chapter 6 friction

8/3/2019 Chapter 6 Friction

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CHAPTER 6

FRICTION


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Lecture Goals

The Laws of Dry Friction

Coefficient of Static Friction

Coefficient of Kinetic Friction

Angles of Friction

angle of static friction

angle of kinetic friction

angle Repose


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Friction

A friction force is a shear force that acts tangent to

the surface of contract between two bodies. This

force opposes sliding motion between bodies.


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Friction

The force of static friction is maximum when the

two bodies in contact are just ready to slip relative

to each other.

The maximum force of friction increases as the

normal force between the bodies increases.


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Friction

There are two force horizontal forces, Pmax, which

is the magnitude of maximum horizontal force

which the object can resist.

where ms is the coefficient of static friction and N is

the normal force.

max sP Nm


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Friction

The other horizontal force, Fk, due to friction is the

kinetic-friction force which the object can resist.

where mkis the coefficient of kinetic friction and Nis the normal force.

k kF Nm


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Friction

The coefficients of friction ms and mkdo not dependupon the area of the surfaces in contact. Both

coefficients depend strongly on the nature of the

surface contact.

metal on metal ms - 0.15 - 0.6

metal on wood ms - 0.20 - 0.6

wood on leather ms - 0.25 - 0.5


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Friction

Four different situation may occur when a rigidbody is in contact with a horizontal surface.

The force applied to the body

do not tend to move it alongthe surface of contact.

The applied force tend to

move body along the surfaceof contact but are not large

enough to set it in motion.

a

b


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Friction

Note

Since there is no evidence that

the maximum value of the

static-friction force has beenreached the equation

can not be used to determine

the friction force.

max sF Nm


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Angle of Friction

k kF Nm

N

FK

ma xm

SS m tan

It is sometimes found convenient to replace thenormal force N and friction force F by their

resultant R.

maxstan

FN

The angle s is known as the angle ofstatic friction.

recall that:

N

FMAX

S

m


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Angle of Friction

Another example that will show how the angle offriction may be used to advantage in the analysis of

certain types of problems. For block on an incline.

The block is in not motion

and friction force is not

overcome.


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Angle of Friction

If we continue to increase the

angle of inclination, motion will

soon become impending. At the

time, the angle between R andthe normal will have reached its

maximum value s. The value

of the angle of inclination

corresponding to impendingmotion is called theangle of

repose.


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Example ProblemFriction I

Determine the value of the friction force required to

maintain equilibrium.

x

y

3

0 100 lb 300 lb5

80 lb

40 300 lb

5240 lb

F F

F

F N

N

The force needed to maintain equilibrium is an 80-lb

force directed up and to the right, the tendency is for

the block to move down the incline.


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Example ProblemFriction I

Maximum friction force is may be computed.

max s

0.25 240 lb 60 lb

F Nm

Since the value of the force required to maintain

equilibrium (80-lb) is larger tan the maximum valuewhich may be obtained (60-lb) equilibrium will not be

maintained and the block will slide down the plane.


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Example ProblemFriction II

Two blocks are connected by a light, flexible cable that

passes over a friction-less pulley. Block A weighs 500

N and block B weighs 200 N. The coefficient of static

friction between block A and ramp is ms = 0.3. Determine whether the

two blocks are in

equilibrium

Determine whether block

A will slide down the

ramp after the cable is cut.


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x

y

30 500 N

5

3200 N 500 N

5

100 N

40 500 N5

400 N

F T F

F

F

F N

N

T

N

F

xy


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T

N

F

xy


Maximum friction force is may be computed.

max s

0.3 400 N 120 N

F Nm

Since the value of the force sign is -100 N assume

sense of F is wrong. The blocks are in equilibrium,

since Fs does not exceed the maximum static frictional

force 120N.


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x

30 500 N

53

0 N 500 N5

300 N

F T F

F

F

The cable is cut and the tension is zero.

The force exceeds the friction resistance 120 N so the

block will slide down the ramp.

T

N

F

xy


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Class ProblemFriction IITwo blocks are connected with a flexible cable thatpasses over a friction-less pulley. The weight of block A

is 25 lb, and the coefficient of friction is 0.20 on both

inclined surfaces. Determine the maximum and

minimum weights for the block B if the system remainsin equilibrium.


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lbsSinlbFx

5.12)30(25'

BLOCK A

lbsCoslbFy

65.21)30(25'

25lb

AT

Ffric30lbsNF 33.4)65.21(2.

max m

BLOCK B

)60(' SinWlbFx

)60(' CosWFy

)60(2.max WCosNF m


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)60(2.)60(sin3.45.12 CosWWFX


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Example ProblemFriction III

The movable bracket shown may be

placed at any height on 3-in.

diameter pipe. If the coefficient of

static friction between the pipe andbracket is 0.25, determine the

minimum distance x at which the

load W can be supported. Neglect

the weight of the bracket.


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x B A

B A

y A B

0

0

F N N

N N

F F F W

FA = mNA and FB=mNB ,which from sum in the x

direction will give FA= FB


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y A B

A B

A

0

0.25 0.25

2

F N N W

N N W

N W

m m

Use the moment about A to determine the distance,x.


