Friction: If it weren't for the normal force, friction would not play a roll in our every day lives, since there would be no reaction force to keep two surfaces together so they could rub.  

Friction is produced when two surfaces rub together.  The amount of friction depends on a constant called the coefficient of friction.  The higher the coefficient of friction, the more friction force that exists between the two surfaces.  This coefficient is given by the ratio of the force of friction to the Normal force between the surfaces:            

m = F_friction/F_N       (F_N = Normal force)  

There are two types of friction; static friction and kinetic friction (sometimes referred to as sliding friction). 

• Static friction is the force of friction which is necessary to overcome in order to start the object moving. 

• Kinetic, or sliding friction, is the force of friction which is necessary to overcome in order to keep the object moving at a constant speed (it is always harder to get an object started moving than it is to keep it moving; try walking on ice - without skates - as long as you step lightly, you might feel a little more sturdy than when you are sliding). 

Since there are two different types of friction, there must be two different coefficients of friction, m_s and m_k.       

The above equation can be rewritten as:            

F_friction = m_kF_N  (kinetic friction)            
or          
F_friction = m_sF_N  (static friction)    

Example #1: How much force is required to just start a 4.0 kg object moving on a surface where the static coefficient of friction is m_s = 0.6?

F_net = ma
F_net = F_f + F_app    (ie.  Fnet = sum of all forces on the object)
F_app = F_net - F_f        (but F_f is a negative number)
F_app = ma + m_sF_N
F_app = ma + m_smg (F_N = mg on level surfaces)
F_app = (4.0 kg)(0 m/s²) - ^- m_sF_N
F_app = (4.0)(0) + (0.6)(4.0kg)(9.8m/s/s)
F_app = 23.5 N

Example #2: How much force is required to keep an object moving at a constant velocity on a surface where the kinetic coefficient of friction is m_k = 0.2?

F_net = ma
F_net = F_f + F_app    (ie.  Fnet = sum of all forces on the object)
F_app = F_net - F_f        (but F_f is a negative number)
F_app = ma + m_kF_N
F_app = ma + m_kmg (F_N = mg on level surfaces)
F_app = (4.0 kg)(0 m/s²) - ^- m_kF_N
F_app = (4.0)(0) + (0.2)(4.0kg)(9.8m/s/s)
F_app = 7.84 N

Try the following problems:   Friction WS #1