13726898 Applied Mechanics Friction

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


3/39

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


4/39

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


5/39

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


6/39

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


7/39

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


8/39

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


9/39

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


10/39

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


11/39

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


12/39

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


13/39

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


14/39

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


16/39

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


17/39

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


18/39

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


19/39

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


20/39

P = 440.6 N


21/39

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


22/39

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


24/39

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


25/39

P = M/r

W

R

> Screw will unwind itself

=> M = rW tan (-)Moment required toprevent unwinding


26/39

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)


27/39


28/39

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


29/39

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


30/39

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


31/39

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


32/39

Couple moment required to overcome frictionresistance, M

Equilibrium conditions and moment equations arenecessary to solve problems


33/39

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


34/39


35/39

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


36/39

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


37/39

( )( ) 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


38/39

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


39/39

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

13726898 Applied Mechanics Friction

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