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STATICS: CE201
Chapter 8
Friction
Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson
Dr M. Touahmia & Dr M. Boukendakdji
Civil Engineering Department, University of Hail
(2012/2013)
Chapter 8: Friction
8. Friction
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Chapter Objectives:





To introduce the concept of dry friction and show how
to analyze the equilibrium of rigid bodies subjected to
this force.
Draw a FBD including friction.
Present specific applications of frictional force analysis
on wedges, screws, belts, and bearings.
Investigate the concept of rolling resistance
Solve problems involving friction.
Chapter 8: Friction
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APPLICATIONS
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In designing a brake system for a bicycle, car, or any
other vehicle, it is important to understand the
frictional forces involved.
For an applied force on the brake pads, how can we
determine the magnitude and direction of the resulting
friction force?
Chapter 8: Friction
APPLICATIONS
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The rope is used to tow the
refrigerator.
In
order
to
move
the
refrigerator, is it best to pull up
as shown, pull horizontally, or
pull downwards on the rope?
What physical factors affect the
answer to this question?
Chapter 8: Friction
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CHARACTERISTICS OF DRY FRICTION
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Friction is defined as a force of resistance
acting on a body which prevents or retards
slipping of the body relative to a second
body.
Experiments show that frictional forces
act tangent (parallel) to the contacting
surface in a direction opposing the
relative motion or tendency for motion.
For the body shown in the figure to
be in equilibrium, the following must
be true: F = P, N = W, and W*x =
P*h.
Chapter 8: Friction
CHARACTERISTICS OF DRY FRICTION
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To study the characteristics of the friction force F, let us
assume that tipping does not occur (i.e., “h” is small or “a”
is large). Then we gradually increase the magnitude of the
force P. Typically, experiments show that the friction force
F varies with P, as shown in the right figure above.
Chapter 8: Friction
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CHARACTERISTICS OF DRY FRICTION
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The maximum friction force is attained
just before the block begins to move (a
situation that is called “impending
motion”). The value of the force is found
using Fs = s N, where s is called the
coefficient of static friction. The value of
s depends on the two materials in
contact.
Once the block begins to move, the
frictional force typically drops and is
given by Fk = k N. The value of k
(coefficient of kinetic friction) is less than
s .
Chapter 8: Friction
CHARACTERISTICS OF DRY FRICTION
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It is also very important to note that the friction force
may be less than the maximum friction force. So, just
because the object is not moving, don’t assume the
friction force is at its maximum of Fs = s N unless you
are told or know motion is impending!
Chapter 8: Friction
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DETERMING s EXPERIMENTALLY
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If the block just begins to slip, the
maximum friction force is Fs = s N,
where s is the coefficient of static
friction.
Thus, when the block is on the
verge of sliding, the normal force N
and
frictional force Fs combine to create
a resultant Rs
From the figure,
tan s = ( Fs / N ) = (s N / N ) = s
Chapter 8: Friction
DETERMING s EXPERIMENTALLY
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A block with weight w is placed on
an inclined plane. The plane is
slowly tilted until the block just
begins to slip.
The inclination, s, is noted. Analysis
of the block just before it begins to
move gives (using Fs = s N):
+  Fy =
= 0
N – W cos s
+  FX = S N – W sin s = 0
Using these two equations, we get s = (W
sin s ) / (W cos s ) = tan s
This simple experiment allows us to find the S
between two materials in contact.
Chapter 8: Friction
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PROBLEMS INVOLVING DRY FRICTION
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Steps for solving equilibrium problems involving dry friction:
1. Draw the necessary free body diagrams. Make sure
that you show the friction force in the correct direction
(it always opposes the motion or impending motion).
2.
Determine the number of unknowns. Do not assume
F = S N unless the impending motion condition is
given.
3.
Apply the equations of equilibrium and appropriate
frictional equations to solve for the unknowns.
Chapter 8: Friction
IMPENDING TIPPING versus SLIPPING
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For a given W and h of the box,
how can we determine if the
block will slide or tip first? In
this case, we have four
unknowns (F, N, x, and P) and
only three E-of-E.
Hence, we have to make an
assumption to give us another
equation
(the
friction
equation!).
Then we can
solve for the unknowns using
the three E-of-E. Finally, we
need to check if our
assumption was correct.
Chapter 8: Friction
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IMPENDING TIPPING versus SLIPPING
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Assume: Slipping occurs
Known: F = s N
Solve:
x, P, and N
Check:
0  x  b/2
Or
Assume: Tipping occurs
Known: x = b/2
Solve:
P, N, and F
Check:
F  s N
Chapter 8: Friction
EXAMPLE 1
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Given: A uniform ladder weighs 30 lb.
