6 friction
TRANSCRIPT
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FRICTIONCHAPTER
Suriati Akmal
6 BMFG 1823Static and Strengthof Material
STATIC & STRENGTH OF MATERIALS
FRICTION (Sections 8.1 - 8.2)Todays Objective:
Students will be able to:
a) Understand the characteristics of
dry friction.
b) Draw a FBD including friction.
c) Solve problems involving friction.
In-Class Activities:
Check homework, if any
Reading quiz
Applications
Characteristics of dry friction
Problems involving dry friction
Concept quiz
Group problem solving
Attention quiz
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STATIC & STRENGTH OF MATERIALS
READING QUIZ
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) sN B) = sN
C) sN
D) = kN
STATIC & STRENGTH OF MATERIALS
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 themagnitude and direction of the
resulting friction force?
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STATIC & STRENGTH OF MATERIALS
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?
STATIC & STRENGTH OF MATERIALS
CHARACTERISTICS OF DRY FRICTION (Section 8.1)
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.
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STATIC & STRENGTH OF MATERIALS
CHARACTERISTICS OF 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 withP, as shown in the left figure above.
STATIC & STRENGTH OF MATERIALS
FRICTION CHARACERISTICS (continued)
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. Thevalue ofs depends on the materials in contact.
Once the block begins to move, the frictional force
typically drops and is given by Fk= kN. The value ofk (coefficient of kinetic friction) is less than s .
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STATIC & STRENGTH OF MATERIALS
DETERMING s
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, s, is noted. Analysis ofthe block just before it begins to move
gives (using Fs = s N):+ Fy = N W cos s = 0
+ FX = S N W sin s = 0
Using these two equations, we get s =(W sin
s
) / (W cos s
) = tan sThis simple experiment allows us to find
the S between two materials in contact.
STATIC & STRENGTH OF MATERIALS
PROCEDURE FOR ANALYSIS (Section 8.2)
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.
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STATIC & STRENGTH OF MATERIALS
IMPENDING TIPPING versus SLIPPING
For a given W and h, 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
EofE.
Hence, we have to make an
assumption to give us another
equation. Then we can solve for
the unknowns using the three
EofE. Finally, we need to check
if our assumption was correct.
STATIC & STRENGTH OF MATERIALS
IMPENDING TIPPING versus SLIPPING (continued)
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
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STATIC & STRENGTH OF MATERIALS
EXAMPLE
Given: A uniform ladder weighs 100 N. The
vertical wall is smooth (no friction).The floor is rough and s = 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 EofE (and friction equations, if appropriate ) to solve for theunknowns.
e) Check assumptions, if required.
STATIC & STRENGTH OF MATERIALS
EXAMPLE (continued)
There are four unknowns: NA, FA, NB, and P. Let usassume that the ladder will tip first. Hence, NB = 0
+ FY = NA 100 = 0 ; so NA = 100 N
+ MA = 100 ( 0.9 ) P( 1.2 ) = 0 ; so P = 75 N
+ FX = 75 FA = 0 ; so FA = 75 N
1.2 m
P100 N
NB
1.2 m
0.9 m 0.9 mNA
FA
A FBD of the ladder
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STATIC & STRENGTH OF MATERIALS
EXAMPLE (continued)
Now check the assumption.
Fmax = s NA = 0.8 * 100 N = 80 N
Is FA = 75 N Fmax = 80 N? Yes, hence our assumption oftipping is correct.
P100 N
NB
1.2 m
0.9 m 0.9 mNA
FA
A FBD of the ladder
STATIC & STRENGTH OF MATERIALS
CONCEPT QUIZ
1. A 100 N box with wide base is pulled by a
force P and s = 0.4. Which force orientationrequires the least force to begin sliding?
A) A B) B
C) C D) Can not be determined
2. A ladder is positioned as shown. Please indicate
the direction of the friction force on the ladder atB.
A) B)
C) D)
P(A)
P(B)
P(C)100 N
A
B
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STATIC & STRENGTH OF MATERIALS
GROUP PROBLEM SOLVING
Given: Drum weight = 500 N,
s = 0.5, a = 0.75 m and b = 1m.Find: The smallest magnitude
of P that will cause impending
motion (tipping or slipping) of
the drum.
Plan:
a) Draw a FBD of the drum.
b) Determine the unknowns.
c) Make friction assumptions, as necessary.
d) Apply EofE (and friction eqn. as appropriate) to solve for theunknowns.
e) Check assumptions, as required.
STATIC & STRENGTH OF MATERIALS
GROUP PROBLEM SOLVING (continued)
There are four unknowns: P, N, F and x.
First, lets assume the drum slips. Then the friction
equation is F = s N = 0.5 N.
A FBD of the drum:
X
3
4
50.375 m 0.375 m
500 N
0
1 m
F
P
N
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STATIC & STRENGTH OF MATERIALS
+ FX = (4 / 5) P 0.5 N = 0
+ FY = N (3 / 5) P 500 = 0
These two equations give:
GROUP PROBLEM SOLVING (continued)
P = 500 N and N = 800 N
+
MO = (3 /5) 500 (0.375) (4 / 5) 500 (1) + 800 (x) = 0Check: x = 0.359 0.375 so OK!
