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# Friction& Dissipative Mechanics

[...]  gobbledygook about friction and molecules.
Ray Bradbury  (1920-2012)  in  Fahrenheit 451

### Related Links (Outside this Site)

Laws of dry friction & coefficients of friction   Engineering Archives.

Wikipedia :   Friction   |   Viscosity.   |   Lubrication.

### Videos :

MIT 8.01 Lecture 8, Fall 1999 (48 minutes)  by  Walter Lewin.
Conceptual Pysics, with  Paul G. HewittCalculating Friction Forces.
The Physics of Waterslides (4:52)  by  Liv Boeree  (2017-12-03).

Phone-Book Friction against Army Tanks   (Mythbusters, 2008-09-18)
Will phone books lift a 1600 kg car?   (Fast, Furious & Funny, with Michael "VSauce" Stevens,   2012-11-23)

## Friction

(2010-11-03)   Newtonian coefficients of sliding friction   (dry friction)
The static coefficient of friction is slightly larger than the kinetic one.

When two dry solids at rest press against each other they won't move as long as the force applied  tangentially  to their common surface of contact does not exceed a certain limit, which is  roughly  proportional to the  normal  force that presses them against each other.

The coefficient of proportionality is called the  static coefficient of friction  and is often denoted by the symbol  m  or  ms  (the Greek letter mu is pronounced  mew ).

As soon as such solids start moving, a lesser friction force appears in the direction which opposes the motion.  It is also proportional to the normal force applied but the coefficient of proportionality  (called  kinetic coefficient of friction  or  sliding coefficient of friction)  is always  less  than in the static case described above.  This means that, unless the normal force changes, a constant tangential force large enough to overcome static friction will impart motion at a positive  (nonzero)  acceleration.

By definition, the "tangential force" is in the same plane as the surface of contact and the "normal force" is perpendicular to it.

The above is a valid approximation for dry surfaces under moderate pressure.  At high pressure,  seizure  can occur, whereby the two solids will not slide at all under any reasonable force.  This happens, in particular, for polished surfaces between solids consisting of the same material:  Pressure makes the interface between the two solids effectively disappear and they may well behave as a single crystal, with large binding forces between nearby atoms.  The solids may well break before they slide.

When a liquid lubricant is used, the above does not apply at all.  Instead, the slightest tangential force will produce sliding and the resistive forces depend on the speed of sliding  (the normal forces are then largely irrelevant).

Friction Coefficients   |   Table of coefficients of friction   |   Engineer's Edge
Friction and Tribology by Roy Beardmore.

Static or kinetic regime?
In the figure at right,  the table is frictionless, the rope is massless, the pulley is both.  The coefficients of static and kinetic friction between m1 and m2 are 0.539 and 0.439.  With m1 = 3.67 kg, m2 = 5.71 kg, m3 = 7.91 kg,  what are the accelerations of m1 and m2?  What's the tension of the string?

Let's use the following notations:

• F  is the horizontal force of friction between m1 and m2.
• a  is the acceleration of  m2.
• b  is the acceleration of  m1  and  m3
• T  is the tension of the string.
• g  =  9.80665 m/s2  is the gravitational field  (standard value).
• u  =  0.439  is the coefficient of  kinetic friction.

Because of Newton's second law, the following  3  equations hold:

• m2 a  =  F
• m1 b  =  T - F
• m3 b  =  m3 g - T

Therefore:

T  =  m3 (g - b)       and      F  =  m3 (g - b) - m1 b  =  m2 a

Under the  static regime  where  m1  and  m2  are at rest with respect to each other, we have  a = b  so that the last of the above equations boils down to:

m3 g = (m1+m2+m3) a     Therefore:
F  =  m2 a  =  g m2 m3 / (m1+m2+m3)

The coefficient of static friction must exceed the ratio of  F  to the normal force  (m1 g).  Thus, in the static regime, we must have:

F / (m1 g)  =  m2 m3 / [m1 (m1+m2+m3)]  <  0.539

With the given values of the masses, that inequality doesn't hold  (since the  left-hand side  is more than 0.711).  We are thus faced with the kinetic regime where  a  and  b  need not be equal.  Instead, motion is determined by the following additional equality, involving the kinetic coefficient  (u).

F  =  u m1 g     [where u = 0.439]

Therefore, the acceleration of  m2  is:

a  =  F/m2  =  g u m1/m2  =  0.282 g  =  2.767 m/s2

The acceleration  b  of  m1  (and also of  m3)  is obtained as follows:

F  =  g u m1  =  m3 (g-b) - m1 b     Therefore:
b  =  g (m3 - u m1) / (m1 + m3)  =  0.544 g  =  5.334 m/s2

Finally, the tension of the string is:

T  =  m3 (g - b)  =  g (1+u) m1 m3 / (m1+m3)  =  35.4 N

Mark Taylor (2010-10-31)   Ladder leaning against a frictionless wall
What's the least angle at which a ladder can rest against a frictionless wall?   When is such a ladder safe to climb?

