Announcements
Dry Friction
Today’s Objectives
• Understand the characteristics of dry friction
• Draw a FBD including friction
• Solve problems involving friction
Class Activities
• Applications
• Characteristics of dry friction
• Problems involving dry friction
• Examples
Engr221 Chapter 8
1
Applications
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?
Applications - continued
Consider pushing a box as
shown here. How can you
determine if it will slide, tilt, or
stay in static equilibrium?
What physical factors affect
the answer to this question?
Engr221 Chapter 8
2
Characteristics of Dry Friction
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 Wx = Ph.
Characteristics of Dry Friction - continued
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.
Engr221 Chapter 8
3
Characteristics of Dry Friction - continued
The maximum friction force is attained just before the
block begins to move (a situation called “impending
motion”). The value of the force is found using Fs = µs N,
where µs is called the coefficient of static friction. The value
of µs depends on the materials in contact.
Once the block begins to move, the frictional force typically
drops and is given by Fk = µk N. The value of µk
(coefficient of kinetic friction) is less than µs .
Determining µs Experimentally
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, θs, is noted. Analysis of
the block just before it begins to move
gives (using Fs = µs N):
+ ∑ Fy = N – W cos θs = 0
+ ∑ FX = µS N – W sin θs = 0
Using these two equations, we get µs = (W
sin θs ) / (W cos θs ) = tan θs
This simple experiment allows us to find
the µS between two materials in contact.
Engr221 Chapter 8
4
Procedure for Analysis
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 that
F = µSN unless the impending motion condition is given.
3. Apply the equations of equilibrium and appropriate
frictional equations to solve for the unknowns.
Impending Tipping vs. Slipping
For a given weight and height,
how can we determine if the
block will slide first 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. Then we can solve
for the unknowns using the
three E-of-E. Finally, we need
to check if our assumption was
correct.
Engr221 Chapter 8
5
Impending Tipping vs. Slipping - continued
Assume:
Known:
Solve:
Check:
Slipping occurs
F = µs N
x, P, and N
0 ≤ x ≤ b/2
Or
Assume:
Known:
Solve:
Check:
Tipping occurs
x = b/2
P, N, and F
F ≤ µs N
Example A
Given: A uniform ladder weighs 20 lb.
The vertical wall is smooth (no
friction).The floor is rough and
µs = 0.8
Find: The minimum force P needed to
move (tip or slide) the ladder.
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
Engr221 Chapter 8
6
Example A - continued
NB
A FBD of the ladder
4 ft
4 ft
P
20 lb
FA
NA
3 ft 3 ft
There are four unknowns: NA, FA, NB, and P. Assume that
the ladder will tip first so NB = 0
+↑ Σ FY = NA – 20 = 0 ;
NA = 20 lb
+ Σ MA = 20(3) – P(4) = 0 ;
P = 15 lb
+ → Σ FX = 15 – FA = 0 ;
FA = 15 lb
Example A - continued
NB
A FBD of the ladder
4 ft
4 ft
P
20 lb
FA
3 ft
3 ft
NA
Now check the assumption.
Fmax = µs NA = 0.8 * 20 lb = 16 lb
Is FA = 15 lb ≤ Fmax = 16 lb? Yes, hence our assumption
of tipping is correct.
Engr221 Chapter 8
7
Example B
Given: Drum weight = 100 lb, µs =
0.5, a = 3 ft, b = 4 ft
Find: The smallest magnitude of P
that will cause impending
motion (tipping or slipping)
of the drum
Plan:
a)
b)
c)
d)
Draw a FBD of the drum
Determine the unknowns
Make friction assumptions, as necessary
Apply E-of-E (and friction equations as appropriate) to
solve for the unknowns
e) Check assumptions, as required
Example B - continued
P
5
3
1.5 ft 1.5 ft
4
100 lb 4 ft
A FBD of the drum
0
F
X
N
There are four unknowns: P, N, F and x
First, let’s assume the drum slips. Then the friction
equation is F = µs N = 0.5 N
Engr221 Chapter 8
8
Example B - continued
P
5
3
1.5 ft 1.5 ft
4
100 lb 4 ft
A FBD of the drum
+ → ∑ FX = (4/5) P – 0.5 N = 0
+ ↑ ∑ FY = N – (3/5) P – 100 = 0
0
F
X
These two equations give:
N
P = 100 lb and N = 160 lb
+ ∑ MO = (3/5) 100 (1.5) – (4/5) 100 (4) + 160 (x) = 0
Check: x = 1.44 ≤ 1.5 so OK!
