Friction (2) [compatibility mode]

Friction

Relative velocity

Cause of dry frictionContact between two surfaces.

Hence first task in a friction problem is correct identification of contact surfaces

Identify the surface, the normal and the tangential vectors.

Also important is to get an idea of probable direction of relative motion

The contact force acts along the normal.Gravity is the most common cause of normal force.

Friction acts along the tangent plane opposite to the direction of relative motion

Normal

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Friction problems are essentially equilibrium problems with one f the

forces being functions of another

Friction

Relative velocity

Normal

N

Fr=f(N,V)N

Fr=f(N,V)




The correct way of writing the dry friction force

r

V ˆF N N VV

N=Normal force vector

V=Relative velocity vector of the bodym= coefficient of dry friction or Coulomb friction




Problem 1

Knowing that the coefficient of friction between the 13.5 kg block and the incline is ms = 0.25, determine

a) the smallest value of P required to maintain the block in equilibrium,

b) the corresponding value of b.




Problem 1

x

y

N cos60 mg f sin60 P sin 0N sin60 f cos60 P cos 0

f Nmg sin60 fP

sin sin60 cos cos60N cos60 mg N sin60 P sin 0

mg P sinNcos60 sin60

N sin60 N cos60 P cos 0mg P sin sin60 cos60 P cos 0

cos60 sin60mg P

sin sin60 cos60 P cos cos60 sin60 0

mg sin60 cos60 P sin sin60 cos60 P cos cos60 sin60 0

P sin60 sin cos60 sin cos60 cos sin60cos mg sin60 cos60

sin60 cos60P mg

cos 60 sin 60

cos 60 sin 61 mgP

o

0sin60 cos60

d 1 d0 cos 60 sin 60 0d P d

d sin60 sin cos60 sin cos60 cos sin60 cos 0dsin60cos cos60 cos cos60 sin sin60 sin 0

sin60 cos60tan 2.614 69cos60 sin60

sinP mg

60 cos60 0.866 0.125mg 0.72mg

cos 60 sin 60 cos 9 sin 9




Problem 2

Knowing that P = 110 N, determine the range of values value of qfor which equilibrium of the 8 kg block is maintained.




x

y

k

P cos N 0P sin mg f 0f N

N P cosP sin mg N 0

P sin mg P cos 0P sin cos mg

mgPsin cos

Hence downward movement will not start before

mgPsin cos

x

y

Problem 2

mg mg

s

P cos N 0P sin mg f 0f N

N P cosP sin mg N 0

P sin mg P cos 0P sin cos mg

mgPsin cos

Hence upward movement will not start before

mgPsin cos




Problem 3

The coefficients of friction are ms = 0.40 and mk = 0.30 between all the surfaces of contact. Determine the force P for which motion of the 27 kg block is impending if cable

a) is attached as shown,

b) is removed




Problem 3

1

1 1

1 1 1

1

1 2

2 1 2

2 1 2 2 1 2

1 2

1 1 1 2

1 2

T N 0N m g 0

T N ,N m gT m g

T N N P 0N N m g 0N N m g 0 N m m gT N N P 0

m g m g m m g P 0P 3 m g m g

TT

mm11ggNN11

ff11

vv

TT

NN11ff11

mm22ggNN22

ff22




Problem 3

1 1

1 1

1 1

1 1

1 2

2 1 2

2 1 2 2 1 2

1 2

1 1 2

1 2

T N m aNow T 0 N m aN m g 0

N m gN N P 0

N N m g 0N N m g 0 N m m g

N N P 0m g m m g P 0

P 2 m g m g

ff11

00

mm11ggNN11

ff11

vv

00

NN11ff11

mm22ggNN22

ff22




Additional Problems

The 8 kg block A and the 16 kg block B are at rest on an incline as shown. Knowing that the coefficient of static friction is 0.25 between all surfaces of contact, determine the value of q for which motion is impending.




Toppling

The magnitude of the force P is slowly increased. Does the homogeneous box of mass m slip or tip first? State the value of P which would cause each occurrence.




Slip or topple?

mgmg

NN11

ff11

NN22

ff22

1 2

1 2

2

1 2 1 2

1 2

2

1 2

P cos 30 N N 0P sin 30 N N mg 0

P cos 30 d mg d N 2d 0P cos 30 N N N N

P sin 30 N N mgP cos 30 mg 2 N

P sin30 N N mg P cos 30 mg

P sin30 P cos 30 mg P sin 30 cos 30 mgmgP

sin30 cos 300.5mgP

0.448mg0.25 0.866




Slip or topple?