Let  m  be the static coefficient of friction between the floor and the ladder  (whose other end is resting against a frictionless wall, consisting of something very slippery like wet ice or glass).

Let  L  be the length of the ladder and  q  be its inclination  (the angle between the ladder and the horizontal).  Let's introduce a parameter  k  that indicates the position of the center of gravity of the loaded ladder.  The combined center of gravity of the ladder and the person who climbs it  (if any)  is located in the vertical line which intersects the ladder at a distance  k L  from its bottom  (for an ordinary ladder with nobody on it, k = 0.5).

At equilibrium, the ladder does not move vertically or horizontally, so the sum of the horizontal components of all existing forces is zero; so is the sum of all vertical components.

• The only downward force is the weight of the whole thing  (ladder + climber)  and is balanced by the upward vertical force exerted by the floor at the bottom of the ladder (since the frictionless wall cannot exert any vertical force).
• The horizontal forces exerted by the floor and the wall have the same magnitude  F  (they have opposite directions).

Furthermore, the ladder doesn't rotate.  Therefore, the torques of all applied forces cancel.  The torques about the point of contact with the floor reduce to the torque of the vertical weight (of magnitude Mg) and the purely horizontal force  (F)  exerted by the frictionless wall.  Therefore:

(k L cos q) Mg   =   (L sin q) F

The ladder will not move as long as the tangential friction force (of magnitude F) exerted by the floor does not exceed its allowed maximum for the applied normal force (of magnitude Mg).

F   <  m Mg

Combining the two relations above, we obtain:

k Mg / tan q   <  m Mg

Which is to say that   q  >  arctg ( k / m )

The unloaded ladder will stay in place as long as this inequality is satisfied for  k = 0.5  but it's safe to climb to the last rung only when the equality is satisfied when  k  is nearly equal to  1.

(2011-04-25)   Rolling Resistance   (rolling friction)
A complicated form of dry friction.

Rolling Friction by Ron Kurtus.
Rolling Resistance (Wikipedia)   |   Railway Industry Overview (© 2003, AREMA® )

(2011-02-11)   Skidding before rolling on an horizontal plane...
Motion of a spinning cylinder of radius R, mass M & moment of inertia J.

(2007-09-25)   Oscillations due to friction
Walter Lewin's method for drawing dotted lines on a blackboard.

(The pseudo-periodicity of earthquakes.)

MIT 8.01  Some of Walter Lewin's best lines  (Fall 1999)

(2010-12-19)   Coefficient of restitution  (e).
The ratio of the initial to final  closing speed.

When a ball dropped from a height  h  bounces off the floor to a height  r h,  the dimensionless coefficient  r  is called the  coefficient of restitution.

Technically, the  coefficient of restitution  is defined as the ratio of the initial to final  closing speed  (with a change of sign that makes  e  positive in the common case where the closing reverses direction after a shock).  It's thus the rato of the  speed of approach  to the  speed of separation.

• e = 1   in an ellastic collision.
• e = 0   if the colliding objects stick together after the collision.

Values of some coefficients of restitution  (e)
e = eNColliding ObjectsSource
0.96Max.Wham-O
Super Ball
0.94Min.
0.94Min.Ping-pong
ball
Hard surface ITTF
0.876Min.Regulation table ITTF rules (2005)
0.87Max. Billiard ballCue (phenolic tip) Dave Alciatore
0.71Min.Cue (leather tip)
0.83Max.Golf ballGolf club US Golfing Association  (1998)
0.78Typ.High-Speed Video
0.76Max.Tennis ballHard floor ATP regulations  (at 7 m/s)
(A. Roux & J. Dickerson)
0.73Min.
0.578Max.BaseballWall of northern white ash wood Baseball regulations (at 85 ft/s)
(D. Kagan  &  D. Atkinson)
0.514Min.
0.1Fur ballWall Street [Legendary]  dead cat bounce
0.0Min.Butter ballFrying pan

The above values correspond to the so-called  normal  coefficients of restitution, which are appropriate for head-on collision with little or no tangential speed.  Otherwise, if the relative velocity of the two objects is not perpendicular to their surface of contact, the ratio of the final to the initial tangential speed is defined to be  another  type of coefficient of restitution whose value is only very loosely related to the normal coefficient of restitution discussed so far.

### Resilience :

Resilience is  the ability to bounce back

The coefficient of restitution for collisions of happy balls, unhappy balls, and tennis balls  by  Rod Cross.
Coefficients of Restitution by  Jamin Bennett & Ruwan Meepagala (Physics Factbook by Glenn Elert)
Thread:  Coefficient of Restitution by  ajinx999  (Physics Forums, 2008-06-30)
Video:  Bouncing Balls by Roger Bowley (University of Nottingham) in Sixty Symbols by Brady Haran.