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Using the moment about A to find x.

A B B0 6 in. 0.25 3 in. 1.5 in.

2 6 in. 0.25 2 3 in. 1.5 in.

12 in. 12 in.

M N N W x

W x W W W

Wx W x

The x=12 in.


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Example ProblemFriction IV

A large rectangular shipping crate ofheight h and width b is at rest on the

floor. It is acted on by a horizontal

force P. Assume that the material in

the crate is uniformly distributed sothat the weights acts at the centroid

of the crate.

(a) Determine the conditions for which the crate is on theverge of sliding.

(b) Determine the conditions under which the crate will

tip about point A.


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Example ProblemFriction IV

If the crate is on the verge of sliding F=msN where thecoefficient of static friction .

If the crate is on the verge ontipping it is on the verge of

rotation about point A, that is the

crate and the floor are in contact

only at A. Therefore x=0

0 0.52

Ph Wbb P

W h

sliding s sP F Nm


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THE WEDGE

A simple machine

Uses mechanical advantage based on the Inclined Plan

Can apply large forces

Can be used to produce small adjustments

a triangular object placed between 2 objects

to either hold them in place, or move onerelative to the other.


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If force P is large enough, the

block will rise

Since before the wedge can move, it must

overcome Fmax, then we can assume:


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The block shown supports a load of 700 lbs and is to be raised by forcing a wedge under it


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The block shown supports a load of 700 lbs and is to be raised by forcing a wedge under it.

The coefficient of friction on the 3 contact surfaces is .25. Calculate the force P needed to

begin moving the wedge. Neglect the weight of the wedge and block.

BEGIN WITH


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BLOCK


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Now consider

the wedge


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The ScrewCombination of an Inclined

Plane wrapped around a

cylinder (forming the path

and pitch) and a Wedgecoiled around the cylinder

Changes rotary motion to

linear motion

Very large mechanical

advantage


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When turned, a

screw converts rotary

motion into linear motion.

It is used in jacks, clamps,

measuring tools, valves, etc.

The Screw


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Nut Splitter


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Converting Linear Motion

Turning the screw one complete revolution

will move it into or out of an object a distance

to the pitch of the screw, or .1 in this case.Therefore, a screw can be used to convert

circular motion into linear motion

1 Revolutions Distance = Pitch of Screw


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Calculating the Mechanical

Advantage of a Screwp = Screw pitch (Load distance)

p = 3.1416rE = Effort arm length

L = Weight of the Load

E = Effort ForceEI

EA

(Ideal)

(Actual)


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Ideal Mechanical Advantage is a function of

distance. It disregards friction, and other

causes that reduce the efficiency of a machine.


Advantage of a Screw

2prEpIMA =


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The Screw

2prEpIMA =


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Ideal Mechanical Advantage can also be

calculated if you know the weight of the Load

and the ideal Effort force (perfect world

scenario) needed to move it.


Advantage of a Screw

LEI

=2prEpIMA

=


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Advantage of a ScrewActual Mechanical Advantage is a function of

weight. It is a measure of the actual performance

of the simple machine and will always be lessthan the IMA due to friction, and other energy

absorbing phenomena.

AMA LEA

=


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Comparison of Mechanical Advantage:


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Other Uses

Because of the mechanical advantage of a screw,

it is also useful for moving heavy loads. Such

screw driven devices include garage door openers

and jacks.

SCREWS


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SCREWS

A bolt and screw in combination

work together as 2 wedges.

What if you could unwrap the

threads off of a screw?

= lead angle of screw


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1. Screw is pushed down by

screwdriver

2. A moment is generated by therotation of the screwdriver


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case 2

The moment dominates and

the screw moves up.

case 1The axial load dominates and

the screw moves down

S l P bl 8 5


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Sample Problem 8.5

A clamp is used to hold two pieces of

wood together as shown. The clamphas a double square thread of mean

diameter equal to 10 mm with a pitch

of 2 mm. The coefficient of friction

between threads is ms

= 0.30.

If a maximum torque of 40 N*m is

applied in tightening the clamp,

determine (a) the force exerted on the

pieces of wood, and (b) the torque

required to loosen the clamp.

SOLUTION

Calculate lead angle and pitch angle.

Using block and plane analogy with

impending motion up the plane, calculate

the clamping force with a force triangle.

With impending motion down the plane,calculate the force and torque required to

loosen the clamp.

Sample Problem 8 5


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Sample Problem 8.5SOLUTION

Calculate lead angle and pitch angle. For the doublethreaded screw, the leadL is equal to twice the pitch.

30.0tan

1273.0mm10

mm22

2tan

ss

r

L

m

pp 3.7

7.16s

Using block and plane analogy with impending

motion up the plane, calculate clamping force with

force triangle.

kN8

mm5

mN40mN40

QrQ

24tan

kN8tan W

W

Qs

kN97.17W

Sample Problem 8 5


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Sample Problem 8.5

With impending motion down the plane, calculate

the force and torque required to loosen the clamp.

4.9tankN97.17tan QW

Qs

kN975.2Q

m105N10975.2mm5kN975.2

33

rQTorque

mN87.14 Torque

chapter 6 friction

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