mgmg

NN11

ff11

NN22

ff22

2

2

2

2

P cos 30 f 0P sin 30 N mg 0

P cos 30 d P sin 30 2d mg d 0

P cos 30 fP sin 30 N mgP cos 30 2P sin30 mg 0

P cos 30 2 sin 30 mgmgP

cos 30 2 sin 30mgP 0.536mg

0.866 1




Additional Problems




Additional Problems




Additional Problems




Wedge

mg

f=mN

N

P

q

P f cos N sin 0f sin N cos mg 0f N

mgN sin N cos mg 0 Nsin cos

P N cos N sin 0cos sinP cos sin N mgsin cos

P tanmg tan 1P tan tan tan

mg tan tan 1




A screw thread is a wedge





M=Pa

W

Q

Nf=mN

Q=Pa/r

= equivalent force





W

Q

Nf=mN

Q f cos N sin 0f sin N cos W 0

f NWN sin N cos W 0 N

sin cosQ N cos N sin 0

cos sinQ cos sin N Wsin cos

Q tanW 1 tan

q

Pa tanWr 1 tan

1 Pif tanW

Screw locks





W

P

Nf=mNq

1 PatanWr

1 Wr 0tan Pa

Screw locks

The

There is a critical value of friction coefficient beyond which the thread

does not move irrespective of the force applied.

This happens when a screw is not maintained properly. Because of dirt

and rust m becomes more than critical.





W

Nf=mN





W

Nf

For no movementf cos N sin 0f sin N cos W 0

f sinN cos

sin tancos

q

Self locking

Therefore after raising the load if we let go of the screw the load will not cause the screw to unscrew by itself.




Terminologies

Lead L npwheren=no. of parallely running threads = starts

Ltan =2 r

Pitch (p)

2pr

Lea

d (L

)




Turnbuckle

T1 T2

2 1

M tanT T r 1 tan

Used to apply tension. Used to apply tension.

The sleeve is rotated to pull the The sleeve is rotated to pull the threads together. threads together.




An improved screw jack

q

q

W

W

T T

T

2T cos q

T




An improved screw jack

f

f

W

W=2T cos f M=Pa

Pa cos sinWr sin cos

M cos sin2Tr cos sin cos




Worm gear

R

MG

GG

G

G

MM WR W

RPa tanWr 1 tan

M tanM 1 tanrR

MR tanM r 1 tan

M0/R

NNM f




Belt drives




x

y

s

s

F 0 T cos F T T cos 02 2

F 0 T sin N T T sin 02 2

F N

T cos F T T cos 02 2

T cos F T cos T cos 02 2 2

F T cos 0 T cos F N2 2

T sin N T T sin 02 2

T sin N T sin2 2

T sin 02

2T sin N T sin 02 2

2T sin N 0 2T sin N2 2

Belt drives




s

s

s

s0 , T 0 0 , T 0

s

T cos N2

2T sin N2

T cos 2T sin2 2

sin1 T 2cos2T 2

sin1 T 2Lim cos Lim2T 2

1 dT 12T d 2

dT dsT

Belt drives




2

1

2

1

T

s 0T

2 1 s

2s

1

T

sT 0

dT dT

lnTlnT lnT

TlnT

dT dsT

T2 seT1

Belt drives

Belt is just about to Belt is just about to slide to the rightslide to the right

22 1

1

2 1 1

T s se T e TTTorque required to drive the pulley

sT T e 1 T




T2 seT1

Belt drives : Important points

Angle bmust be expressed in radians

Larger tension occurs at the end of the belt where motion is about to begin

or is already moving T2 is used to denote the larger

tension




Belt drives

q

q

q

q

A B

A B




Belt drives

q

q

2

1

Texp 2 ???

T

T2

T1

T2

T1 1

2

Texp 2 ???

T

A

B

Which expression is correct???? Is T2>T1

Or T1>T2.One pulley must slip.

Friction force is larger for the larger pulley since

angle of wrap is larger. Hence smaller pulley

slips and determines the tension

Always check for mb value for each pulley in a system. The one

with the smallest mb value will

determine the tensions.




Band brakes

T2

T1

Important!!!In pulleys the belt

is mobile and tries to drive the pulley. Here the drum moves and tries to move the belt. Hence the sense of slack is opposite to that

for a pulley.




Band brakes

T2

T1

The drum rotates towards the right. If this were a pulley the slack side

would be on the left. But here as the drum is the moving part, the slack

side is on the left.

x x 2

x 2

y y 2 1

y 2 1

B 2 1

1 2

F 0 B T sin( ) 0B T sinF 0 B T cos( ) T P 0

B T cos T P 0

M 0 cT bT a b P 0bT cT

Pa b

a -p

y

x

Ry

Rx Mext

1 2 k

ext 1 2

ext 1 2

x x 2

y y 1 2

T T exp( )M 0 M r T T 0

M T T rF 0 R T sin( )F 0 R T T cos( )




Band brakes

1 2 k ext 1 2

1 2x 2 y 2 1

T T exp( ), M T T rbT cTB T sin , B T cos T P 0, P

a b

ext 1 2 2 k 2 k 2

ext2

k

2 k 2 k1 22

k ext

k

k

k

M T T r T exp( ) T r exp( ) 1 T rM

Texp( ) 1 r

bT exp( ) cT bexp( ) cbT cTP T

a b a b a bbexp( ) c M

Pa b exp( ) 1 r

c1 exp( )bP

1 exp( )

ext

s kk s

Ma1 rb

cNow 1 exp( ) 0 P 0 if 1 exp( ) 0 k kbP 0 iff b > ce > ce

Otherwise P is negativeOtherwise P is negative. Which means no force is required for braking!

