friction is an experiment

conducted by Mythbusters involving the pages from two

telephone books. The pages were carefully interwoven with

each page of one book between two pages of the other (see the

photo at the side). Cords were placed though holes in the

bindings of each book and tied to a rope. One rope was tied to a

strong post and four men pulled on the other rope trying to pull

the books apart—they were unsuccessful. Two cars could not

pull the books apart. Finally they attached the ropes to World

War II Sherman tanks. Only after the tanks exerted a force on

each rope of more than 6000 lb were they able to pull the books

apart. This experiment depended on the friction forces that

adjacent pages exerted on each other. If you try the experiment,

be sure to keep the books flat on a table when trying to pull

them apart.

Kinetic friction

If we repeat the previous friction experiments with a block but instead pull the block so

that it moves at constant velocity, we find a similar relationship between the resistive friction force

exerted by the surface on the block and the normal force exerted by the surface on the block. There

are two differences however: (1) to keep the block sliding at constant velocity, we don’t have to

pull quite as hard as the maximum static friction needed to start the block sliding; and (2) the

resistive force exerted by the surface on the moving object does not vary but has a constant value.

The magnitude of this kinetic friction force k

f depends on the roughness of the contacting


surfaces (indicated by a coefficient of kinetic friction k

3 ) and on the magnitude N of the normal

force exerted by one of the surfaces on the other, but not on the surface area of contact. The word

kinetic indicates that the surfaces in contact are moving relative to each other.

Kinetic friction force When an object slides along a surface, the surfaces exert kinetic

friction forces on each other. These forces are exerted parallel to the surfaces and oppose

the motion of one surface relative to the other surface. The kinetic friction force depends

on the surfaces themselves (on the coefficient of kinetic friction k

3 ) , and on the

magnitude of the normal force N exerted by one surface on the other:

k k

f N3" !"3.11)

Tip! Although in our experimental setup we pulled the block so that it moved at constant velocity,

this expression for the kinetic friction force is also valid for accelerated motion.

As with any mathematical model, the above expression for kinetic friction has its limitations. First,

it has the same assumption about the rigidity of the surfaces as we had for static friction. Second,

predictions based on Eq. (3.11) fail for objects moving at high speed. Thus, this equation does not

have general applicability, but is simple and useful for rigid surfaces and everyday speeds.

Table 3.4 The coefficients of kinetic and static friction for two different surfaces.

Contacting surfaces Coefficient of static friction Coefficient of kinetic

friction

Rubber on concrete 0.6 – 1.0 0.8

Steel on steel 0.74 0.57

Aluminum on steel 0.61 0.47

Glass on glass 0.94 0.4

Wood on wood 0.25 – 0.5

Ski wax on wet snow 0.1

Ski wax on dry snow 0.04

Teflon on Teflon 0.04 0.04

Greased metals 0.1 0.06

Surfaces in healthy joint 0.015

What causes friction?

Although friction is something that we encounter every second of our lives, it is one of the

least understood physical phenomena. There is at present no accepted scientific explanation for

how friction works. We know that rougher surfaces usually cause greater friction than smooth

surfaces (for example, the coefficient of friction for car tires on dry asphalt is greater than for the

same tires on an icy surface). On a microscopic level, surfaces are bumpy and less than 1 percent

of the surfaces are in direct contact. Velcro is an extreme example of the phenomenon of friction.

For Velcro, the hooks and loops of the two surfaces connect with each other making it almost


impossible to slide one surface relative to the other. Maybe real objects have this hooking Velcro

effect but to a much smaller extent. Perhaps real surfaces have tiny bumps that can hook onto tiny

bumps on another surface (Fig.3.17). If this explanation is correct, then smoothing the surfaces

should reduce friction. This is what happens. For example, when you send a book with a rough

matte finish sliding across a table and then send a book with a glossy smooth finish sliding across

the same table, the glossy book slides farther before it stops—the friction force exerted by the table

on the glossy book is less that on the other book.

Figure 3.17 Microscopic view of contacting surfaces

However, if you examine Table 3.4, you find that the coefficient of friction for rubber

against concrete (both surfaces are rough) is almost the same as for glass on glass or steel on steel

(the surfaces are very smooth). Why? As we will learn later, substances are made of particles that

attract each other. This attraction between particles keeps a solid object in whatever shape it has.

But this attraction between particles works only at very short distances. That is why you cannot fix

a broken vase by bringing the pieces closer together. When the vase broke, tiny pieces of the

material were lost, and now the distances between the parts of the vase when you bring them

together is too large for the particles to connect again. However, if the two surfaces are very

smooth so that the microscopic particles are close enough to attract to each other strongly (almost

close enough to form chemical bonds with each other), it becomes more difficult to slide one

surface relative to the other. This is what happens to the smooth glass on smooth glass and the

smooth steel on smooth steel.

