Friction

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© 1998-2008 by Glenn Elert -- A Work in Progress
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Discussion

The force between surfaces in contact that resists their relative tangential motion (slipping).

Types: static & kinetic

Classical Approximations

microscopic description

miscellaneous stuff

Coefficients of Friction for Selected Interfaces
(in order of generally decreasing value)
μs μk interface 
  1.16 rubber – rubber
  1.02 rubber – concrete
  0.72 car tire – asphalt
  0.35 car tire – grass
0.8–1.0   skin – metals
0.9–1.0   glass – glass
  0.9 sheep – steel mesh
  0.7 sheep – plastic batten (⊥)
  0.6 sheep – plastic batten (∥)
  0.6 sheep – wood batten (⊥)
  0.5 sheep – wood batten (∥)
0.58   steel – steel
0.4   brakes – cast iron
0.6   wood – brick
0.2–0.6   wood – metals
0.29 0.22 wood – felt
0.28 0.17 wood – wood
0.3   snow – nylon
0.04–0.4 0.04–0.4 snow – hickory, waxed
0.1   graphite – graphite
0.1   graphite – steel
  0.03 ice – steel
0.05–0.5 0.02–0.09 ice – ice
0.2   teflon – steel
0.04   teflon – teflon
  0.0044–0.0057 ankle cartillage – synovial fluid
  0.0013 tendon – sheath

Summary

Problems

practice

  1. Push a load with enough force to overcome dry friction. What happens after it starts moving?
    • Answer it.
  2. Determine the following quantities for a car driving on a level surface with a coefficient of static friction of 0.75 (¾) and a coefficient of kinetic friction of 0.67 (⅔).
    1. Determine the car's maximum starting acceleration with and without "burning rubber". How do these two methods of starting a car compare?
    2. Determine the car's minimum braking distance with normal brakes and antilock brakes as a function of initial speed. How do these two methods of stopping a car compare?

    Solutions …

    1. The net external force propelling a car comes from the friction force between tires and pavement. When a driver starts a car by "flooring it" (pressing the accelerator to the floor) the tires grind on the road producing a smoke of burning rubber and pavement. Since the tires are slipping, the coefficient of kinetic friction determines the maximum acceleration. Under normal circumstances, however, most drivers are not willing to subject their tires to such extreme punishment. Typical car tires rotate over the surface of the road without slipping, thus the coefficient of static friction determines a car's maximum acceleration in most situations.

      To solve this problem, set the frictional force on level ground equal to the net force of the second law of motion.
                   
          ∑ F  =  m a    
          f = μN = μmg  =  ma    
          a  =  μg    
           
      aburnout  =  ⅔ (9.8 m/s2)   anormal  =  ¾ (9.8 m/s2)
      aburnout  =  6.54 m/s2   anormal  =  7.35 m/s2
                         
      aburnout  =  μkg  =  μk  =  ⅔  =  8  = 88.9%
      anormal μsg μs ¾ 9
                         
      Contrary to popular belief, flooring the accelerator is not an effective method of starting a car. Burning rubber is only about 90% as effective as accelerating a car normally from rest.
    2. The net external force stopping a car comes from the friction force between tires and pavement. Stopping a car with ordinary brakes may result in wheel lock; that is, the wheels lock in position and are not able to rotate. When this happens, the tires skid and the coefficient of kinetic friction determines the braking distance. Cars equipped with an antilock braking system (ABS) have a sensor that releases the brake pads the instant the wheel locks up. After a brief pause the brakes are then quickly re-engaged. If they don't lock up again, all is well. If they do, the ABS releases the brake pads again. This processes can repeat many times a second. In any case, the tires are not allowed to lock for more than a few milliseconds. The car is then stopped using the force of static friction alone.

      To solve this problem, determine acceleration using the displacement-velocity formula of kinematics. Set this equation equal to the formula for acceleration due to friction derived above.
                   
          v2 = 2aΔs = 2μgΔs     
          Δs  =  v2        
          2μg        
                               
      Δsnormal  =  v2     Δsantilock  =  v2  
      2(⅔)(9.8 m/s2)     2(¾)(9.8 m/s2)  
      Δsnormal  ∝  v2       Δsantilock  ∝  v2    
      13.1       14.7    
                         
      Δsantilock  =  v2 / 2μsg  =  μk  =  ⅔  =  8  = 88.9%
      Δsnormal v2 / 2μkg μs ¾ 9
                         
      Antilock brakes need 90% of the distance of regular brakes to stop a car traveling at the same speed. This decrease in distance is certainly significant, but doesn't really seem all that great given the high cost of an ABS. In addition to reduced braking distance, however, antilock braking systems also increase performance during extreme braking. Locked brakes are useless for steering. ABS ensures that the wheels retain their static frictional grip on the road, which allows for maneuvering while braking in an emergency.
  3. The critical angle problem. At what angle of inclination will an object held in place by friction just begin to slip?

    Solution …

    The component of the weight parallel to the incline pulls the object down the incline while the frictional force tries to keep it from sliding.Since nothing is going anywhere, these two forces must balance each other.

    ∑ F  =  m a
    W∥ − f  =  0
    mg sin θ − μmg cos θ  =  0
    sin θ − μ cos θ  =  0

    As the angle increases, friction decreases. Eventually the static friction force won't be strong enough to hold the object and it will slip. The critical angle at which this transition takes place is …

    tan θ = μ

  4. Write something completely different.
    • Answer it.

conceptual

  1. Note to self: work this idea into a problem of some sort. "The horizontal force component of the heel as it strikes the ground when a person is walking has been measured and found to be approximately 15% of a person's weight."

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