The vertical wall is smooth (no
friction). The floor is rough
and s = 0.2.
Find:
Whether it remains in this position
when it is released.
Plan:
a) Draw a FBD.
b) Determine the unknowns.
c) Make any necessary friction assumptions.
d) Apply E-of-E (and friction equations, if appropriate ) to solve for the
unknowns.
e) Check assumptions, if required.
Chapter 8: Friction
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EXAMPLE 1
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NB
FBD of the ladder
12 ft
12 ft
30
lb
FA
5 ft
5 ft
NA
There are three unknowns: NA, FA, NB.
  FY = NA – 30 = 0 ;
so NA = 30 lb
+  MA = 30 ( 5 ) – NB( 24 ) = 0 ;
+   FX = 6.25 – FA = 0 ;
so NB = 6.25 lb
so FA = 6.25 lb
Chapter 8: Friction
EXAMPLE 1
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NB
FBD of the ladder
12 ft
12 ft
30 lb
FA
5 ft
5 ft
NA
Now check the friction force to see if the ladder slides or stays.
Fmax = s NA = 0.2 * 30 lb = 6 lb
Since FA = 6.25 lb
 Ffriction max = 6 lb,
the pole will not remain stationary. It will move.
Chapter 8: Friction
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GROUP PROBLEM SOLVING
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Given: Refrigerator weight = 180
lb,
s = 0.25
Find: The smallest magnitude of
P
that will cause impending
motion (tipping or slipping)
of the refrigerator.
Plan:
a) Draw a FBD of the refrigerator.
b) Determine the unknowns.
c) Make friction assumptions, as necessary.
d) Apply E-of-E (and friction equation as appropriate) to solve for the
unknowns.
e) Check assumptions, as required.
Chapter 8: Friction
WEDGES AND FRICTIONAL FORCES ON FLAT BELTS
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Today’s Objectives:
In-Class Activities:
Students will be able to:
• Check Homework, if any
a) Determine the forces on a
wedge.
• Reading Quiz
b) Determine the tensions in a belt.
• Analysis of a Wedge
• Applications
• Analysis of a Belt
• Concept Quiz
• Group Problem Solving
• Attention Quiz
Chapter 8: Friction
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WEDGES AND FRICTIONAL FORCES ON FLAT BELTS
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APPLICATIONS
Wedges are used to
adjust the elevation or
provide stability for heavy
objects such as this large
steel pipe.
How can we determine the
force required to pull the
wedge out?
When there are no applied forces on the wedge, will it
stay in place (i.e., be self-locking) or will it come out on its
own? Under what physical conditions will it come out?
Chapter 8: Friction
WEDGES AND FRICTIONAL FORCES ON FLAT BELTS
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APPLICATIONS
Belt drives are commonly used for transmitting the
torque developed by a motor to a wheel attached to a
pump, fan or blower.
How can we decide if the belts will function properly,
i.e., without slipping or breaking?
Chapter 8: Friction
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WEDGES AND FRICTIONAL FORCES ON FLAT BELTS
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APPLICATIONS
In the design of a band brake,
it is essential to analyze the
frictional forces acting on the
band (which acts like a belt).
How can you determine the
tensions in the cable pulling
on the band?
Also from a design
perspective, how are the belt
tension, the applied force P
and the torque M, related?
Chapter 8: Friction
ANALYSIS OF A WEDGE
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W
A wedge is a simple machine in which a small
force P is used to lift a large weight W.
To determine the force required to push the wedge
in or out, it is necessary to draw FBDs of the wedge
and the object on top of it.
It is easier to start with a FBD of the wedge since
you know the direction of its motion.
Note that:
a) the friction forces are always in the direction
opposite to the motion, or impending motion, of the
wedge;
b) the friction forces are along the contacting
surfaces; and,
c) the normal forces are perpendicular to the
contacting surfaces.
Chapter 8: Friction
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ANALYSIS OF A WEDGE
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Next, a FBD of the object on top of the
wedge is drawn. Please note that:
a) at the contacting surfaces between the
wedge and the object the forces are equal
in magnitude and opposite in direction to
those on the wedge; and, b) all other
forces acting on the object should be
shown.
To determine the unknowns, we must apply EofE,  Fx = 0
and
 Fy = 0, to the wedge and the object as well as the
impending motion frictional equation, F = S N.
