Transcript ENGR-36_Lec-21_Flat-Friction

Engineering 36
Chp08:
Flat Friction
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
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Outline - Friction
 The Laws of Dry Friction
• Coefficient of Static Friction
• Coefficient of Kinetic (Dynamic) Friction
 Angles of Friction
• Angle of Static Friction
• Angle of Kinetic Friction
• Angle of Repose
 Wedge & Belt Friction
• Self-Locking & Contact-Angle
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Friction Physics
 When Two Bodies in Contact Attempt to Move
Laterally (Sideways) Opposing Tangential
Forces Develop Between The two bodies
• The Tangential Force is Called FRICTION
– Friction Forces Caused Primarily by
Surface MicroRoughness
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Coefficient of Friction
 Consider the Block of Weight W, Balanced by the
Normal Reaction Force N.
 A Lateral Push, P, is Applied to the Block, The Push will
Be Balanced, Up to a Point, By The Friction Force, F
 The Friction Force Rises With P Until The Block Reaches
the “Break-Away” Condition and Motion Ensues
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Coefficient of Friction cont.
 After Break-Away, The Block Accelerates per
a m
F
x
m
 P  Fk 
 Experiment Shows That The Resisting Friction Force
Follows a General Profile as Noted in Fig.c Below
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Coefficient of Friction cont.2
 Experiments Also Show that the MAXIMUM Resisting
Force Just Prior to Break Away, Fm, is LINEAR With
The Normal Contact Force, N
• The Constant of (Linear) Proportionality is Called the
Coefficient of STATIC Friction and is Defined by
 s  Fm / N
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Fm   s N
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Coefficient of Friction cont.3
 Similarly After Break NOTE: Before BreakAway, The Coefficient of
Away the Fiction Force
Friction Under Moving, or
Does NOT = Fm
KINETIC, Conditions
• Before Impending Motion
F friction   P
 k  Fk / N
 Thus if µs or µk is
Known, These Friction
Forces Can Be
Calculated a-Priori
Fm   s N
Fk   k N
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C o e ffic ie n t o f F rictio n
S u rfa c e s
µs
µk
S te e l o n s te e l (d ry)
0 .6
0 .4
S te e l o n s te e l (g re a s y)
0 .1
0 .0 5
T e flo n o n s te e l
0 .0 4 1
0 .0 4
B ra k e lin in g
o n c a s t iro n
0 .4
0 .3
R u b b e r tire s o n
d ry p a ve m e n t
0 .9
0 .8
M e ta l o n ic e
0 .0 2 2
0 .0 2
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Rigid Body Friction
 The Actions of Friction Forces
Divide into 4 Distinct Cases
1. NO Lateral Forces to Generate
Resisting Tangential Forces →
NO Friction Forces (Fig.a)
2. The applied force tends to move
body along the surface of contact
but are NOT large enough to set
it in motion (Fig.b)

NOT At BreakAway so
F friction  F m   s N
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R.B. Friction cont.
 The Actions of Friction Forces
Divide into 4 Distinct Cases
3. The applied forces are such that
the body is just about to slide,
MOTION IS IMPENDING (Fig.c)

The Static Case Where The
Friction Equation CAN Be Applied
F  Fm   s N
4. The body Slides under the action
of the applied forces (Fig.d)

The equations of Static equilibrium
no Longer Apply. (Kinetic case)
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Angle of Friction
 Consider the Situation
Depicted at Right
s
• Block of Mass M
• Angle of Inclination s
• Impending Motion
 Thus
s
 Summing Forces:
F
• Static Equilibrium Applies
• Anti-Sliding Friction
Force Described by
F friction  F m   s N
 Apply Equilibrium
Analysis
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y
 0  N  Mg cos  s
 N  Mg cos  s
F
or
so
x
 0   s N  Mg sin  s
0   s Mg cos  s  Mg sin  s
s 
Mg sin  s
Mg cos  s
 tan  s
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Angle of Friction cont.
 Thus The CoEfficient of
Friction is EASILY
Measured with a Simple
Inclined Plane
 Once Motion Begins
Experiment Shows That
The Angle of Inclination
can be REDUCED
without Halting the Slide
 Reducing The Angle to
Where Motion Stops
Defines the Kinetic
Coefficient of Friction
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 k  tan  k
 For Angles of
Inclination, , Greater
than s The Body Slides
per μk and
 k N  Fk  Mg sin 
 So the block
accelerates per
Newton’s Eqn
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Angle of Friction – 4 Cases
 The Angle of Friction Also Divides into 4 Cases
1. Angle of Inclination,  = 0 → NO Friction (Fig.a)
2.  <s → Below BreakAway so the The block is in not
motion and friction force is not overcome (Fig.b)
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Angle of Friction – 4 Cases cont.
 The Angle of Friction Also Divides into 4 Cases
3. With increasing angle of inclination, motion will soon
become impending. At that time, the angle between
R and the normal will have reached its maximum
value s (Fig.c)

The value of the angle
of inclination corresponding
to impending motion is called
the ANGLE OF REPOSE
 S  Angle of Repose
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Angle of Friction – cont.2
 The Angle of Friction Also Divides into 4 Cases
4. With Further increases in the angle of inclination,
motion occurs and the Resultant force, R, Applied by
the Inclined plane on the Body no Longer Balances
the Gravity Force (Fig.d).

The Body is not in Equilibrium
so This case Will NOT be
Considered in this STATICs
Course.

