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Sliding friction is a reaction force generated by the surfaces in contact. The microscopic view of the surfaces shows ridges and valeys which cause a bumpy ride as the surfaces slide across each other.


Friction always opposes motion, or the tendency of an object to move if it is at rest. The force of friction is parallel to the plane of the surfaces and depends on the following:

• the types of surfaces in contact
  
• the normal force (FN).
  

A fact surprising to most students is that the amouont of frictional force does not depend on the amount of surface in contact. Larger surface areas do not create more friction than smaller areas. The reason for this comes from the reasoning that any section of the larger surface is pressed into the surface less than a similarly sized region of the smaller area. As a result the product of the amount they are pressed together and the surface area are the same.


There are two types of friction:

• static (Fs) - frictional force opposing the starting of relative motion between two surfaces.
  
• kinetic (Fk) - frictional force opposing the continuation of relative motion between two surfaces.
  

The static force is larger than the kinetic force. There is more opposition to starting the sliding of surfaces than to keep them sliding.

At the right is a graph of the force applied to a stationary object until it moves at a constant rate. You can see a peak force is reached at the point where the object begins to slide. At this peak the applied force would be equal to the maximum force of static friction.
After the object is moving a smaller force is needed to keep it moving at a constant rate. Since the object is in equilibrium the applied force would be equal to the kinetic frictional force.


One of the factors determining the frictional force is the Normal force.
<BLOCKQUOTE>The Normal force (N) is the force applied by a surface on an object.</BLOCKQUOTE>The Normal force is always directed away from the surface and is perpendicular to the surface. The Normal force is also a reaction force. The harder the object pushes on the surface the harder the surface pushes back.




EX. What is the normal force on a 2.4 Kg book sitting on a table?


The roughness of the surfaces in contact is indicated by a coefficient of friction (m). The value of the coefficient is determined by the ratio of the frictional force to the Normal force. A higher value for the coefficient indicates a larger frictional force exists between the surfaces.


Since there are two distinct frictional forces, there are two different coefficients for the surfaces. As you might suspect the kinetic coefficient is smaller than its corresponding static coefficient.
The coefficients depend only on the roughness of the surfaces, not on the Normal force. If the normal force is increased, the frictional force is increased, but the coefficient remains the same. The only way to change the coefficient is to change the surfaces in contact. A limited table of coefficients is given below.




EX 1. A 30 Kg wooden crate sits on a cement floor where its m_s = 0.7 and a m_k = 0.4.
<BLOCKQUOTE>a) What is the maximum force of static friction?
b) What is the kinetic force of friction?
c) If a 70 Kg person sits on the crate,
1) would the coefficients of friction change, and
2) what minimum force, parallel to the floor, would it take to make the box move?</BLOCKQUOTE>Solution:
a) Fs = m_s (N) where N = F_g = 9.8 N/Kg (30 Kg) = 294 N
= 0.7 (294 N)
= 206 N
b) F_k = m_k (N) where N = F_g = 9.8 N/Kg (30 Kg) = 294 N
= 0.4 (294 N)
= 118 N
c) 1) No. The coefficient depends on the types of surfaces and they haven't changed.
2) The minimum force to move the crate would be equal to the static frictional force.
Fs = m_s (N) where N = F_g = 9.8 N/Kg (100 Kg) = 980 N
= 0.7 (980 N) (the weight of both the person and the crate together)
= 686 N
The minimum force to move the crate would be 686 N

EX 2. A 1.7 Kg book lying on a table has a m_s = 0.6 and a m_k = 0.2. If it is pushed by a horizontal force of 9.0 N, will the book move?
What you need to know is, "Is the 9.0 N push more than the maximum force of static friction?"
<BLOCKQUOTE>Fs = m_s (N) where N = F_g = 9.8 N/Kg (1.7 Kg) = 16.7 N
= 0.6 (16.7 N)
= 10.0 N</BLOCKQUOTE>The maximum force of static friction is more than the 9.0 N push so the book will not move.