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Contents
Introduction
Laws of Dry Friction. Coefficients of Friction.
Angles of Friction
Problems Involving Dry Friction
Sample Problem 8.1
Sample Problem 8.3
Wedges
Square-Threaded Screws
Sample Problem 8.5
Journal Bearings. Axle Friction.
Thrust Bearings. Disk Friction.
Wheel Friction. Rolling Resistance.
Sample Problem 8.6
Belt Friction.
Sample Problem 8.8
School of Mechanical Engineering 8 - 2
Introduction
Laws of Dry Friction. Coefficients of Friction.
Angles of Friction
Problems Involving Dry Friction
Sample Problem 8.1
Sample Problem 8.3
Wedges
Square-Threaded Screws
Sample Problem 8.5
Journal Bearings. Axle Friction.
Thrust Bearings. Disk Friction.
Wheel Friction. Rolling Resistance.
Sample Problem 8.6
Belt Friction.
Sample Problem 8.8
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Introduction• In preceding chapters, it was assumed that surfaces in contact were
either frictionless (surfaces could move freely with respect to each other) or rough (tangential forces prevent relative motion between surfaces).
• Actually, no perfectly frictionless surface exists. For two surfaces in contact, tangential forces, called friction forces, will develop if one attempts to move one relative to the other.
School of Mechanical Engineering 8 - 3
• However, the friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied.
• The distinction between frictionless and rough is, therefore, a matter of degree.
• There are two types of friction: dry or Coulomb friction and fluid friction. Fluid friction applies to lubricated mechanisms. The present discussion is limited to dry friction between nonlubricated surfaces.
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The Laws of Dry Friction. Coefficients of Friction
• Block of weight W placed on horizontal surface. Forces acting on block are its weight and reaction of surface N.
• Small horizontal force P applied to block. For block to remain stationary, in equilibrium, a horizontal component F of the surface reaction is required. F is a static-friction force.
School of Mechanical Engineering 8 - 4
• Small horizontal force P applied to block. For block to remain stationary, in equilibrium, a horizontal component F of the surface reaction is required. F is a static-friction force.
• As P increases, the static-friction force Fincreases as well until it reaches a maximum value Fm.
NF sm m=
• Further increase in P causes the block to begin to move as F drops to a smaller kinetic-friction force Fk.
NF kk m=
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The Laws of Dry Friction. Coefficients of Friction
• Maximum static-friction force:NF sm m=
• Kinetic-friction force:
sk
kk NFmm
m75.0@
=
School of Mechanical Engineering 8 - 5
• Maximum static-friction force and kinetic-friction force are:- proportional to normal force- dependent on type and condition of
contact surfaces- independent of contact area
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The Laws of Dry Friction. Coefficients of Friction
• Four situations can occur when a rigid body is in contact with a horizontal surface:
School of Mechanical Engineering 8 - 6
• No friction,(Px = 0)
• No motion,(Px < Fm)
• Motion impending,(Px = Fm)
• Motion,(Px > Fm)
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Angles of Friction
• It is sometimes convenient to replace normal force N and friction force F by their resultant R:
School of Mechanical Engineering 8 - 7
• No friction • Motion impending• No motion
ss
sms N
NN
F
mf
mf
=
==
tan
tan
• Motion
kk
kkk N
NNF
mf
mf
=
==
tan
tan
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Angles of Friction
• Consider block of weight W resting on board with variable inclination angle q.
School of Mechanical Engineering 8 - 8
• No friction • No motion • Motion impending
• Motion
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Problems Involving Dry Friction
School of Mechanical Engineering 8 - 9
• All applied forces known
• Coefficient of static friction is known
• Determine whether body will remain at rest or slide
• All applied forces known
• Motion is impending
• Determine value of coefficient of static friction.
• Coefficient of static friction is known
• Motion is impending
• Determine magnitude or direction of one of the applied forces
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Sample Problem 8.1SOLUTION:• Determine values of friction force
and normal reaction force from plane required to maintain equilibrium.
• Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide.
School of Mechanical Engineering 8 - 10
A 100 N force acts as shown on a 300 N block placed on an inclined plane. The coefficients of friction between the block and plane are ms = 0.25 and mk = 0.20. Determine whether the block is in equilibrium and find the value of the friction force.
• Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide.
• If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force.
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Sample Problem 8.1SOLUTION:• Determine values of friction force and normal
reaction force from plane required to maintain equilibrium.
:0=å xF ( ) 0N 300 - N 100 53 =- F
N 80-=F
:0=å yF ( ) 0N 300 - 54 =N
School of Mechanical Engineering 8 - 11
:0=å yF ( ) 0N 300 - 54 =N
N 240=N
• Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide.
( )0.25 240 N 60 Nm s mF N Fm= = =
The block will slide down the plane.
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Sample Problem 8.1
• If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force.
( )N 240200N
.FF kkactual
=== m
N 48=actualF
School of Mechanical Engineering 8 - 12
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Sample Problem 8.3
SOLUTION:
• When W is placed at minimum x, the bracket is about to slip and friction forces in upper and lower collars are at maximum value.
• Apply conditions for static equilibrium to find minimum x.
School of Mechanical Engineering 8 - 13
The moveable bracket shown may be placed at any height on the 3-cm diameter pipe. If the coefficient of friction between the pipe and bracket is 0.25, determine the minimum distance x at which the load can be supported. Neglect the weight of the bracket.
• Apply conditions for static equilibrium to find minimum x.
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Sample Problem 8.3SOLUTION:• When W is placed at minimum x, the bracket is about to
slip and friction forces in upper and lower collars are at maximum value.
BBsB
AAsANNFNNF
25.025.0
====
mm
• Apply conditions for static equilibrium to find minimum x.
School of Mechanical Engineering 8 - 14
• Apply conditions for static equilibrium to find minimum x.:0=å xF 0=- AB NN AB NN =
:0=å yF
WNWNN
WFF
A
BA
BA
==-+
=-+
5.0025.025.0
0
WNN BA 2==
:0=å BM ( ) ( ) ( )( ) ( )
( ) ( ) ( ) 05.1275.02605.125.036
0cm5.1cm3cm6
=---=---
=---
xWWWxWNNxWFN
AA
AA
cm12=x
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Wedges
• Wedges - simple machines used to raise heavy loads.
• Force required to lift block is significantly less than block weight.
• Friction prevents wedge from sliding out.
• Want to find minimum force P to raise block.
• Block as free-body • Wedge as free-body
School of Mechanical Engineering 8 - 15
• Wedges - simple machines used to raise heavy loads.
• Force required to lift block is significantly less than block weight.
• Friction prevents wedge from sliding out.
• Want to find minimum force P to raise block.
• Block as free-body
0
:00
:0
21
21
=+--
==+-
=
å
å
NNW
FNN
F
s
y
s
x
m
m
or
021 =++ WRRvrr
( )
( ) 06sin6cos
:00
6sin6cos:0
32
32
=°-°+-
==+
°-°--=
å
å
s
y
ss
x
NN
FP
NNF
m
mm
• Wedge as free-body
or032 =+- RRP
rrr
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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 L vertically.
• Moment of force Q is equal to moment of force P. rPaQ =
School of Mechanical Engineering 8 - 16
• Impending motion upwards. Solve for Q.
• Self-locking, solve for Q to lower load.
,qf >s • Non-locking, solve for Q to hold load.
,qf >s
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Sample Problem 8.5
A clamp is used to hold two pieces of wood together as shown. The clamp has a double square thread of mean diameter equal to 10 mm with a pitch of 2 mm. The coefficient of friction between threads is ms = 0.30.
If a maximum torque of 40 N*m is applied in tightening the clamp, determine (a) the force exerted on the pieces of wood, and (b) the torque required to loosen the clamp.
SOLUTION
• Calculate lead angle and pitch angle.
• Using block and plane analogy with impending motion up the plane, calculate the clamping force with a force triangle.
• With impending motion down the plane, calculate the force and torque required to loosen the clamp.
School of Mechanical Engineering 8 - 17
A clamp is used to hold two pieces of wood together as shown. The clamp has a double square thread of mean diameter equal to 10 mm with a pitch of 2 mm. The coefficient of friction between threads is ms = 0.30.