You’ll Take up This Subject
in The DYNAMICS Course
at The Transfer Institution
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ME104 Dynamics @ UCB
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Classes of Friction Problems

I.
II.
III.
Static Force Problems Involving Friction Tend to
Divide into Three Classes
All of the applied forces are given and the
coefficients of friction are known; need to
determine whether the body considered will
REMAIN AT REST or SLIDE.
All applied forces are given and the motion is
known to be impending; need to determine the
value of the COEFFICIENT OF STATIC FRICTION.
The static friction coefficient is known, and it is
known that motion is impending in a given
direction; need to determine the MAGNITUDE OR
DIRECTION OF ONE OF THE APPLIED FORCES
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Example: Class I
 Check Equilibrium
  36.87 
• Determine the Value of the
Force REQUIRED for
Equilibrium. Assuming That
Friction Directly Opposes
Sliding, Draw the F.B.D.
 A 100-lb force acts on a 300-lb
block on an inclined plane.
The coefficient of friction
between the block and the
plane are µs = 0.25 and
µk= 0.2.
 Determine whether or not the
block is in equilibrium and find
the value of the friction force.
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Example: Class I cont.
 For the F.B.D. Write
Eqns of Equilibrium
F

0 N 
300 lb
5
N  240 lb
F

y
4
x
 0  100 lb 
3
300 lb  F
5
F   80 lb
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 Thus To Maintain
Equilibrium. the Friction
Forces MUST Add 80lb to
the Existing 100lb Push
 Now Given µs, Find MAX
possible Value for F
Fm   s N  0 . 25  240 lb

Fm  60 lb
(only)
 Since The Block Can
Only Generate 60lbs of
Frictional Resistance
When it Needs 80lbs, The
Block WILL SLIDE
Bruce Mayer, PE
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Example: Class I cont.2
 To Find The ACTUAL
Value for the Friction
Force, Note that the
Block is in Kinetic motion
(Sliding) so µk Applies
F friction  F k   k N  0 . 2  240 lb

F friction  48 lb
 The Actual Situation
Displayed in Diagram at
Right
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 Note that the Forces are
UNBALANCED.
• The Block will Accelerate
Downward due to the Net
Lateral Force of
32lbs (180-148)
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Example – Class III
 A large rectangular shipping crate
of height h and width b rests on the
floor. A Dock Worker Applies a
force P to the Upper-Right Edge of
the Crate. Assume that the material
in the crate is uniformly distributed
so that the weights acts at the
Geometric centroid of the crate.

Determine
a) the conditions for which the crate is on the verge of sliding
b) the conditions under which the crate will tip about point A
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Example – Class III cont
 Draw a Free-Body-Diagram of the
Crate, noting that the Pressure
Applied by the Floor Decreases at
the Right-Bottom Edge as The
Worker Applies a Greater Push.
 From The FBD the Eqns of
Equilibrium Including the Friction
Force F:
 Fx  0  F  P
F
M
y
 0  N W
A
 0  N x  W  0.5 b   P h
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Example – Class III cont.2
 In Equilbrium
• F=P
• N=W
 Substituting These Values in the
moment equation Yields The
Location for the Application of the
Resultant Normal Force. By ∑MA=0
0  W x  W  0.5 b   P h  x  0.5 b 
 If the crate is on the
verge of sliding F=µsN
where µs is the
coefficient of static friction .
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P
Ph
W
W
Psliding  Fs   s N
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Example – Class III cont.3
 Now, if the crate is on the verge on
tipping it is just about to rotate
about point A, so the crate and the
floor are in contact ONLY at PointA. Therefore the Normal-Resultant
Application Point has moved to
Point-A, and Hence x=0
 Setting x to Zero in the Moment
Equation Yields the TIPPING
Condition of ∑MA = 0:
W
b
 Ph
2
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
P 
Wb
2h
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Example – Class III cont.5
 Which will Happen FIRST;
Tipping or Sliding?
 Note that tipping will occur before
sliding, provided that Psliding >
Ptipping. So if P increases until some
Sort of motion occurs Tipping will
occur BEFORE Sliding by:
b
Pslide   sW
Ptip 
W
2h

TIPPING

 sW 
b
2h
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W
Pslide  Ptip

 s
b
2h
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Example – Class III cont.6
 Run The Numbers. Make Some
Realistic Assumptions
•
•
•
•
b = 3 feet
h = 5 feet
W = 300 lb
µs = 0.5 for Wood on ConCrete
http://www.adtdl.army.mil
/cgi-bin/atdl.dll/fm/334.343/apph.pdf
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Bruce Mayer, PE
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Example – Class III cont.
 ReCall the
Tipping Criteria
 s , min ,tip  b
2h
 In this case b 2 h  3 /  2  5   0 . 3
 So Since The Actual Friction Factor
of 0.5 EXCEEDS this value, then
the Crate WILL, in fact, TIP OVER
 Calc The Overturning and Sliding
Pushes
b
3
Ptip 
W 
2h
2 5
300 lb  90 lb
Pslide   sW  0 . 5  300 lb  150 lb
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CoEffs
of
Friction
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WhiteBoard Work
Let’s Work
This Nice
Problem

Two blocks A and B have a weight of 10 lb and 6 lb, respectively. They are
resting on the incline for which the coefficients of static friction are µA = 15%
and µB = 25%. Determine the incline angle for which both blocks begin to
slide. Also find the required stretch or compression in the connecting spring
for this to occur. The spring has a stiffness of k = 2 lb/ft.
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Engineering 36
Appendix
dy
dx
 sinh
µx
T0

µs
T0
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
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Fun with Friction
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Measure Coeff ofDynamic Friction
 Use concept
of SpringMass
Damped
Harmonic
Motion as
studied in
Physics and
Engineering25
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3 kN
5m
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3 kN
7m
5m
Bruce Mayer, PE
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