Measuring static friction for everyday objects

Walking on a flat horizontal sidewalk would not be possible if there were no static

friction. When one foot swings forward and contacts the sidewalk, static friction prevents it from

continuing forward and slipping (the way it might on ice). Similarly, your other foot pushes

backward on the sidewalk, which means the sidewalk pushes forward on your foot (Newton’s third

law) causing your body to move forward. If there were no static friction, your foot would slip

backward when you tried to push backward on the sidewalk.

How could we determine the coefficient of static friction between a running shoe and, say,

a kitchen floor tile? Let’s try two independent methods. If only one method is used, it is difficult to


decide if the result is reasonable. If we use at least two independent experiments, we can evaluate

the consistency of the results.

Experiment 1

In the first experiment, a tile is secured on a horizontal tabletop, and a shoe is

placed on top of it. The shoe is pulled horizontally with a spring scale, which exerts an

increasingly greater force on it until the shoe begins to slide (3.18a). A force diagram for the

moment just before it slides is shown in Fig. 3.18b. Since the shoe is not accelerating in the

vertical direction, the upward normal force exerted on it by the tile balances the downward

gravitational force exerted on it by Earth. Thus, the magnitudes of these forces are equal:

T on S SN m g" .

Figure 3.18 One way to determine µs for shoe-tile surface

For the horizontal x -direction, only the tension force exerted by the spring scale on the

shoe and the static friction force exerted by the tile on the shoe are included in the x -scalar

component of Newton’s second law. The other two forces exerted on the shoe do not have x -

components. Thus, the x -scalar component form of Newton’s second law is:

Scale on S s T on S x x xma T f" #

Just before the shoe starts to slide, its acceleration is zero, and the scale reads the maximum force

of static friction that the tile exerts on the shoe:

max Scale on S s max T on S0 T f" %

Since max Scale on S s max T on S s T on S,T f N3" " we can determine the coefficient of static friction:

max T on S max Scale on S

T on S S

s

s

f T

N m g3 " "

All we need to do is measure the mass of the shoe and record the reading on the spring scale at the

moment the shoe starts to slide.

The measured shoe mass is 0.37 kg, and the maximum scale reading is 2.6 N just before

the shoes starts to slide. Thus, the coefficient of static friction is:

max T on S max Scale on S

T on S S

2.6 N0.72.

(0.37 kg)(9.8 N/kg)

s

s

f T

N m g3 " " " "

ALG

3.4.5;

3.4.6


We reported just two significant digits since the measured quantities have just two significant

digits. Thus, based on this experiment, the coefficient of static friction is 0.72 0.01s

3 " 5 .

Experiment 2

For the second experiment, we place the shoe on the tile and tilt the tile until the shoe

starts to slide. How does this experiment allow us to determine the maximum coefficient of static

friction? Consider three sketches, one for each of three different tilt angles of the tile. Draw

corresponding force diagrams for the shoe (the system of interest). When on a flat surface, there is

no force of static friction (Fig. 3.19a). For a moderately tilted tile, the relatively small static

friction force can prevent the shoe from sliding (Fig. 3.19b). For the tilted tile just before the shoe

slides, the static friction has its maximum possible value (Fig. 3.19c). The static friction force that

the tile exerts on shoe increases as the tilt angle increases. Notice that the x -axis of the coordinate

system is parallel to the tilted tile and y -axis is perpendicular.

Figure 3.19 Another way to determine µs for shoe-tile surface

We can now use the force diagram in Fig. 3.19c to help apply Newton’s second law in

component form for the shoe just before it starts sliding. The magnitude of the gravitational force

that Earth exerts on the shoe is E on S SF m g" , and the shoe’s acceleration is zero just before it

starts to slide.

y -component equation: 0 0

S T on S S s Max T on S• 0 sin 90 cos sin 0m N m g f$" % #

x -component equation: 0 0

S T on S S s Max T on S• 0 cos 90 sin – sin 0m N m g f$" #

Computing the values of the sines and cosines and inserting the expression for the maximum static

friction force max T on S T on S s s

f N3" , we get:


T on S S0 cosN m g $" %

S T on S0 sin .s

m g N$ 3" %

This is another system of two equations with two unknowns, T on SN and s

3 . Since our interest is

in the coefficient of static friction, we solve the first equation for the normal force

( T on S S cosN m g $" ) and substitute this into the second equation:

Rearrange the above equation to get an expression for s

3 :

sin = = tan

cos s

$3 $

$.

This is an amazing result —the coefficient of static friction between the shoe and the tile can be

determined just from the angle of the tile’s tilt when the shoe starts sliding! It doesn’t depend on

any other factors! But if you go back to Figure 3.15 you would see that is gives us exactly the

same result – the ratio of friction force over normal force is equal to the tan$ .

