EGR 277

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Chapter 8 - Friction

• Chapter 8 Friction
• Previously we have only considered two types of
  surfaces
• Frictionless surfaces no resistance to movement
• Rough surfaces no movement assumed
•  
• Now we will consider friction forces in greater
  detail.
• There are two types of friction
• Dry (Coulomb) friction covered in this chapter
• Fluid friction covered in a Fluid Mechanics
  course

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Dry Friction Consider the diagram shown below on
the left. The block rests on a surface where
friction will be present between the block and
the surface. A force P is applied to attempt to
move the block. The FBD on the right shows the
familiar reaction, including a normal force, N,
and a frictional force, F. Note that F always
opposes the direction of motion.
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N
-  If a force of P 10 lb is applied, the
friction force pushes back with a F 10
lb -  If a force of P 20 lb is applied, the
friction force pushes back with a F 20
lb -  If a force of P 30 lb is applied, the
friction force pushes back with a F 30 lb ? ?
? -   If P increases until the block is about to
move, then F Fm (maximum friction force) -  
Once the block begins moving, the friction force
decreases as the kinetic friction force, Fk, is
less than the static friction force (perhaps 25
less). -   The graph on the following page
illustrates the situation described above.
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Coefficients of friction The maximum friction
force, Fm, is controlled by the normal force, N,
and the roughness of the material. Fm ?sN
where ?s the coefficient of static
friction Similarly, Fk ?kN where ?k the
coefficient of kinetic friction  Some typical
values for ?s are listed below
Some notes regarding ?s 1) ?s is approximately
independent of surface area 2) ?s is unitless and
can be used to calculate friction forces with
either SI or US units 3) 0 lt ?s lt 1
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Angles of friction For impending motion, Fm and N
are related by Fm ?sN. Fm and N are also the
components of the reaction, R, where the block
rests on the surface as shown below on the left.
The force triangle on the right further
illustrates the relationship between R, N, and
Fm.
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Relationship between the angle of incline and
?s Will the block shown below slide down the
incline? It turns out that the weight of the
block is not a factor and the answer depends on
the relative values of ? and ?s. Develop this
relationship in class.
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Demonstration Raise the angle of an incline
until a block begins to slide down the incline.
Use this information to determine the coefficient
of friction.
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Impending upward motion versus impending downward
motion In the diagram shown below, there is a
range of values of P for which the block will be
in equilibrium (not moving uphill or downhill).

• The minimum value of P is when the block is about
  to begin sliding down the incline (impending
  downward motion). In this case the friction
  force is directed up the incline (opposing
  motion).
• The maximum value of P is when the block is about
  to begin moving up the incline (impending upward
  motion). In this case the friction force is
  directed down the incline (opposing motion).

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Example Determine the range of values of P in
the problem shown below for which the block will
remain in equilibrium. Assume that ?s 0.25.
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Example Determine the minimum coefficient of
static friction if the block shown below is to
move upward.
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Example A 60 kg cabinet is moved by applying a
force P as shown. If the casters are locked the
cabinet will slide (or tip) rather than roll. If
?s 0.30, determine the value of P to move the
cabinet if 1) all casters are locked 2) the
casters at B only are locked 3) the casters at A
only are locked In each case also check to see if
the cabinet tips.
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• Wedges
• Wedges are simple machines that can be used to
  lift heavy loads or to make minor adjustments in
  load positions.
• Notes
• The reactions are equal and opposite between
  blocks, so be sure to change the direction of F
  and N as they are transferred from one block (or
  wedge) to the next.
• It is easy to make a mistake in the direction of
  the friction force in wedge problems. It is
  easiest to begin by determining the direction of
  the friction force in a moving part (such as a
  wedge) and then transfer the force (reverse
  direction) to an adjacent object.

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Example Determine the force P required to begin
moving the 200 lb block upward if ?s 0.25 for
all surfaces in contact. Neglect the weight of
the wedge.
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Square-Threaded Screws

• Square-threaded screws frequently used in jacks,
  presses, etc. Analysis similar to block on
  inclined plane. Recall friction force does not
  depend on area of contact.

