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8/6/2019 13726898 Applied Mechanics Friction
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Friction
Earlier we assumed action and reaction forces at contactingsurfaces are normal
Seen as smooth surface not practically true
Normal & tangential forces are important
Tangential forces generated near contacting surfaces areFRICTIONAL FORCES
Sliding of one contact surface to other friction occurs and it isopposite to the applied force
Reduce friction in bearings, power screws, gears, aircraftpropulsion, missiles through the atmosphere, fluid flow etc.
Maximize friction in brakes, clutches, belt drives etc.
Friction dissipated as heat loss of energy, wear of parts etc.
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Friction
Dry friction
(coulomb friction)
Fluid friction
Occurs when un-lubricated surfacesare in contact during sliding
friction force always oppose thesliding motion
Occurs when the adjacent layers ina fluid (liquid, gas) are moving atdifferent velocities
This motion causes friction betweenfluid elements
Depends on the relative velocitybetween layers
No relative velocity no fluid friction depends on the viscosity of fluidmeasure of resistance to shearingaction between the fluid layers
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Dry friction: Laws of dry frictionW
N
W weight; N Reaction of the surface
Only vertical component
P applied load F static friction force : resultant of many forces actingover the entire contact area
Because of irregularities in surface & molecular attraction
A
W
P
N
F
A
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P
W
N
F
A B
P is increased; F is also increased and continue to oppose P
This happens till maximum Fm is reached Body tend to move till Fm isreached
After this point, block is in motion
Block in motion: Fm reduced to Fk lower value kinetic friction force andit remains same related to irregularities interaction
N reaches B from A Then tipping occurs abt. B
Fm
Fk
F
p
Equilibrium Motion
More irregularities
interaction
Less irregularities
interaction
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EXPERIMENTAL EVIDENCE:Fm proportional to N
Fm = s N; s static friction co-efficient
Similarly, Fk = k N; k kinetic friction co-efficient
s and k depends on the natureof surface; not on contact areaof surface
k = 0.75 s
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Four situations can occur when a rigid body is in contact with a
horizontal surface:
We have horizontal and vertical force equilibrium equns. and
F = N
No motion,
(Px
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No motion Motion No friction Motion impending
It is sometimes convenient to replace normal forceNand friction
forceFby their resultant R:
ss
sms
N
N
N
F
=
==
tan
tan
kk
kkk
N
N
N
F
=
==
tan
tan
s angle of staticfriction maximumangle (like Fm)
k angle of kinetic
friction; k < s
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Consider block of weight Wresting on board with variable
inclination angle .
Angle of inclination =
angle of repose; = s
R Not vertical
ANGLE OF INCLINATION IS INCREASING
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Three categories of problems
All applied forces are given, co-effts. of friction are known
Find whether the body will remain at rest or slide
Friction force F required to maintain equilibrium is unknown
(magnitude not equal to sN)
Determine F required for equilibrium, by solving equilibrium equns;
Also find N
Compare F obtained with maximum value Fm i.e., from Fm = sN
F is smaller or equal to Fm, then body is at rest
Otherwise body starts moving
Actual friction force magnitude = Fk= kN
Solution
First category: to know a body slips or not
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A 100 N force acts as shown on a 300 N blockplaced on an inclined plane. The coefficients of
friction between the block and plane are s =
0.25 and k= 0.20. Determine whether the
block is in equilibrium and find the value of the
friction force.
Beer/Johnston
:0=F ( ) 0N300-N100 53 =F
80=:0= yF ( ) 0N300- 5
4 =N
N240=
The block will slide down the plane.
Fm < F
Fm = s N = 0.25 (240) = 60 N
= 36.9 DEG
= 36.9
DEG
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If maximum friction force is less than friction
force required for equilibrium, block will slide.
Calculate kinetic-friction force.
( )N240200N
.FF kkactual
===
N48=actualF
Fm
Fk
F
p
Equilibrium Motion
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Meriam/Kraige; 6/8
M
30
Cylinder weight: 30 kg; Dia: 400 mmStatic friction co-efft: 0.30 between cylinder and surface
Calculate the applied CW couple M which cause thecylinder to slip
30 x 9.81
NA
FA = 0.3 NANB
FB = 0.3 NB
M
C
Fx = 0 = -NA+0.3NB Cos 30-NB Sin 30 = 0
Fy = 0 =>-294.3+0.3NA+NBCos 30-0.3NB Sin 30 = 0
Find NA & NB by solving these two equns.
