Chapter 4B. Friction and
Equilibrium
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
Equilibrium: Until motion
begins, all forces on the mower
are balanced. Friction in wheel
bearings and on the ground
oppose the lateral motion.
Objectives: After completing this
module, you should be able to:
• Define and calculate the coefficients of
kinetic and static friction, and give the
relationship of friction to the normal force.
• Apply the concepts of static and kinetic
friction to problems involving constant
motion or impending motion.
Friction Forces
When two surfaces are in contact, friction forces
oppose relative motion or impending motion.
P
Friction forces are parallel to
the surfaces in contact and
oppose motion or impending
motion.
Static Friction: No
relative motion.
Kinetic Friction:
Relative motion.
Friction and the Normal Force
12 N n
8N n
4N
n2N
4N
6N
The force required to overcome static or kinetic
friction is proportional to the normal force, n.
fs = msn
fk = mkn
Friction forces are independent of area.
4N
4N
If the total mass pulled is constant, the same
force (4 N) is required to overcome friction
even with twice the area of contact.
For this to be true, it is essential that ALL
other variables be rigidly controlled.
Friction forces are independent of
temperature, provided no chemical or
structural variations occur.
4N
4N
Heat can sometimes cause surfaces to become
deformed or sticky. In such cases, temperature
can be a factor.
Friction forces are independent of speed.
5 m/s
2 N
20 m/s
2 N
The force of kinetic friction is the same at
5 m/s as it is for 20 m/s. Again, we must
assume that there are no chemical or
mechanical changes due to speed.
The Static Friction Force
When an attempt is made to move an
object on a surface, static friction slowly
increases to a MAXIMUM value.
fs
n
P
W
f s  ms
n
In this module, when we use the following
equation, we refer only to the maximum
value of static friction and simply write:
fs = msn
Constant or Impending Motion
For motion that is impending and for
motion at constant speed, the resultant
force is zero and SF = 0. (Equilibrium)
fs
P
Rest
P – fs = 0
fk
P
Constant Speed
P – fk = 0
Here the weight and normal forces are
balanced and do not affect motion.
Friction and Acceleration
When P is greater than the maximum fs
the resultant force produces acceleration.
fk
a
P
Constant Speed
This case will be
discussed in a
later chapter.
fk = mkn
Note that the kinetic friction force remains
constant even as the velocity increases.
EXAMPLE 1: If mk = 0.3 and ms = 0.5,
what horizontal pull P is required to
just start a 250-N block moving?
1. Draw sketch and freebody diagram as shown.
n
fs
P
+
W
2. List givens and label
what is to be found:
mk = 0.3; ms = 0.5; W = 250 N
Find: P = ? to just start
3. Recognize for impending motion: P – fs = 0
EXAMPLE 1(Cont.): ms = 0.5, W = 250 N. Find
P to overcome fs (max). Static friction applies.
n
For this case: P – fs = 0
P
fs
4. To find P we need to
know fs , which is:
+
fs = msn
250 N
5. To find n:
SFy = 0
W = 250 N
n=?
n–W=0
n = 250 N
(Continued)
EXAMPLE 1(Cont.): ms = 0.5, W = 250 N. Find P
to overcome fs (max). Now we know n = 250 N.
6. Next we find fs from:
fs = msn = 0.5 (250 N)
7. For this case: P – fs = 0
P = fs = 0.5 (250 N)
P = 125 N
n
fs
P
+
250 N
ms = 0.5
This force (125 N) is needed to just start motion.
Next we consider P needed for constant speed.
EXAMPLE 1(Cont.): If mk = 0.3 and ms = 0.5,
what horizontal pull P is required to move with
constant speed? (Overcoming kinetic friction)
SFy = may = 0
mk = 0.3
fk
n
n-W=0
P
n=W
Now: fk = mkn = mkW
+
mg
P = (0.3)(250 N)
SFx = 0;
P - fk = 0
P = fk = mkW
P = 75.0 N
The Normal Force and Weight
The normal force is NOT always equal to
the weight. The following are examples:
n
m
P
Here the normal force is
less than weight due to
upward component of P.
300
W
P
n
W
Here the normal force is
equal to only the component of weight perpendicular to the plane.
Review of Free-body Diagrams:
For Friction Problems:
• Read problem; draw and label sketch.
• Construct force diagram for each object,
vectors at origin of x,y axes. Choose x or y
axis along motion or impending motion.
• Dot in rectangles and label x and y components opposite and adjacent to angles.
