Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • ENGR-36_Lec-21_Flat-Friction.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp08:Flat

Friction



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



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



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



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



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



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



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



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)



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



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



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)



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



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



ME104 D

ynamics @

UC

B



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



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



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



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



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



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



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



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



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



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



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



CoEffsof

Friction



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.



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix 00

sinhTµs

Tµx

dxdy



Fun with Friction



Measure Coeff ofDynamic Friction Use concept

of Spring-Mass Damped Harmonic Motion as studied in Physics and Engineering-25



3 kN 3 kN

5 m 5 m7 m

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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