MC = 0 = > 0.3 NA (0.2)+0.3 NB (0.2) - M = 0
Put NA & NB; Find M
NA = 237 N & NB = 312 N; M = 33 Nm
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Meriam/Kraige; 6/5
Wooden block: 1.2 kg; Paint: 9 kgDetermine the magnitude and direction of (1) thefriction force exerted by roof surface on the woodenblock, (2) total force exerted by roof surface on thewooden block
= tan-1 (4/12) = 18.43
Paint
Wooden
block
12
4
Roofsurface
(2) Total force = 10.2 x 9.81 = 100.06 N UP
N
F
10.2x 9.81
X
Y
(1)Fx = 0 => -F+100.06 sin 18.43 => F = 31.6 N
Fy = 0 => N = 95 N
Second category: Impending relative motion when
two or three bodies in contact with each other
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Beer/Johnston
20 x 9.81 = 196.2 N
N1
F1
T
30 x 9.81 = 294.3 N
F2
For 20 kg block For 30 kg block
F1P
N1
N2
(a)
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(B)490.5 N
N
P
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Beer/Johnston
A
B
6 m
2.5 m
A 6.5-m ladder AB of mass 10 kg leans against a wall as shown.
Assuming that the coefficient of static friction on s is the same
at both surfaces of contact, determine the smallest value of sfor which equilibrium can be maintained.
A
B
FB
NB
FANA
W
1.25 1.25
O
Slip impends at both A and B, FA= sNA, FB= sNB
Fx=0=> FANB=0, NB=FA=sNA
Fy=0=> NAW+FB=0, NA+FB=W
NA+sNB=W; W = NA(1+s2)
Mo = 0 => (6) NB - (2.5) (NA) +(W) (1.25) = 0
6sNA - 2.5 NA + NA(1+s2) 1.25 = 0
s = -2.4 2.6 = > Min s = 0.2
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Wedges
Wedges - simple machines used to raiseheavy loads like wooden block, stone etc.
Loads can be raised by applying force P towedge
Force required to lift block is significantlyless than block weight
Friction at AC & CD prevents wedge fromsliding out
Want to find minimum force P to raise block
A wooden block
C, D Wedges
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0
:0
0
:0
21
21
=+
=
=+
=
NNW
F
NN
F
s
ys
x
FBD of block
( )
( ) 06sin6cos
:0
0
6sin6cos
:0
32
32
=+
=
=+
=
s
y
ss
x
NN
F
P
NN
F
FBD of wedge
N3
6
F36
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Two 8 wedges of negligible weight are used tomove and position a 530-N block. Knowing that thecoefficient of static friction is 0.40 at all surfaces ofcontact, determine the magnitude of the force P forwhich motion of the block is impending
Beer/Johnston
s = tan1 s = tan1 (0.4) = 21.801
21.8
R1
FBD of block
20
21.8
530
R2530
R2
R141.8
91.8 46.4 (R2/Sin 41.8) = (530/sin 46.4)
R2 = 487.84 N
Using sine law,
slip impends at wedge/block
wedge/wedge and block/incline
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P = 440.6 N
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Beer/Johnston
A 6 steel wedge is driven into the end of an axhandle to lock the handle to the ax head. Thecoefficient of static friction between the wedge andthe handle is 0.35. Knowing that a force P ofmagnitude 60 N was required to insert the wedge
to the equilibrium position shown, determine themagnitude of the forces exerted on the handle bythe wedge after force P is removed.
P = 60 N s = tan1
s= tan1
(0.35 ) = 19.29
19.29
3
19.29
36
By symmetry R1= R2; in EQUILIBRIUM
Fy = 0: 2R1 sin 22.29 60 N =0
R1 = R2 = 79.094 N
WHAT WILL HAPPEN IF P IS REMOVED ?
R1R2
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Vertical component of R1, R2 will be eliminated
Hence, H1 = H2 = 79.094 N cos22.29 = 73.184 N
Final force = 73.184 N
Since included angle is 3(< s) from the normal,the wedge is self-locking and will remain in place.
No motion
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Screws
Used for fastening, transmitting power or motion, lifting body
Square threaded jack - screw jack V-thread is also
possible
W- AXIAL LOAD
M APPLIED MOMENT ABOUT AXIS OF SCREW
M = P X r
L LEAD DISTANCE Advancement per
revolution
HELIX ANGLE
M
Upward
motion
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2r
L
W
RP = M/r
One full threadof screw
To raise load
M
F
angle of friction
R
P
w +
tan (+) = P/W = M/rW
=> M = rW tan (+) = tan-1 (L/2r)
To lower load unwinding condition
P = M/r
W
R
< Screw will remain in place self locking
=> M = rW tan (-)
= In verge of un-winding
Moment required tolower the screw
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P = M/r
W
R
> Screw will unwind itself
=> M = rW tan (-)Moment required toprevent unwinding
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Beer/Johnston A clamp is used to hold two pieces of woodtogether as shown. The clamp has a double
square thread of mean diameter equal to 10 mmwith a pitch of 2 mm. The coefficient of friction
between threads is s = 0.30.
If a maximum torque of 40 Nm is applied in
tightening the clamp, determine (a) the forceexerted on the pieces of wood, and (b) the torque
required to loosen the clamp.