When we do the experiment, the shoe starts sliding when the tile is at an angle of about

036$ " . Thus, the coefficient of static friction determined from this experiment is:

0sin = = tan = tan 36 0.73.

cos s

$3 $

$"

Again, because of the number of significant digits, this is equivalent to 0.73 0.01s

3 " 5 . This is

consistent with the result from the first experiment, since the ranges of possible values overlap.

The coefficient of static friction between the shoe and the tile is between 0.72 and 0.73.

Tip! Notice that the magnitude of the normal force that a surface exerts on an object does not

necessarily equal the magnitude of the gravitational force that Earth exerts on the object—

especially when the object is on an inclined surface!

Now, let’s consider a real-world situation that involves kinetic friction.

Skid Marks Used for Evidence

Police stations have charts listing the kinetic friction coefficients of various brands of car

tires on different types of road surfaces. Why do police stations have these charts? Normally, when

a car stops, it rolls to a stop, and the tires do not skid across the road surface. However, if the

driver presses really hard on the brakes, the tires can lock up, causing the car to skid. Police

officers use the length of skid marks to estimate the speed of the vehicle at the time the driver

applied the brakes, as in the next example.

Example 3.8 Was the car speeding? A car involved in a minor accident left 18.0-m skid marks

on a horizontal road. Inspecting the car and the road surface, the police officer decided that the

S S0 sin coss

m g m g$ 3 $" %


coefficient of kinetic friction was 0.80. The speed limit was 35 mph on that street. Was the car

speeding?

Sketch and Translate A sketch of the process and all known information are shown in Fig. 3.20a

The coordinate system consists of a horizontal x -axis pointing in the direction of the velocity and

a vertical y -axis points upward. Choose the car as the system.

Figure 3.20(a) Use skid marks to estimate car’s initial speed

Simplify and Diagram Assume that the car can be modeled as a point-like object, that its

acceleration while stopping was constant, and that the resistive force exerted by the air on the car

is small compared with the other forces exerted on it. A motion diagram for the car is shown in

Fig. 3.20b. Three forces exerted on the car are shown in a force diagram in Fig. 3.20c. Earth exerts

a downward gravitational force on the car E on CF

!, the road exerts an upward normal force on the

car R on CN

! (perpendicular to the road surface), and the road also exerts a backward kinetic friction

force on the car R on Ckf

! (parallel to the road’s surface and opposite the car’s velocity). This

friction force causes the car’s speed to decrease.

Figure 3.20(b)(c)

Represent Mathematically Use the force diagram in Fig. 3.20c to help apply Newton’s second law

in component form. Use the expression for the kinetic friction force ( Ron C R on Ck kf N3" ), the

expression for the gravitational force ( E on C CF m g" ), and one of the kinematics equations. The

car remains in contact with the road surface, so the y -component of its acceleration y

a is zero.

y -component equation: R on C, E on C, R on C, 0y y k y

N F f" # #

0 0 0

R on C C R on C 0 sin 90 sin 90 sin 0k

N m g f" % #

Note that 0sin 90 1.0" and

0sin 0 0" . Thus,


R on C CN m g" .

The magnitude of the kinetic friction force is then:

R on C R on C Ck k kf N m g3 3" " !#

x -component equation: C R on C, E on C, R on C, x x x k xm a N F f" # #

0 0 0

C R on C C R on C cos90 cos90 cos 0x k

m a N m g f" # %

Substitute 0cos90 0" and

0cos 0 1.0" into the above to get:

C R on C x km a f" % #

Combine the two equations for S on C kf to get:

C Cx km a m g3" %

or !

x ka g3" % .

Now use kinematics to determine the car’s velocity 0,xv before the skid started:

2 2

0, 02( )x x x

v v x x a% " %

Solve and Evaluate The car’s acceleration while stopping was:

2 2(0.80)(9.8 m/s ) 7.84 m/s .x k

a g3" % " % " %

The kinematics equation is used to determine the initial speed of the car before the skid started

(recall that the final speed x

v = 0 and that the stopping distance was 18 m):

2 2 2 2 2

0, 00 – = 2( – ) = 2(18 m – 0)(–7.84 m/s ) = –282 m /sx x

v x x a

0, = 16.8 m/s = 16.8 m/s(3600 s/h)(1 mile/1609 m) = 38 mphx

v6 .

This is slightly over the 35 mph speed limit, but probably not enough for a speeding conviction.

Note also that the answer had the correct units.

Try It Yourself: Your car is moving at a speed 16 m/s on a flat icy road when you see a

stopped vehicle ahead. Determine the distance needed to stop if the effective coefficient of

friction between your car tires and the road is 0.40.

Answer: 33 m.

Other types of friction

There are other types of friction besides static and kinetic friction, such as rolling friction

(due to the tires indenting slightly as the tires turn—this friction is less if the tires have been

inflated to a higher pressure). In Chapter 11 we learn about another type of friction, the friction

that air or water exerts on a solid object moving through the air or water—a so-called drag force.

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3.4 Friction - SharpSchool