ch.6.2_dry friction

8/11/2019 CH.6.2_Dry friction

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Dry Friction6

APPLIED Mechanics

MET152

MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.

5.2 Problems Involving Dry Friction

2

Types of Friction Problems

Equilibrium

Total number of unknowns = Total number of available

equilibrium equations Frictional forces must satisfy F sN; otherwise,

slipping will occur and the body will not remain in

equilibrium

We must determine the frictional

forces at A and C to check

for equilibrium

4/7/2013MET152 Applied Mechanics Sem.121/ by Dr. Miloud S.


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5.2 Problems Involving Dry Friction

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Equilibrium Versus Frictional Equations

Frictional force always acts so as to oppose the

relative motion or impede the motion of the body

over its contacting surface

Assume the sense of the frictional force that require F

to be an equilibrium force

Correct sense is made after solving the equilibrium

equations

If F is a negative scalar, the sense of F is the reverse of

that assumed


Example 2

4

The uniform crate has a mass of 20kg. If a force P = 80N

is applied on to the crate, determine if it remains in

equilibrium. The coefficient of static friction is = 0.3.



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Solution

5

Resultant normal force NC act a distance x from the crates

center line in order to counteract the tipping effect caused

by P.

3 unknowns to be determined by 3 equations of equilibrium.


Solution

6

Solving

mmmxNNF

xNmNmN

M

NNN

F

FN

F

NC

C

O

C

y

x

08.900908.0,3.69

0)()2.0(30cos80)4.0(30sin80

;0

02.19630sin80

;0

030cos80

;0

236



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Solution

7

Since x is negative, the resultant force acts (slightly) to theleft of the crates center line.

No tipping will occur since x 0.4m

Max frictional force which can be developed at the surfaceof contact

Fmax = sNC = 0.3(236N) = 70.8N

Since F = 69.3N < 70.8N, the crate will not slip though itis close to doing so.


AnglesofFriction

It is sometimes convenient to replace normal force

Nand friction force Fby their resultant R:

No friction Motion impending No motion

ss

sms

N

N

N

F

tan

tan

Motion

kk

kkk

N

N

N

F

tan

tan

4/7/2013 MET152 Applied Mechanics Sem.121/ by Dr. Miloud S. 8


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Angles

of

Friction Consider block of weightWresting on board with

variable inclination angle

No friction No motion Motion

impending

Motion


ProblemsInvolvingDryFriction

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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6.3 Frictional Forces on Screws

11

Screws used as fasteners

Sometimes used to transmit power or motion from

one part of the machine to another

A square-ended screw is commonly used for the

latter purpose, especially when large forces are

applied along its axis

A screw is thought as an inclined plane or wedgewrapped around a cylinder


12

A nut initially at A on the screw will move up to B

when rotated 360 around the screw

This rotation is equivalent to translating the nut up an

inclined plane of height l and length 2r, where r is the

mean radius of the head

Applyingtheforce equationsofequilibrium, we have

srWM tan




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13

Downward Screw Motion

If the surface of the screw is very slippery, the screw

may rotate downward if the magnitude of the moment

is reduced to say M < M

This causes the effect of M to become S

M = Wr tan( )



Example 3

14

The turnbuckle has a square thread with a mean radius of

5mm and a lead of 2mm. If the coefficient of static friction

between the screw and the turnbuckle is s = 0.25,

determine the moment M that must be applied to draw

the end screws closer together. Is the turnbuckle self-

locking?



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Solution

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Since friction at two screws must be overcome, this

requires

Solving

When the moment is removed, the turnbuckle will be

self-locking

64.352/2tan2/tan

04.1425.0tantan,5,2000

tan2

11

11

mmmmr

mmrNW

WrM

ss

mNmmN

mmNM

.37.6.7.6374

64.304.14tan520002


5.4 Frictional Forces on Flat Belts

16

It is necessary to determine the frictional forces

developed between the contacting surfaces

Consider the flat belt which passes over a fixed curved

surface

Obviously T2 > T1

Consider FBD of the belt

segment in contact with the surface

N and F vary both in

magnitude and direction



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17

Consider FBD of an element having a length ds

Assuming either impending motion or motion of the

belt, the magnitude of the frictional force

dF = dN

Applying equilibrium equations

02

sin2

sin)(

;0

02

cos)(2

cos

;0

dT

ddTTdN

F

ddTTdNdT

F

y

x



18

We have

eTT

T

TIn

dT

dT

TTTT

dT

dT

TddN

dTdN

T

T

12

1

2

0

21

2

1

,,0,



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Example 4

19

The maximum tension that can be developed In the cord

is 500N. If the pulley at A is free to rotate and the

coefficient of static friction at fixed drums B and C is s =

0.25, determine the largest mass of cylinder that can be

lifted by the cord. Assume that the force F applied at the

end of the cord is directed vertically downward.


Example 5

20

Weight of W = mg causes the cord to move CCW over

the drums at B and C.

Max tension T2 in the cord occur at D where T2 = 500N

For section of the cord passing over the drum at B180 = rad, angle of contact between drum and cord

= (135/180) = 3/4 rad

NN

e

NT

eTN

eTT s

4.27780.1

500500

500

;

4/325.01

4/325.0

1

12



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Example 5

21

For section of the cord passing over the drum at C

W < 277.4N

kgsm

N

g

Wm

NW

We

eTT s

7.15/81.9

9.153

9.153

4.277

;

2

4/325.0

12


Example 6

22

The 100mm diameter pulley fits loosely on a 10mm

diameter shaft for which the coefficient of static friction is

s = 0.4. Determine the minimum tension T in the belt

needed to (a) raise the 100kg block and (b) lower the

block. Assume that no slipping occurs between the beltand the pulley and neglect the weight of the pulley.



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Example 6

23

Part (a)

FBD of the pulley is shown.

As tension T is increased, the pulley will roll around the

shaft to point before motion P2 impends.

Friction circles radius, rf= r sins.

Using the simplification,

8.204.0tanand06.11063

0)48()52(981;0

2)4.0)(5(

)(tansin

1

2

s

P

sf

sss

kNNT

mmTmmNM

mmmmrr


Example 6

24

Part (a)

For radius of friction circle,

Therefore,

kNNT

mmmmTmmmmN

M

mmrr

P

sf

06.11057

0)86.150()86.150(981

;0

86.18.21sin5sin

2



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Example 6

25

Part (b)

When the block is lowered, the resultant force R acting

on the shaft passes through the point P3.

Summing moments about this point,

NTmmTmmN

MP

9060)52()48(981

;03


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