If a maximum torque of 40 N*m is applied in tightening the clamp, determine (a) the force exerted on the pieces of wood, and (b) the torque required to loosen the clamp.
• With impending motion down the plane, calculate the force and torque required to loosen the clamp.
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Sample Problem 8.5
SOLUTION
• Calculate lead angle and pitch angle. For the double threaded screw, the lead L is equal to twice the pitch.
( )
30.0tan
1273.0mm 10mm22
2tan
==
===
ss
rL
mfpp
q °= 3.7q
°= 7.16sf
• Using block and plane analogy with impending motion up the plane, calculate clamping force with force triangle.
School of Mechanical Engineering 8 - 18
• Using block and plane analogy with impending motion up the plane, calculate clamping force with force triangle.
kN8mm5
mN 40mN 40 =×
=×= QrQ
( )°
==+24tan
kN8tan WWQ
sfq
kN97.17=W
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Sample Problem 8.5
• With impending motion down the plane, calculate the force and torque required to loosen the clamp.
( ) ( ) °==- 4.9tankN97.17tan QWQ
s qf
kN975.2=Q
( )( )( )( )m105N10975.2
mm5kN975.233 -´´=
== rQTorque
School of Mechanical Engineering 8 - 19
( )( )( )( )m105N10975.2
mm5kN975.233 -´´=
== rQTorque
mN87.14 ×=Torque
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Journal Bearings. Axle Friction
• Journal bearings provide lateral support to rotating shafts. Thrust bearings provide axial support
• Frictional resistance of fully lubricated bearings depends on clearances, speed and lubricant viscosity. Partially lubricated axles and bearings can be assumed to be in direct contact along a straight line.
• Forces acting on bearing are weight W of wheels and shaft, couple M to maintain motion, and reaction Rof the bearing.
School of Mechanical Engineering 8 - 20
• Forces acting on bearing are weight W of wheels and shaft, couple M to maintain motion, and reaction Rof the bearing.
• Reaction is vertical and equal in magnitude to W.
• Reaction line of action does not pass through shaft center O; R is located to the right of O, resulting in a moment that is balanced by M.
• Physically, contact point is displaced as axle “climbs” in bearing.
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Journal Bearings. Axle Friction
School of Mechanical Engineering 8 - 21
• Angle between R and normal to bearing surface is the angle of kinetic friction jk.
k
kRrRrMm
f»= sin
• May treat bearing reaction as force-couple system.
• For graphical solution, R must be tangent to circle of friction.
k
kf
r
rr
m
f
»
= sin
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Thrust Bearings. Disk Friction
Consider rotating hollow shaft:
( )21
22 RR
APr
AAPrNrFrM
k
kk
-
D=
D=D=D=D
pm
mm
( )
21
22
31
32
32
2
0
221
22
2
1
RRRRP
drdrRR
PM
k
R
R
k
-
-=
-= ò ò
m
qp
m p
School of Mechanical Engineering 8 - 22
For full circle of radius R,
PRM km32=
( )
21
22
31
32
32
2
0
221
22
2
1
RRRRP
drdrRR
PM
k
R
R
k
-
-=
-= ò ò
m
qp
m p
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Wheel Friction. Rolling Resistance
School of Mechanical Engineering 8 - 23
• Point of wheel in contact with ground has no relative motion with respect to ground.
Ideally, no friction.
• Moment M due to frictional resistance of axle bearing requires couple produced by equal and opposite P and F.
Without friction at rim, wheel would slide.
• Deformations of wheel and ground cause resultant of ground reaction to be applied at B. P is required to balance moment of W about B.
Pr = Wbb = coef of rolling resistance
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Sample Problem 8.6
A pulley of diameter 0.1m can rotate about a fixed shaft of diameter .05 m. The coefficient of static friction between the pulley and shaft is 0.20.
Determine:• the smallest vertical force P
required to start raising a 500 N load,
• the smallest vertical force Prequired to hold the load, and
• the smallest horizontal force P required to start raising the same load.
SOLUTION:
• With the load on the left and force P on the right, impending motion is clockwise to raise load. Sum moments about displaced contact point B to find P.
• Impending motion is counter-clockwise as load is held stationary with smallest force P. Sum moments about C to find P.
