Flat Friction - Engineering Mechanics Statics - Lecture Slides, Slides of Mechanical Engineering

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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

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

a  m  Fx  m  P  Fk 

 Experiment Shows That The Resisting Friction Force Follows a General Profile as Noted in Fig.c Below

Coefficient of Friction cont.

• Similarly After Break-Away, The Coefficient of Friction Under Moving, or KINETIC, Conditions

 k  Fk / N

 Thus if μs or μk is Known, These Friction Forces Can Be Calculated a-Priori

F N

F N

k k

m s









 NOTE: Before Break- Away the Fiction Force Does NOT = Fm

• Before Impending Motion F (^) friction   P Coefficient of Friction Surfaces μs μk Steel on steel (dry) 0.6 0. Steel on steel (greasy) 0.1 0. Teflon on steel 0.041 0. Brake lining on cast iron

0.4 0. Rubber tires on dry pavement

0.9 0. Metal on ice 0.022 0.

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  Fm   sN

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

 Fy ^0  N  Mg cos^  s

 Apply Equilibrium Analysis

F (^) friction  Fm   sN

 Summing Forces:

 s

 s

 N  Mg cos  s  Fx ^0   sN  Mg sin^  s

or 0  s Mg cos s  Mg sin  s

s s

s s (^) Mg

Mg  

  tan cos

sin so  

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

 k  tan  k

 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

 (^) k N  Fk  Mg sin

 So the block accelerates per Newton’s Eqn

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

Angle of Friction – cont.

• 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

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   36.87 Sliding, Draw the F.B.D.

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

F lb

Fx lb lb F

80

300 5

3 0 100

  

    

60 (only)

1. 25 240 F lb

F N lb m

m s  

 Since The Block Can Only Generate 60lbs of Frictional Resistance When it Needs 80lbs, The Block WILL SLIDE

N lb

Fy N lb

240

300 5

0 4

 

   

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

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.

F F P

F N W

M Nx W b Ph

  

  

   





