PLANE FRICTION

Climber on ''Spud Boy''Clark Canyon, California

FRICTION Can’t live with it and sure can’t live without it

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Friction

What are the benefits and pitfalls of high or low friction?

Disadvantages Low friction means slipping Uses energy – (loss in engines, etc.) Creates wear on surfaces (abrasion)

(i.e. worn out tires)

Advantages Friction means traction or stability (tires) Allows for life as we know it (sit/stand) Makes a predictable model of how to use it

(i.e.: striking a match)

Forces of Friction Consider a block of weight 'W' resting on another block. A force 'P' is

applied such that motion is impending The application of the equations of equilibrium dictate:

A force 'N' opposite and equal to 'W' exists. This is called the normal* force and is always perpendicular to the surface supporting weight 'W'

A friction force, Ff, opposes 'P'. This force always opposes the direction of impending motion. It is also parallel to the surface supporting weight 'W'

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* In this context, 'normal' is defined as a geometric perpendicular

P

Ff

W

N

(Impending Motion)

Coefficient of Static Friction The first law of friction states the force of friction is

proportional to the normal force. That is:

Notice this is a linear relationship. The larger the normal force, the greater the force of friction by an equal factor

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P

Ff

W

N

(Impending Motion)

F f = ⋅N

ExamplePush a 200 lbf box across the floorPush a 20 lbf box across the floor

The force P to overcome developed friction Ff will be 10 times greater

Coefficient of Static Friction Vectors 'Ff' and 'N' form resultant 'R'. The resultant vector 'R'

is at angle φ with the normal, the friction angle The tangent of the friction angle is known as the coefficient of

static friction and is denoted by the symbol µs

A low value of µs denotes a smooth surface. It is also affected by whether or not the surfaces are wet, and if so, wetted with what

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P

Ff

W

N

(Impending Motion)

φR

s=F f

N

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Coefficients of Friction Values of µs can be found in the appendix to this online text.

They are also found in a number of handbooks and on various Internet websites. However, it is important to understand the following Values of friction coefficients are approximate at best. They

are intended only for guidance Exercise care when using friction coefficients. Multiple

independent references should be used For any specific application the ideal method of determining

the coefficient of friction is through testing

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Coefficients of Friction Coefficients of friction are sensitive to atmospheric dust and

humidity, oxide films, surface finish, velocity of sliding, temperature, vibration, and extent of contamination. In many cases the degree of contamination is perhaps the most important single variable

A short table is included to illustrate how the coefficient of friction is affected by surface films. When a metal surface is perfectly clean in a vacuum , the friction is much higher than the normal accepted value and seizure can easily occur

Effect of films on coefficient of static friction

Material Clean Dry Thick Oxide Film Sulfide Film

Steel-Steel 0.78 0.27 0.39

Copper-Copper 1.21 0.76 0.74

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Solving a Friction Problem Friction problems can be solved in one of three ways

Analytically Trigonometrically Graphically

The analytical method is a simple sum of forces analysis. It is the preferred method in most cases

The trigonometric method requires the construction of a force triangle. All unknowns are solved using basic trigonometric identities. This method can be easier to apply for certain types of friction problems than the analytical method

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Solving a Friction Problem Graphical solutions also require the construction of a force

triangle. However, it MUST be drawn to scale Such solutions will generally have a single vector with a

known magnitude and direction. The other two vectors will have known directions but unknown magnitudes

Start by drawing the vector with known magnitude and direction to scale

Then draw the lines of action for the vectors with unknown magnitiude. Make sure they are drawn in a proper direction and tip-to-tail

Where the two unknowns intersect defines the magnitude of the two unknowns. These magnitudes can be measured with a ruler. Although not as accurate as the other two methods, it is well within typical engineering tolerances

FBD of M2

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What must be the mass of M1 for impending motion to occur?µs = 0.3M2 = 10 kg(assume pulley is frictionless)

M2

M1

W = M 2⋅g = 10 kg⋅9.807m/ s2= 98.1N

P = M 1⋅g

F x = 0 = P−F f

F y = 0 = N−98.1

... P = F f and N = 98.1 NF f = ⋅N = 0.3⋅98.1 = 29.4 N

P = 29.4 N

M 2 =Pg

=29.4 N

9.807m/ s2 = 3.0 kg

P

Ff

N

W

Solving a Friction ProblemAnalytically

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Solving a Friction ProblemTrigonometrically

To solve this same problem trigonometrically, we need to resolve N and Ff into vector R and determine angle φ.

