1. FRICTIONAL FORCES
• There are two types of frictional forces:
Type of friction Definition Equation
Static Friction A force that acts parallel to the two surfaces & Fmax = μsFN
(used for stationary objects) keeps an object from moving.
Kinetic Friction A force that acts opposite to the direction of an
(used for moving objects) Ff = μkFN
object’s motion.
Fmax is the maximum force
that can be applied to an Ff is the frictional force
object before it begins to
move
The greek symbol “μ” is pronounced “mu”

2. NEW FREE-BODY DIAGRAM
• Now we include the force of friction!
FN
Ff Fa
mg

3. HOW TO CALCULATE FRICTIONAL FORCES
• When an object is moving at a CONSTANT SPEED,
we can find the force of friction due to the coefficient
of kinetic friction.
Constant speed means that acceleration = 0
FN
So Fnet = 0
SO…. Fa – Ff = 0
Ff Fa
Ff = Fa
mg
• When a force is applied that causes an object to
JUST BEGIN TO MOVE, we can find the force of
friction due to the coefficient of static friction
Once again…. Ff = Fa

4. Example: Net force = 0
A 10-kg box is being pulled across the table to the right at a constant speed with a force
of 50N.
a) Calculate the Force of Friction Fa = Ff = 50N
a) Calculate the Force Normal mg Fn (10 )(9.8) 98 N
a) Calculate the coefficient of kinetic friction
Ff = mk FN
FN
Fa
50 = m k (98)
Ff
50
mg m k = = 0.51
98

5. Example: Net force = 0
Suppose the same box is now pulled at an angle of 30 degrees above the horizontal.
a) Calculate the Force of Friction
Fax Fa cos 50 cos30 43.3N
Ff Fax 43.3N
a) Calculate the Force Normal
FN mg!
FN Fay mg
FN Fa FN mg Fay (10)(9.8) 50 sin 30
Fay FN 73N
Ff 30
Fax
mg

6. EXAMPLE: NET FORCE = MA
A 50 N applied force drags an 8.16 kg log to the right across a horizontal
surface. What is the acceleration of the log if the force of friction is 40.0 N?
Fn a FNET = ma
50 N
40 N Fa - Ff = ma
mg 50 - 40 = 8.16a
10 = 8.16a
a= 1.23 m/s/s

7. EXAMPLE: NET FORCE = MA
A sled is being accelerated to the right at a rate of 1.5 m/s/s by a rope at a 33
degree angle above the + x . Calculate the acceleration of the sled if the
Frictional Force is 26.8 N, the mass of the sled is 66 kg and the tension in
the rope is 150 N.
a
FN
Tsin
Tcos FNET = ma
T cosq - Ff = ma
T cosq - ma = Ff
Ff
150 cos33- (66)(1.5) = Ff
mg Ff =
1.5 m/s/s