• Label all components; choose positive
direction.
For Friction in Equilibrium:
• Read, draw and label problem.
• Draw free-body diagram for each body.
• Choose x or y-axis along motion or impending
motion and choose direction of motion as positive.
• Identify the normal force and write one of
following:
fs = msn or fk = mkn
• For equilibrium, we write for each axis:
SFx = 0
SFy = 0
• Solve for unknown quantities.
Example 2. A force of 60 N drags a 300-N
block by a rope at an angle of 400 above the
horizontal surface. If uk = 0.2, what force P
will produce constant speed?
W = 300 N
fk
n
P=?
400
m
W
The force P is to be
replaced by its components Px and Py.
1. Draw and label a sketch
of the problem.
2. Draw free-body diagram.
P sin 400 Py
n
fk
W
P
400
Py
Px
P cos 400
+
Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.
3. Find components of P:
Px = P cos 400 = 0.766P
Py = P sin 400 = 0.643P
Px = 0.766P; Py = 0.643P
P sin 400
n
P
400
fk
P cos 400
mg
+
Note: Vertical forces are balanced, and for
constant speed, horizontal forces are balanced.
F
x
0
F
y
0
Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.
Px = 0.766P
Py = 0.643P
4. Apply Equilibrium conditions to vertical axis.
SFy = 0
0.643P
n
fk
300 N
P
400
0.766P
+
n + 0.643P – 300 N= 0 [Py and n are up (+)]
n = 300 N – 0.643P; Solve for n in terms of P
n = 300 N – 0.643P
Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.
n = 300 N – 0.643P
0.643P
n
5. Apply SFx = 0 to con-
stant horizontal motion.
SFx = 0.766P – fk = 0
fk
300 N
P
400
0.766P
+
fk = mk n = (0.2)(300 N - 0.643P)
fk = (0.2)(300 N - 0.643P) = 60 N – 0.129P
0.766P – fk = 0;
0.766P – (60 N – 0.129P) = 0
Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.
0.643P
n
P
400
fk
0.766P
300 N
+
0.766P – (60 N – 0.129P )=0
6. Solve for unknown P.
0.766P – 60 N + 0.129P =0
0.766P + 0.129P = 60 N
0.766P + 0.129P = 60 N
0.895P = 60 N
P = 67.0 N
If P = 67 N, the
block will be
dragged at a
constant speed.
P = 67.0 N
Example 3: What push P up the incline is
needed to move a 230-N block up the
incline at constant speed if mk = 0.3?
P
Step 1: Draw free-body
including forces, angles
and components.
y
n
W sin 600
fk
P
x
W cos 600
600
230 N
W =230 N
Step 2:
SFy = 0
n – W cos 600 = 0
n = (230 N) cos 600
n = 115 N
Example 3 (Cont.): Find P to give
move up the incline (W = 230 N).
n
y
W sin 600
x
P
fk
W cos 600
600
W
n = 115 N
W = 230 N
Step 3. Apply SFx= 0
P - fk - W sin 600 = 0
fk = mkn = 0.2(115 N)
fk = 23 N, P = ?
P - 23 N - (230 N)sin 600 = 0
P - 23 N - 199 N= 0
P = 222 N
Summary: Important Points to Consider
When Solving Friction Problems.
• The maximum force of static friction is
the force required to just start motion.
fs
n
P
W
f s  ms
n
Equilibrium exists at that instant:
SFx  0;
SFy  0
Summary: Important Points (Cont.)
• The force of kinetic friction is that force
required to maintain constant motion.
n
fk
P
W
f k  mk
n
• Equilibrium exists if speed is constant,
but fk does not get larger as the
speed is increased.
SFx  0;
SFy  0
Summary: Important Points (Cont.)
• Choose an x or y-axis along the direction
of motion or impending motion.
mk = 0.3
fk
n
P
+
W
The SF will be zero
along the x-axis and
along the y-axis.
In this figure, we have:
SFx  0;
SFy  0
Summary: Important Points (Cont.)
• Remember the normal force n is not
always equal to the weight of an object.
n
m
P
It is necessary to draw
the free-body diagram
and sum forces to solve
300
W
P
n
W
for the correct
SFx  0;
n value.
SFy  0
Summary
Static Friction: No
relative motion.
Kinetic Friction:
Relative motion.
fs ≤ msn
fk = mkn
Procedure for solution of equilibrium
problems is the same for each case:
SFx  0 SFy  0
CONCLUSION: Chapter 4B
Friction and Equilibrium