Drum slips as assumed at P = 500 N
A FBD of the drum:
X
3
4
50.375 m 0.375 m
500 N
0
1 m
F
P
N
STATIC & STRENGTH OF MATERIALS
1. A 10 N block is in equilibrium. What is
the magnitude of the friction force
between this block and the surface?
A) 0 N B) 1 N
C) 2 N D) 3 N
ATTENTION QUIZ
2. The ladder AB is postioned as shown. What is the
direction of the friction force on the ladder at B.
A) B)
C) D) A
B
S = 0.3
2 N
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STATIC & STRENGTH OF MATERIALS
FRICTION IN PRACTICALENGINEERING
STATIC & STRENGTH OF MATERIALS
READING QUIZ
1. A wedge allows a ______ force P to lift
a _________ weight W.
A) (large, large) B) (small, small)
C) (small, large) D) (large, small)
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 e
W
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STATIC & STRENGTH OF MATERIALS
APPLICATIONS
Wedges are used to adjust the
elevation or provide stability forheavy 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 onits own? Under what physical
conditions will it come out?
STATIC & STRENGTH OF MATERIALS
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, theapplied force P and the torque M,
related?
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STATIC & STRENGTH OF MATERIALS
ANALYSIS OF A WEDGE
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.
W
STATIC & STRENGTH OF MATERIALS
WEDGE ANALYSIS (continued)
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 impendingmotion frictional equation, F = S N.
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.
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STATIC & STRENGTH OF MATERIALS
WEDGE ANALYSIS (continued)
If the object is to be lowered, then the wedgeneeds 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 wedgeangle is large.
W
STATIC & STRENGTH OF MATERIALS
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.
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STATIC & STRENGTH OF MATERIALS
BELT ANALYSIS (continued)
Detailed analysis (please refer to your textbook) shows that
T2 = T1 e
where is the coefficient of static frictionbetween the belt and the surface. Be sure to use radians whenusing this formula!!
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.
Consider a flat belt passing over a fixedcurved surface with the total angle of
contact equal to radians.
STATIC & STRENGTH OF MATERIALS
EXAMPLEGiven: The load weighs 100 N and the
S between surfaces AC and BDis 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 EofE to wedge B. Why do B first?
4. Apply the EofE to wedge A.
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STATIC & STRENGTH OF MATERIALS
EXAMPLE (continued)
The FBDs of wedges A and B are shown
in the figures. Applying the EofE towedge B, we get
+ FX = N2 sin 10 N3 = 0
+ FY = N2 cos 10 100 0.3 N3 = 0
Solving the above two equations, we get
N2 = 107.2 N and N3 = 18.6 N
Applying the EofE to the wedge A, we get
+ FY = N1 107.2 cos 10 = 0; N1 = 105.6 N
+ FX = P 107.2 sin 10 0.3 N1 = 0; P = 50.3 N
10N2
AP
N1F1= 0.3N1
N2 10
B
F3= 0.3N3
N3
100 N
STATIC & STRENGTH OF MATERIALS
CONCEPT QUIZ
2. The boy (hanging) in the picture weighs
500 N and the woman weighs 750 N. The
coefficient of static friction between hershoes 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.
1. Determine the direction of the friction
force on object B at the contact point
between A and B.
A) B)
C) D)
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STATIC & STRENGTH OF MATERIALS
GROUP PROBLEM SOLVING
Given: Blocks A and B weigh 50 N and
30 N, respectively.
Find: The smallest weight of cylinder D
which will cause the loss of static
equilibrium.
STATIC & STRENGTH OF MATERIALS
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.
GROUP PROBLEM SOLVING (continued)
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STATIC & STRENGTH OF MATERIALS
GROUP PROBLEM SOLUTION
Case a:
+ FY = N 80 = 0
N = 80 N
+ FX = 0.4 (80) P = 0
P = 32 N
PB
A
N
F=0.4 N
30 N
50 N
STATIC & STRENGTH OF MATERIALS
GROUP PROBLEM SOLUTION
+ Fy = N cos 20 + 0.6 N sin 20 30 = 0N = 26.20 N
+ Fx = P + 0.6 ( 26.2 ) cos 20 26.2 sin 20 = 0P = 5.812 N
Case b has the lowest P and will occur first. Next, using the
frictional force analysis of belt, we get
WD = P e = 5.812 e 0.5 ( 0.5 ) = 12.7 N
A Block D weighing 12.7 N will cause the block B to slide over
the block A.
20
30 N
0.6 NP
N
Case b:
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STATIC & STRENGTH OF MATERIALS
ATTENTION QUIZ
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
2. In the analysis of frictional forces on a flat belt, T2 = T1 e .
In this equation, equals ______ .
A) angle of contact in degrees B) angle of contact in radians
C) coefficient of static friction D) coefficient of kinetic friction
W
STATIC & STRENGTH OF MATERIALS