Hence self lockingself locking occurs.Since Mext is equal in magnitude to the torque required

for braking it is also called the braking torque.




Clutches

2

2

R2

0

Assume pressure is uniform over the contact zonePp=R

Consider a small element of width dr at a radial distance r and spanning an angle d dA r dr ddF pdA prdrddM rdF r prdrd pr drd

M pr dr

2R323 3

0 0 0

32

r p 2d p 2 R pR3 3 3

2 P 2M R PR3 R 3




Clutches

1

0

2 21 0

2

r22

0 r

Assume pressure is uniform over the contact zone

Pp=r r

Consider a small element of spanning angle d at a radial distance rdA r dr ddF pdA prdrddM rdF r prdrd pr drd

M pr drd p

1

0

2r33 3 3 3

1 0 1 0r 0

3 31 02 2

1 0

3 31 0

2 21 0

r p 2r r 2 p r r3 3 3

2 PM r r3 r r

r r2M P3 r r




Disc Brake

2

2 2

2 20 i

Assume pressure is uniform over the contact zoneArea of a whole circle (spanning 2 radians) = r

r rArea of a sector = = 2 2

2Pp=R R

Consider a small element of spanning angle d at a radi

oo

i i

2

RR 32 3 3 3 3

o i o i0 R R 0

3 3 3 3o i o i 2 2

0 i

3 3o i

2 20 i

al distance rdA r dr ddF pdA prdrddM rdF r prdrd pr drd

r p pM pr drd p R R R R3 3 3

p 2PM R R R R3 3 R R

R R2M3 R R

dq




Thrust bearing

PP

qq qq

dd11

qqdd22

dd11

dd11

dd22

qqqq

qq

PP




Thrust bearing

dN

y

xO

dl

Xdf




Thrust bearing

2

1

2

1

2 22 1 2 2

2 1

R 2

R 0

R 2

R 0

Assume pressure is uniform over the contact zonePP= R R p p=

R R

M rdf

dN pdA prd dl , dl drcosecdN prd dl prd dr cosec pcosec d rdrdf dN pcosec d rdr

N pcosec d rdr

N p c

22 2

1 1 1

2 2

1 1

22 2

1 1 1

RR R 22

0R R R

2 22 1

R R2 2

R 0 R 0

RR R 32 2 20

R R R

32 1

rosec rdr 2 pcosec rdr 2 pcosec2

N pcosec R R

M rdf r rpcosec d rdr

rM pcosec r dr 2 pcosec r dr 2 pcosec3

2 pM R R3sin

3 32 13

2 22 1

R R2 P3sin R R

dN

y

xO

dl

X

r dr

df

What about q

= 0 ?




Torque wrench

Smallest m for self locking ?Smallest m for self locking ?

T1

T1

T2

N

N

mN

mN

T2

1 2T T exp 2

1 2

2

T N cos T N sin 0

P N sin T N cos 0

1 2r T T aP




Torque wrench

Condition for self locking ?Condition for self locking ?

1 2

1 2

1 2

2

2 1

1 1

1 1

1

T T exp 2

r T T aP

T N cos T N sin 0

P N sin T N cos 0

T T sin cos P cos sin

T T sin cos P cos sin exp 2

T T sin cos exp 2 P cos sin exp 2

T sin co

1

1 2 1

s exp 2 1 P cos sin exp 2

P cos sin exp 2 P cos sinT

sin cos exp 2 1 sin cos exp 2

aPr T T aP Tr 1 exp 2

P cos sinaPr 1 exp 2 sin cos e

xp 2

r cos sin a sin cosexp 2

r cos sin a

r cos sin aexp 2

r cos sin a sin cos

acos sin1 rln a2 cos sin sin cosr




Band brakes




Band brakes

T1 T3

T2

T4

2 1 1

2 1

3 4 4

4 3

0 3 4 2 1

x 1 3 1 3

y 1 3 B 1

75 75T T exp T exp 0.25180 180

T 1.39T135 135T T exp T exp 0.25180 180

T 0.55TM T T T T RConsider the pin B

F 0 T cos45 T cos 45 T T

F 0 T cos 45 T cos45 T 2T cos45

B 1 B

2 1 3 1 4 1

2 1 3 4 B

T 2T T

T 1.39T ,T T ,T 0.55TThus the largest tension is T 5.6 T T 4.03,T 2.22,T 5.7

M 5.6 2.22 R 3.38 0.16 0.54KNm 540 Nm

B

T1 T3

TB




Problems




Tip or Slide




Problems




Problems




Date post:	28-Nov-2014
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Friction (2) [compatibility mode]

Education