Lead distance = 2 x pitch = 2 x 2 = 4 mm
r = 5 mm
( )
30.0tan
1273.0mm10mm22
2tan
==
===
ss
rL
= 3.7
= 7.16s
(double square thread)
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The position of the automobile jack shown is
controlled by a screw ABC that is single-threaded at each end (right-handed thread atA, left-handed thread at C). Each thread hasa pitch of 2 mm and a mean diameter of 7.5
mm. If the coefficient of static friction is 0.15,determine the magnitude of the couple Mthat must be applied to raise the automobile.
Beer/Johnston
FBD jointD
:
Fy = 0 => 2FADsin254 kN=0
FAD = FCD = 4.73 kN
By symmetry:
4 kN
FAD FCD
25 25D
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FBD joint A:
4.73 kN
FAC
FAE = 4.73
25
25AFx = 0 => FAC2(4.73) cos25=0FAC = 8.57 kN
L = Pitch = 2 mm
W = FAC = 8.57
R
Joint A
P = M/r
(7.5)
Here is used instead of used earlier
MA = rW tan (+) = (7.5/2) (8.57) tan (13.38) = 7.63 Nm
Similarly, at C, Mc = 7.63 Nm (by symmetry); Total moment = 7.63 (2) = 15.27 Nm
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Journal & Thrust bearing
Journal bearings provide lateral support to rotating shafts
Thrust bearings provide axial support
Journal bearing - Axle frictionThrust bearing - Disc friction
shaft
bearing
shaft
bearing
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Friction between two
ring shaped areas
Friction in full circular area
- DISK FRICTION (Eg., Disc clutch)
Consider Hollow shaft (R1, R2)
M Moment required for shaftrotation at constant speed
P axial force which maintains
shaft in contact with bearing
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Couple moment required to overcome frictionresistance, M
Equilibrium conditions and moment equations arenecessary to solve problems
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A .178 m-diameter buffer weighs 10.1 N. The
coefficient of kinetic friction between the buffingpad and the surface being polished is 0.60.Assuming that the normal force per unit areabetween the pad and the surface is uniformly
distributed, determine the magnitudeQ
of thehorizontal forces required to prevent motion ofthe buffer.
Beer/Johnston
O
M
Q - Q0.2 m
Mo = 0 => (0.2) Q M = 0; Q = M / 0.2
M = 2/3 (0.6) (10.1) (0.178/2) = 0.36 Nm
Q = M / 0.2 = 0.36/0.2 = 1.8 N
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V- Belt
T2/T1 = eS /sin (/2)
ln (T2/T1) = S ; T2/T1 = e S
Applicable to belts passing over fixed drums; ropes wrapped around a post;belt drives
T2 > T1
This formula can be used only if belt, rope are about to slip;Angle of contact is radians; rope is wrapped n times - 2n rad
In belt drives, pulley with lesser value slips first, with S remaining same
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Beer/Johnston
A flat belt connects pulleyA to pulleyB. The
coefficients of friction are s = 0.25 and k= 0.20
between both pulleys and the belt.
Knowing that the maximum allowable tension in
the belt is 600 N, determine the largest torque
which can be exerted by the belt on pulleyA.
Since angle of contact is smaller, slippage will occur on pulleyB first.
Determine belt tensions based on pulleyB; = 120 deg = 2/3 rad
( )
N4.355
1.688
N600
688.1N600
1
3225.0
11
2 s
==
===
T
eT
eTT
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( )( ) 0N600N4.355mc8:0 =+= MM
mcN8.1956 =M
Check for belt not sliping at pulley A:
ln (600/355.4) = x 4/3 => = 0.125 < 0.25
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A 120-kg block is supported by a rope which iswrapped 1.5 - times around a horizontal rod.Knowing that the coefficient of static frictionbetween the rope and the rod is 0.15, determine
the range of values ofP
for which equilibrium ismaintained.
Beer/Johnston
PW = 9.81 X 120
= 1177.2 N
= 1.5 turns = 3 rad For impending motion ofWupP= W e s = (1177.2 N) e (0.15)3
= 4839.7 N
For impending motion ofWdownP= W es = (1177.2 N) e(0.15)3
= 286.3 N
For equilibrium: 286 N P 4.84 kN
B /J h t
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In the pivoted motor mount shown, the weight W of
the 175-N motor is used to maintain tension in thedrive belt. Knowing that the coefficient of staticfriction between the flat belt and drumsA and B is0.40, and neglecting the weight of platform CD,
determine the largest couple which can betransmitted to drum B when the drive drumA isrotating clockwise.
Beer/Johnston
For impending belt slip: CW rotation = radians
Obtain FBD of motor and mount; MD = 0 => find T1 and T2
Obtain FBD of drum at B; MB in CCW; MB = 0; Find MB
T1 = 54.5 N, T2 = 191.5 N
MB=10.27 N.m
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