Draw a force triangle using W, P and R and identify angle φ.

P

Ff

N

W=98.1 N

φR

R98.1

P

φ

α

As illustrated on the previous page...W = M 2⋅g = 10 kg⋅9.807m/ s2

= 98.1NP = M 1⋅g

Sketch the force triangle. It need not be drawn to scale.Solve the force triangle using trigonometric identities.

= tan−10.3 = 16.7o

P = 98.1⋅tan 16.7 = 29.4N

R = 98.12 29.42

= 102.4N

M 2 =Pg

=29.4 N

9.807m/ s2 = 3.0 kg

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Solving a Friction Problem Graphically

A graphic solution starts similarly to a trigonometric solution. A major exception is the force triangle is drawn to scale. As such, select a scale in force

For this example, select a scale of 1” = 10 N and draw the force triangle. This means angle φ must also be measured correctly

R

98.1

P

φ

Draw the vertical weight vector 9.81 inches long.

Vector 'P' has a known direction but unknown length. Draw vector'P', placing its tail on the tip of 'W', some arbitrary length to the right.Vector R has a known direction but unknown length. Draw thisvector, placing the tip of 'R' on the tail of 'W', until it intersects 'P'

With a ruler, measure vectors 'P' and 'R'. They should be about2-7/8 and 10-1/4 in respectively. Apply your scale to arrive at:Apply your scale of 1 in = 10 N to arrive at:

P = 29 N and R= 102N

M 2 =Pg

=29 N

9.81m / s2 = 2.96 kg

10 N

Forces of Friction The previous discussion and example were contingent upon a

force 'P' of sufficient magnitude to cause impending motion. What happens if 'P' is less than that value and motion is not impending?

In this case, the equations of equilibrium simply dicatate that . However, F will be less than Ff as previously determined. As such, the definition and application of µs does not apply

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P

F

W

N

W = N and F = P

Example 2 – Motion Not Impending Assume a wooden crate filled with parts has a weight of

200 lbf and rests on a concrete floor. A force of P = 62 lbf is applied Is motion impending? What is the apparent coefficient of static friction? How does it compare to tabulated values? Why?

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P

F

W

N

Example 2 – Motion Not Impending By inspection, we see that N = 200 lbf and F = 62 lbf. From the appendix, we know that µs = 0.62. Since:

and P must equal Ff for motion to be impending, then:

Obviously, motion is not impending We can also calculate an apparent coefficient of friction

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62 lbf

Ff

200 lbf

N

s =F f

N

P = F f = s⋅N = 0.62⋅200 lb f = 124 lb f

s =FN

=62 lb f200 lb f

= 0.31

This apparent discrepency is because the coefficient of static friction is valid ONLY if motion is impending. Under circumstances defined in this example, µs cannot be applied or calculated.

Example 3 – Static Friction The Stillson wrench is used to

assemble and dissassemble pipe Given the dimensions shown, what

are the minimum values for coefficient of friction at each of the two jaws?

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P

20”

2”

2.5”

1”

A

B C

DE

F

Example 3 – Static Friction Assume members AB and DE are rigidly attached A friction force and its corresponding normal force develops at point 'A' Recognize the jaw assembly as a two-force member with forces acting

through points 'A' and 'D'. Then equilibrium requires the forces at 'A' to be balanced by the forces at 'D'

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A

B

DE

F

N

φR

4.5”

1”

Since this is a two-force member, thegeometry provides direction. The slopeof the resultant is 4.5:1.