Drum slips as assumed at P = 100 lb
Questions
1. A friction force always acts _____ to the contact surface.
A) Normal
B) At 45°
C) Parallel
D) At the angle of static friction
2. If a block is stationary, then the friction force acting on it is
________
A) ≤ µs N
B) = µs N
C) ≥ µs N
D) = µk N
Engr221 Chapter 8
9
Question
P(A)
100 lb
P(B)
P(C)
A 100 lb box with wide base is pulled by a force P and µs = 0.4
Which force orientation requires the least force to begin sliding?
A) A
B) B
C) C
D) Can not be determined
Questions
1. A 10 lb block is in equilibrium. What is
the magnitude of the friction force
between this block and the surface?
A) 0 lb
B) 1 lb
C) 2 lb
D) 3 lb
µ S = 0.3
2 lb
2. The ladder AB is positioned as shown. What is the
direction of the friction force on the ladder at B?
A)
B)
C) ←
D) ↑
B
A
Engr221 Chapter 8
10
Example Problem
The refrigerator has a weight of 180 lb and rests on a tile floor for
which µs = 0.25 Also, the man has a weight of 150 lb and the
coefficient of static friction between the floor and his shoes is µs = 0.6
If he pushes horizontally on the refrigerator, determine if he can move
it. If so, does the refrigerator tip or slip?
Tips: P = 67.5 lb
Slips: P = 45 lbs
Man slips at 90 lbs
F = 45 lbs
Yes, he can move it
Textbook Problem 8.53
The 50-lb board is placed across the channel and a 100-lb boy
attempts to walk across. If the coefficient of static friction at A and B
is µs = 0.4, determine if he can make the crossing; and if not, how far
will he get from A before the board slips?
NA = 60.3 lb
NB = 86.2 lb
d = 6.47 ft
No, the board will slip
Engr221 Chapter 8
11
Example Problem
The 5-kg cylinder is suspended from two equal-length cords. The
end of each cord is attached to a ring of negligible mass, which
passes along a horizontal shaft. If the coefficient of static friction
between each ring and the shaft is µs = 0.5, determine the greatest
distance d by which the rings can be separated and still support
the cylinder.
Textbook Problem 8.44 (HW)
The crate has a weight of 300 lb and a center of gravity at G. If the
coefficient of static friction between the crate and floor is µs = 0.4,
determine the smallest weight of the man so he can push the crate to the
left. The coefficient of static friction between his shoes and the floor is
µs = 0.4. Assume the man exerts only a horizontal force on the crate.
Wman = 171 lb
Engr221 Chapter 8
12
Textbook Problem 8.50 (HW)
Determine the angle φ at which P should act on the block so that the
magnitude of P is as small as possible to begin pulling the block up
the incline. What is the corresponding value of P? The block weighs
W and the slope α is known.
φ= tan-1 µ
P = Wsin(φ + α)
Textbook Problem 8.58
The carpenter slowly pushes the uniform board horizontally over
the top of the saw horse. The board has a uniform weight of 3
lb/ft, and the saw horse has a weight of 15 lb and a center of
gravity at G. Determine if the saw horse will stay in position, slip,
or tip if the board is pushed forward when d = 14 ft. The
coefficients of static friction are shown in the figure.
NA = 60.3 lb
NB = 86.2 lb
d = 6.47 ft
No, the board will slip
Engr221 Chapter 8
13
Summary
• Understand the characteristics of dry friction
• Draw a FBD including friction
• Solve problems involving friction
Announcements
Engr221 Chapter 8
14
Wedges and Belts
Today’s Objectives
• Determine the forces on a wedge
• Determine the tensions in a belt
Class Activities
• Applications
• Analysis of a wedge
• Analysis of a belt
• Questions
Applications
Wedges are used to adjust the
elevation or provide stability for
heavy objects such as this large
steel vessel.
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?
Engr221 Chapter 8
15
Applications - continued
Belt drives are commonly
used for transmitting power
from one shaft to another.
How can we decide that the
belts will function properly,
i.e., without slipping or
breaking?
Applications - continued
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 we determine the
tensions in the cable pulling on
the band?
How are these tensions, the
applied force P and the torque
M, related?
Engr221 Chapter 8
16
Analysis of a Wedge
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:
1) The friction forces are always in the
direction opposite to the motion, or
impending motion, of the wedge
2) The friction forces are along the contacting
surfaces
3) The normal forces are perpendicular to the
contacting surfaces
Analysis of a Wedge - continued
Next, a FBD of the object on top of the wedge
is drawn. 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
b) all other forces acting on the object should
be shown
To determine the unknowns, we must apply E-of-E, ∑ Fx = 0 and
∑ Fy = 0, to the wedge and the object as well as the impending
motion frictional equation, F = µSN
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 equations.