Chapter 8: Friction
ANALYSIS OF A WEDGE
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Now of the two FBDs, which one should we start
analyzing first?
We should start analyzing the FBD in which the
number of unknowns are less than or equal to the
number of EofE and frictional equations.
Chapter 8: Friction
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ANALYSIS OF A WEDGE
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W
NOTE:
• If the object is to be lowered, then
the wedge needs to be pulled out.
• If the value of the force P needed
to remove the wedge is positive,
then the wedge is self-locking, i.e., it
will not come out on its own.
BELT ANALYSIS
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Consider a flat belt passing over a
fixed curved surface with the total
angle of contact equal to  radians.
If the belt slips or is just about to
slip, then T2 must be larger than T1
and the motion resisting friction
forces. Hence, T2 must be greater
than T1.
Detailed analysis (please refer to your textbook) shows
that T2 = T1 e   where  is the coefficient of static
friction between the belt and the surface. Be sure to
use radians when using this formula!!
Chapter 8: Friction
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EXAMPLE 2
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Given: The crate weighs 300 lb and
S at all contacting surfaces is
0.3. Assume the wedges have
negligible weight.
Find: The smallest force P needed
to pull out the wedge.
Plan:
1. Draw a FBD of the crate. Why do the crate
first?
2. Draw a FBD of the wedge.
3. Apply the E-of-E to the crate.
4. Apply the E-of-E to wedge.
Chapter 8: Friction
EXAMPLE 2
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NC
FC=0.3NC
300 lb
FB=0.3NB
P
NB
15º
FD=0.3ND
FC=0.3NC
15º
NC
FBD of Crate
ND
FBD of Wedge
The FBDs of crate and wedge are shown in the figures.
Applying the EofE to the crate, we get
+  FX =
NB – 0.3NC = 0
+  FY = NC – 300 + 0.3 NB = 0
Solving the above two equations, we get
NB = 82.57 lb = 82.6 lb,
NC = 275.3 lb = 275 lb
Chapter 8: Friction
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EXAMPLE 2
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NC
FB=0.3NB
FC=0.3NC
300 lb
P
NB = 82.6 lb
15º
FD=0.3ND
FC=0.3NC
15º
NC = 275 lb
FBD of Crate
ND
FBD of Wedge
Applying the E-of-E to the wedge, we get
+  FY = ND cos 15 + 0.3 ND sin 15 – 275.2= 0;
ND = 263.7 lb = 264 lb
+  FX = 0.3(263.7) + 0.3(263.7)cos 15 – 0.3(263.7)cos 15 – P = 0;
P = 90.7 lb
CONCEPT QUIZ
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1. Determine the direction of the friction
force on object B at the contact point
between A and B.
A) 
B) 
C)
D)
2. The boy (hanging) in the picture
weighs 100 lb and the woman weighs
150 lb. The coefficient of static friction
between her shoes and the ground is
0.6. The boy will ______ ?
A) Be lifted up
B) Slide down
C) Not be lifted up D) Not slide down
Chapter 8: Friction
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GROUP PROBLEM SOLVING
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Given: Blocks A and B weigh 50 lb
and 30 lb, respectively.
Find: The smallest weight of
cylinder D which will cause the
loss of static equilibrium.
Chapter 8: Friction
GROUP PROBLEM SOLVING
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Plan:
1. Consider two cases: a) both blocks slide together, and,
b) block B slides over the block A.
2. For each case, draw a FBD of the block(s).
3. For each case, apply the EofE to find the force needed to
cause sliding.
4. Choose the smaller P value from the two cases.
5. Use belt friction theory to find the weight of block D.
Chapter 8: Friction
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GROUP PROBLEM SOLVING
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Case a (both blocks sliding
together):
P
B
 +  FY = N – 80 = 0
N = 80 lb
A
+  FX = 0.4 (80) – P = 0
30 lb
50 lb
F=0.4 N
N
P = 32 lb
Chapter 8: Friction
GROUP PROBLEM SOLVING
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Case b (block B slides over A):
 +  Fy = N cos 20 + 0.6 N sin 20 – 30 = 0
N = 26.20 lb
 +  Fx = – P + 0.6 ( 26.2 ) cos 20 – 26.2 sin 20 = 0
P = 5.812 lb
Case b has the lowest P (case a was 32 lb) and thus will occur first. Next,
using a frictional force analysis of belt, we get:
W D = P e   = 5.812 e 0.5 ( 0.5  ) = 12.7 lb
A Block D weighing 12.7 lb will cause the block B to slide over the block A.
Chapter 8: Friction
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