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Friction Notes

 

 

No Motion (figure 1):

µS  =  Coefficient of static friction

 


Σ FX  =  FT  +  (-FS)  =  0      (translational equilibrium)

 

FT  =  FS

 

Σ FY  =  FWT  +  (-Fn)  =  0

 

FWT  =  Fn

 

In this case, FT  <  FS(MAX), so FS/Fn < µS 

 

 

 

Body just about to move (figure 2):

 

In this case, FT = FS(MAX).

This is a critical value for FT, where

FS is at its maximum value.  Any effort

above FT’s current value will cause the

box to move.

 

FS(MAX) / Fn  =  µS

 

The coefficient of static friction,  µS,

depends on the surface features of the box

and the material it rests upon.

 

FS is found using this equation only at the

point beyond which FT will cause motion.

µS 

rubber on concrete  =  1.0

steel on steel  =  0.74

Teflon on Teflon  =  0.04

 

 

 

 

Sliding motion begins (figure 3):

µK  =  The coefficient of kinetic friction. 

 

 

In this case, FT > FS.

(FS(MAX) < FT)

Also, FK(MOTION) < FS(MAX STATIONARY)

         FT(M0TION) < FT(MAX STATIONARY)

         FT(MOTION) = FK(CONSTANT VELOCITY)    

 

FK / Fn  =  µK                µK  < µS, because FK < FS

 

  µK 

rubber on concrete (dry) = 0.8

rubber on concrete (wet = 0.25

steel on steel = 0.57

 

 

 

 

 

 

 


Rolling Motion (figure 4):

µr  = The coefficient of rolling friction.

 

 

The ratio of the horizontal force needed to make a body move with constant speed on a flat surface, to the upward normal force exerted by the surface, is the coefficient of rolling friction.  It is also referred to as “tractive resistance”.

 

FT/Fn =  µr        

 

µr 

Rubber on concrete = 0.02

Steel on steel = 0.003

 

 

 

 

 

 

 

 

 

 

 

Motion of a body when FT is at an angle

with the horizontal (figure 5):

 

In this case,

FH  =  FK

FH  =  FT Cos θ

 

Fn  =  FWt  +  (-FV)

Fn  =  FWT – FT Sin θ

 

 

 

µK =  FK / Fn

 

     = FH / [FWt + (-FV)]

    

µK = FT Cos θ / [FWT – FT Sin θ]

 

µK ( FWT – FT Sin θ) = FT Cos θ

 

µK ·FWT – µK ·FT Sin θ  =  FT Cos θ

 

µK ·FWT  =  FT (µK Sin θ   +  Cos θ)

 

µK ·FWT  / (µK Sin θ   +  Cos θ)  = FT

 

The equation above allows you to solve for an unknown force needed to pull an object across the specified surface at constant velocity, at the given angle.

 

Compare FT(HORIZONTAL)  and FT (ANGLE θ).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Motion of a body on an inclined surface:

 

 

                                   

µK =  FK / Fn  =  FP / FN  =  FWT Sin θ /FWT Cos θ  =  Tan θ

In Figure 6, the block will begin to slide down an inclined plane with constant speed when the Tan θ of the slope angle equals its µK.  Notice, the equation is independent of the body’s weight.   As the angle of the slope is slowly increased, the block will probably have to be lightly tapped to encourage it to begin the steady downward motion observed at angle Ř.  If the block is not disturbed while the slope of the incline is gradually increased, a point is reached above angle Ř  where the block “breaks free” and moves downward with accelerated motion.  This greater angle of inclination corresponds to μS, which is greater in magnitude than μK, due to the greater magnitude of FS in FS/Fn, compared to FK in FK/Fn.

The following questions should be answered:

(1)  What happens to the magnitude of the normal contact force (Fn) as the angle of inclination increases?

(2)  When is FK established?

(3)  When is μS determined?

(4)  Once the block begins to move, what is the net force causing its motion in each case:

      (A)  At the point where µK is established?

      (B)  At the “break free” point where accelerated motion is noted?

 

 

 

 

Online Links to Additional Information: 

This section is under construction, beginning 10-8-01, and will appear as a semi-organized list of links.  Click on a particular link as the material it contains becomes pertinent following our discussions in class.

The University of Guelph, Canada, offers three problems with linked solutions involving a sled, a crate on an incline, and a baseball player sliding into home plate.

Syracuse University, PHY 211 Tutorial for Free Body Diagrams:  Use this interactive tutorial as a guide to solve problems involving a friction force vector.  Click on "to proceed:" button as you work through Jeff and Nancy's problem.  This is a continuation of an inclined plane problem, taking the force of friction and the normal (contact) force into consideration.  Once you conclude this problem, go to the "Table of Contents" listed in the left window (black) and click on "friction", "normal", and "tension".  Keep the "Glossary" of terms in mind.  Scroll to the bottom of this list to "Practice #1" and "Quiz #1".

University of Kentuky:  Read this short vector analysis of a wood block on a horizontal surface versus an inclined surface when its behavior is influenced by friction.

University of Illinois, Urbana-Champaign:  Follow the lecture on friction.