• Thread of base has been unwrapped and shown as
  straight line. Slope is 2pr horizontally and
  lead (or pitch) L vertically. So,

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Two important features of a threaded device that
relate to friction calculations are the mean
radius of the screw and the pitch or advance or
lead of the screw.
Note the radius shown at the right. Although the
advance that a screw makes in one revolution may
be stated in terms of pitch or lead, the
fundamental measurement of interest is how far
the screw advances in one revolution.
The next slide examines the forces acting on such
a particle.
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This is a model of a screw jack that might be
used to lift a heavy load. Our analyses will
include two cases (a) When the screw is
advancing toward the applied load, and (b) when
the screw is moving away from the applied load.
SFX 0 S R sin(? ?)
SFY 0 R cos(? ?) - W
Then M Sr Wr tan(? ?)
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When the motion is downward (the screw retreating
from the load, the friction resultant is on the
opposite side of the normal index and
M W r tan(? F)
For downward impending motion
If fS is greater than ? the screw is said to be
self-locking, which means that the screw will not
back down without a moment applied to turn it in
that direction.
Conversely, if fS is less than ? the screw will
retreat under the load automatically and must be
held in place by application of a moment or by a
brake.
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Example The car jack shown below is being used
to lift a load of F 6.5 kN. The jack is
operated by turning the threaded shaft at A,
causing points B and D to move closer together
which causes point C to move upward vertically.
The pitch of the threaded shaft is 5 mm and the
mean radius of the thread is 10 mm, and the
coefficient of kinetic friction is 0.15, what
couple M is necessary to turn the shaft at a
constant rate moving point C upward?
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Axle Friction (Journal Bearings) Journal bearings
provide lateral support for rotating axles and
shafts. Journal bearings may be found on
wheelbarrows, wagons, carts, hinges, and with
many other types of axles. Some examples are
shown below.
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Axle Friction (Journal Bearings) If the journal
bearings fit the axle tightly and are
well-lubricated, then the laws of fluid mechanics
are used to determine the frictional resistance.
However, if the bearing is somewhat loose fitting
and is not well-lubricated, then the laws of dry
friction apply. This is the approach used here.

• Assumptions
• The axle is loose fitting
• The bearing is not well-lubricated
• As the axle rotates, it climbs up the wall of
  the support and there is a single point of
  contact between the axle and the support.

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Analysis Axle Friction It is assumed that the
axle and support only make contact at one point,
A. The reaction force, R, has a normal component,
N, and a tangential frictional component, F, as
shown below.
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Example Determine the force, P, needed to raise
the 100 lb load shown below at a constant
velocity if the coefficient of kinetic friction
between the axle and the support is 0.25. The
diameter of the pulley is 4 inches and the
diameter of the pulley shaft is 0.75 inches.
Neglect belt friction.
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Example Determine the force, P, needed to raise
the 100 lb load shown below at a constant
velocity if the coefficient of kinetic friction
between the axle and the support is is 0.25. The
diameter of the pulley is 4 inches and the
diameter of the pulley shaft is 0.75 inches.
Neglect belt friction.
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Belt Friction Frictional forces on belts, bands,
or ropes are considered in this section. We have
assumed that the tension is the same on each side
of an ideal pulley. However, if belt friction is
considered, a larger tension will exist on one
side in order to overcome the friction that
opposed the motion of the belt.
The text presents a derivation of the
relationship provided below.
Where T1 and T2 are belt tensions. Note that T2
gt T1 since T2 opposed friction. ? coefficient
of static or kinetic friction between the belt
and the surface of contact ? angle of contact
(in radians) e 2.71828 base of the natural
log
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Example A hawser is wrapped around a fixed
capstan to secure a ship for docking. If the
tension in the rope, caused by the ship, is 1500
lb, determine the least number of complete turns
the rope must be wrapped around the capstan in
order to prevent slipping of the rope. The
greatest horizontal force that a longshoreman can
exert on the rope is 50 lb. Assume that ?s 0.3
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Example - A flat belt is used to transmit a
torque from drum B to drum A. Knowing that the
coefficient of static friction is 0.40 and that
the maximum allowable belt tension is 450N,
determine the largest torque that can be exerted
on drum A.