bruce mayer, pe licensed electrical & mechanical engineer [email protected]
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Engineering 36. Chp08: Flat Friction. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]. Outline - Friction. The Laws of Dry Friction Coefficient of Static Friction Coefficient of Kinetic (Dynamic) Friction Angles of Friction Angle of Static Friction - PowerPoint PPT PresentationTRANSCRIPT
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 36
Chp08:Flat
Friction
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Coefficient of Friction cont. After Break-Away, The Block Accelerates per
kx FPmFma Experiment Shows That The Resisting Friction Force
Follows a General Profile as Noted in Fig.c Below
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
NFms / NF sm
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Coefficient of Friction cont.3 Similarly After Break-
Away, The Coefficient of Friction Under Moving, or KINETIC, Conditions
NFkk / Thus if µs or µk is
Known, These Friction Forces Can Be Calculated a-Priori
NFNF
kk
sm
NOTE: Before Break-Away the Fiction Force Does NOT = Fm
• Before Impending Motion
PF frictionCoefficient of Friction
Surfaces µs µk Steel on steel (dry) 0.6 0.4
Steel on steel (greasy) 0.1 0.05
Teflon on steel 0.041 0.04 Brake lining on cast iron
0.4 0.3
Rubber tires on dry pavement
0.9 0.8
Metal on ice 0.022 0.02
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
NFF smfriction
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
NFF sm 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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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Angle of Friction Consider the Situation
Depicted at Right• Block of Mass M• Angle of Inclination s
• Impending Motion Thus
• Static Equilibrium Applies• Anti-Sliding Friction
Force Described by
sy MgNF cos0
Apply Equilibrium Analysis
NFF smfriction
Summing Forces:
s
s
sMgN cos
ssx MgNF sin0
sss MgMg sincos0or
ss
ss MgMg
tan
cossinso
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
kk tan
Reducing The Angle to Where Motion Stops Defines the Kinetic Coefficient of Friction
For Angles of Inclination, , Greater than s The Body Slides per μk and
sinMgFN kk So the block
accelerates per Newton’s Eqn
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
Repose of Angle S
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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 beConsidered in this STATICsCourse. You’ll Take up This Subject
in The DYNAMICS Courseat The Transfer Institution
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
ME104 D
ynamics @
UC
B
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Classes of Friction Problems Static Force Problems Involving Friction Tend to
Divide into Three ClassesI. 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.
II. All applied forces are given and the motion is known to be impending; need to determine the value of the COEFFICIENT OF STATIC FRICTION.
III. 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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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Class I
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.
Check Equilibrium• Determine the Value of the
Force REQUIRED for Equilibrium. Assuming That Friction Directly Opposes Sliding, Draw the F.B.D.36.87
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Class I cont.
For the F.B.D. Write Eqns of Equilibrium
Thus To Maintain Equilibrium. the Friction Forces MUST Add 80lb to the Existing 100lb Push
Now Given µs, Find MAX possible Value for F
lbF
FlblbFx
80
300531000
(only)6024025.0
lbFlbNF
m
sm
Since The Block Can Only Generate 60lbs of Frictional Resistance When it Needs 80lbs, The Block WILL SLIDE
lbN
lbNFy
240
300540
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
lb
lbNFF
friction
kkfriction
48
2402.0
F
Note that the Forces are UNBALANCED.• The Block will Accelerate
Downward due to the Net Lateral Force of 32lbs (180-148)
The Actual Situation Displayed in Diagram at Right
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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.
Determinea) the conditions for which the crate is on the verge of slidingb) the conditions under which the crate will tip about point A
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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:
x
y
A
0
0
0 0.5
F F P
F N W
M Nx W b Ph
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
If the crate is on the verge of sliding F=µsN where µs is the coefficient of static friction .
0 0.5 0.5 PhWx W b Ph x bW
sliding s sP F N
W
P
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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 Point-A. 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:
hWbPPhbW22
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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:
WhbPWP tipsslide 2
tipslide PP TIPPING
hbW
hbW ss 22
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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/3-34.343/apph.pdf
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example – Class III cont. ReCall the
Tipping Criteria hb
,tips 2min,
In this case 3.052/32 hb 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
lblbWhbPtip 90300
523
2
lblbWP sslide 1503005.0
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
CoEffsof
Friction
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
WhiteBoard Work
Let’s WorkThis NiceProblem
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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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
Engineering 36
Appendix 00
sinhTµs
Tµx
dxdy
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Fun with Friction
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Measure Coeff ofDynamic Friction Use concept
of Spring-Mass Damped Harmonic Motion as studied in Physics and Engineering-25
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
3 kN 3 kN
5 m 5 m7 m