Since only friction and normal forces actat point 'A', this slope is the tangent ofand , in this case, is the friction angle.The minimum coefficient of friction is:

A= tan =1

4.5= 0.222

Example 3 – Static Friction

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P

20”

2”

1”

C

D

F

Fc

Nc

φ

To determine the coefficient of friction of the lower jaw, we will need to develop the FBD of the handle-jaw assembly

We know the direction of vector D from the previous FBD is 4.5:1 (or 12.52o)

It is important to note that Nc and Fc are not equal to N and F from the previous FBD. This part of the assembly is not a two-force member; the applied force 'P' must be considered

This requires we write the equations of equilibirum for this FBD before we can do the friction analysis

Example 3 – Static Friction

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P

20”

2”

1”

C

D

F

Fc

Nc

φ

From previous geometry: = tan−1 14.5 = 12.52o

M c=−cos12.52o⋅D ⋅1 sin 12.52o⋅D⋅2 P⋅22

F y = cos 12.52o⋅D − N

F x= P − F c sin 12.52o⋅D

c=F cN c

-------------------------------------From M c : P = 0.0246⋅D

Substituting into F x : F c= 0.0246⋅D 0.217⋅D= 0.242⋅D

From F y : N = 0.976⋅D

. . . =0.242⋅D0.976⋅D

= 0.247

Kinetic Friction

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We looked at a situation where force 'P' caused impending motion. This is the only time when the coefficient of static friction can be applied

Then we looked at a scenario that did not result in impending motion. It was obvious the coefficient of friction cannot be used to aid in the solution of such problems

But what happens when the value of 'P' exceeds the value required for impending motion?

Kinetic Friction

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To answer this, consider pushing a heavy wooden box across a concrete floor. Experience tells us it can take quite some effort to get the box moving, but once moving, we no longer need to apply as much force to keep it moving

This illustrates the concept of sliding or kinetic friction. When the box is in motion, the coefficient of static friction does not apply. We must now introduce the concept of kinetic friction

A coefficient of kinetic friction is:

Common values are tabulated in the appendix

Let's investigate further the transition from static to sliding friction

k =F kN

P

Fk

W

N

Motion

Kinetic Friction

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Consider a weight resting on a flat surface with a normal force 'N' between the weight and the surface.

Let's apply force Fapp gradually, starting at zero. A friction force, 'f', will develop and increase linearly while the block remains static. This is indicated by the sloping red line

When 'f' reaches fs, motion will be impending. At this point, frictional force fs = µs N

As soon as 'f' exceeds fs, the weight will go in motion. The amount of force, Fapp required to maintain motion will decrease.

Friction Force = -Applied Force

Static Friction Dynamic Friction

Applied Force (Fapp)

fs = µsN fk = µkN

Ne

t F

orc

e(C

au

sin

g M

otio

n)

Fric

tion

Fo

r ce

Net F

orce

ff,k = µ N

fnet = Fapp - f

Fapp = Applied force

N = Normal force

Kinetic Friction

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The decrease in Fapp implies two very important concepts First, since the normal force remains

constant, the coefficient of kinetic friction must decrease

Second, if force Fapp is not reduced, the mass will accelerate. Acceleration of a body is covered in a dynamics course

If the body does go in motion and if Fapp is reduced to maintain a constant velocity, then the system can be analyzed using static methods

Friction Force = -Applied Force

Static Friction Dynamic Friction

Applied Force (Fapp)

fs = µsN fk = µkN

Ne

t F

orc

e(C

au

sin

g M

otio

n)

Fric

tion

Fo

r ce

Net F

orce

ff,k = µ N

fnet = Fapp - f

Fapp = Applied force

N = Normal force

Example 4 - Kinetic Friction

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A pallet of oak was delivered to a carpenter shop. The floor of the shop is also of oak. The pallet has a weight of 1000 lbf and needs to be pushed from one end of the shop to the other. How much force is required to start the pallet moving? How much force is required to keep it moving?

For motion to begin:

s = 0.62 (From appendix)

F y = N − 1000 . . . N = 1000 lb f

F x = F s − P . . . P =−F s

F s = 0.62⋅1000 = 620 lb f

. . . P = 620 lb f

Ps

Fs

1000 lb

N

Motion

Example 4 - Kinetic Friction

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A pallet of oak was delivered to a carpenter shop. The floor of the shop is also of oak. The pallet has a weight of 1000 lbf and needs to be pushed from one end of the shop to the other. How much force is required to start the pallet moving? How much force is required to keep it moving?

To sustain motion:

Pk

Fk

1000 lb

N

Motion

k = 0.48 (From appendix)

F y = N − 1000 . . . N = 1000 lb f

F x = F k − P . . . P =−F k

F k = 0.48⋅1000 = 480 lb f

. . . P = 480 lb f