Engr221 Chapter 8
17
Analysis of a Wedge - continued
W
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.
However, if the value of P is negative, or
zero, then the wedge will come out on its
own unless a force is applied to keep the
wedge in place. This can happen if the
coefficient of friction is small or the wedge
angle θ is large.
Belt Analysis
Belts are used for transmitting
power or applying brakes.
Friction forces play an important
role in determining the various
tensions in the belt. The belt
tension values are then used for
analyzing or designing a belt
drive or a brake system.
Engr221 Chapter 8
18
Belt Analysis - continued
Consider a flat belt passing over a fixed
curved surface with the total angle of
contact equal to β radians.
If the belt slips or is just about to slip,
then T2 must be larger than T1 and the
friction forces. Hence, T2 must be
greater than T1.
Detailed analysis (please refer to your textbook) shows that
T2 = T1 e µ β where µ is the coefficient of static friction
between the belt and the surface. Be sure to use radians
when using this formula!
Example A
Given: The load weighs 100 lb and the
µS between surfaces AC and BD
is 0.3 Smooth rollers are placed
between wedges A and B.
Assume the rollers and the
wedges have negligible weights.
Find: The force P needed to lift the
load.
Plan:
1. Draw a FBD of wedge A. Why do A first?
2. Draw a FBD of wedge B.
3. Apply the E-of-E to wedge B. Why do B first?
4. Apply the E-of-E to wedge A.
Engr221 Chapter 8
19
Example A - continued
100 lb
The FBDs of wedges A and B are shown in
the figures. Applying the E-of-E to wedge
B, we get
F3= 0.3N3
B
→+ ∑ FX = N2 sin 10° – N3 = 0
N2
↑+ ∑ FY = N2 cos 10° – 100 – 0.3 N3 = 0
10º
N2 10º
Solving the above two equations, we get
N2 = 107.2 lb and N3 = 18.6 lb
N3
A
P
F1= 0.3N1
Applying the E-of-E to the wedge A, we get
↑+ ∑ FY = N1 – 107.2 cos 10° = 0;
N1 = 105.6 lb
→+ ∑ FX = P – 107.2 sin 10° – 0.3 N1 = 0;
P = 50.3 lb
N1
Example B
Given: Blocks A and B weigh 50 lb and 30 lb, respectively.
Find:
Engr221 Chapter 8
The smallest weight of cylinder D which will cause
the loss of static equilibrium.
20
Example B - continued
Plan:
1. Consider two cases: a) both blocks slide together, and
b) block B slides over block A
2. For each case, draw a FBD of the block(s).
3. For each case, apply the E-of-E 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.
Example B - continued
P
Case a: blocks slide together.
B
A
30 lb
50 lb
↑ + ∑ FY = N – 80 = 0
N = 80 lb
F=0.4 N
N
→+ ∑ FX = 0.4 (80) – P = 0
P = 32 lb
Engr221 Chapter 8
21
Example B - continued
30 lb
P
0.6 N
20º
N
Case b:
→ + ∑ Fy = N cos 20° + 0.6 N sin 20° – 30 = 0
N = 6.20 lb
→ + ∑ Fx = – P + 0.6 ( 26.2 ) cos 20° – 26.2 sin 20° = 0 P = 5.812 lb
Case b has the lowest P and will occur first. Next, using the
frictional force analysis of belt, we get
WD = P e µ β = 5.812 e 0.5 ( 0.5 π ) = 12.7 lb
Block D weight of 12.7 lb will cause block B to slide over block A.
Questions
1. A wedge allows a ______ force P to lift
a _________ weight W.
A) (large, large) B) (small, small)
C) (small, large) D) (large, small)
W
2. Considering friction forces and the
indicated motion of the belt, how are belt
tensions T1 and T2 related?
A) T1 > T2 B) T1 = T2
C) T1 < T2 D) T1 = T2 eµ
Engr221 Chapter 8
22
Questions
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
Questions
1. When determining the force P needed to lift
the block of weight W, it is easier to draw a
FBD of ______ first.
A) the wedge
B) the block
C) the horizontal ground D) the vertical wall
W
2. In the analysis of frictional forces on a flat belt, T2 = T1 e µ β
In this equation, β equals ______
A) angle of contact in degrees
B) angle of contact in radians
C) coefficient of static friction D) coefficient of kinetic friction
Engr221 Chapter 8
23
Summary
• Determine the forces on a wedge
• Determine the tensions in a belt
Engr221 Chapter 8
24