School of Mechanical Engineering 8 - 24
• the smallest vertical force Prequired to start raising a 500 N load,
• the smallest vertical force Prequired to hold the load, and
• the smallest horizontal force P required to start raising the same load.
• Impending motion is counter-clockwise as load is held stationary with smallest force P. Sum moments about C to find P.
• With the load on the left and force P acting horizontally to the right, impending motion is clockwise to raise load. Utilize a force triangle to find P.
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Sample Problem 8.6
SOLUTION:
• With the load on the left and force P on the right, impending motion is clockwise to raise load. Sum moments about displaced contact point B to find P.
The perpendicular distance from center O of pulley to line of action of R is
( ) m005.020.0m 025.sin =»»= fssf rrrr mj
School of Mechanical Engineering 8 - 25
( ) m005.020.0m 025.sin =»»= fssf rrrr mj
Summing moments about B,( ) ( ) ( ) 0m .0450N500 m .0550:0 =-=å PM B
N611=P
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Sample Problem 8.6
The perpendicular distance from center O of pulley to line of action of R is again 0.20 cm. Summing moments about C,
• Impending motion is counter-clockwise as load is held stationary with smallest force P. Sum moments about C to find P.
School of Mechanical Engineering 8 - 26
The perpendicular distance from center O of pulley to line of action of R is again 0.20 cm. Summing moments about C,
( )( ) ( ) 0m .0550N500m 0.045:0 =-=å PM C
N409=P
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Sample Problem 8.6• With the load on the left and force P acting
horizontally to the right, impending motion is clockwise to raise load. Utilize a force triangle to find P.Since W, P, and R are not parallel, they must be concurrent. Line of action of R must pass through intersection of W and P and be tangent to circle of friction which has radius rf = .005 m.
School of Mechanical Engineering 8 - 27
Since W, P, and R are not parallel, they must be concurrent. Line of action of R must pass through intersection of W and P and be tangent to circle of friction which has radius rf = .005 m.
( )°=
===
1.4
0707.02mc2
20.0sin
q
q cmODOE
From the force triangle,
( ) ( ) °=-°= 9.40cotN50045cot qWP
N577=P
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Belt Friction• Relate T1 and T2 when belt is about to slide to right.
• Draw free-body diagram for element of belt
( ) 02
cos2
cos:0 =D-D
-D
D+=å NTTTF sx mqq
( ) 02
sin2
sin:0 =D
-D
D+-D=åqq TTTNFy
• Combine to eliminate DN, divide through by Dq,
School of Mechanical Engineering 8 - 28
• Combine to eliminate DN, divide through by Dq,( )
22sin
22cos
qqmq
q DD
÷øö
çèæ D
+-D
DD TTT
s
• In the limit as Dq goes to zero,
0=- TddT
smq• Separate variables and integrate from bqq == to0
bmbm seTT
TT
s ==1
2
1
2 orln
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Sample Problem 8.8
SOLUTION:
• Since angle of contact is smaller, slippage will occur on pulley B first. Determine belt tensions based on pulley B.
• Taking pulley A as a free-body, sum moments about pulley center to determine torque.
School of Mechanical Engineering 8 - 29
A flat belt connects pulley A to pulley B.The coefficients of friction are ms = 0.25and mk = 0.20 between both pulleys andthe belt.
Knowing that the maximum allowabletension in the belt is 600 N, determine thelargest torque which can be exerted bythe belt on pulley A.
• Taking pulley A as a free-body, sum moments about pulley center to determine torque.
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Sample Problem 8.8
SOLUTION:
• Since angle of contact is smaller, slippage will occur on pulley B first. Determine belt tensions based on pulley B.
( )
N4.3551.688
N600
688.1N600
1
3225.0
11
2 s
==
===
T
eT
eTT pbm
School of Mechanical Engineering 8 - 30
( )
N4.3551.688
N600
688.1N600
1
3225.0
11
2 s
==
===
T
eT
eTT pbm
• Taking pulley A as free-body, sum moments about pulley center to determine torque.
0 m) (0.2 N) (355.4 m) (0.2 N) (600 - M:0 A =+=å AM
mM A ×= N 9.48