lecture slides statics 2011 lectures 14 15 [compatibility mode]

8/7/2019 Lecture Slides Statics 2011 Lectures 14 15 [Compatibility Mode]

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Determine the location x of the two supports so as to minimize the

maximum bending moment in the beam. Specify the maximumbending moment.



ME101: Engineering Mechanics

Lecture 14

01st February 2011

Friction



• 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 incontact, tangential forces, called friction forces, will develop if one attempts

to move one relative to the other.

Introduction

,

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.



Dry friction: Laws of dry frictionW

N

• W – weight; N – Reaction of the surface

• Only vertical componentA

W

• P – applied load

• F – static friction force : resultant of many forces actingover the entire contact area

• Because of irregularities in surface & molecular attraction

P

N

F

A



• 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 re uired. F is a static- riction orce.

The Laws of Dry Friction. Coefficients of Friction

• As P increases, the static-friction force F

increases as well until it reaches a maximum

value F m.

N F sm

µ =

• Further increase in P causes the block to begin

to move as F drops to a smaller kinetic-friction

force F k .

N F k k =



• Maximum static-friction force:

N F sm =

• Kinetic-friction force:

sk

k k N F

µ µ 75.0≅

=


• 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



EXPERIMENTAL EVIDENCE:Fm proportional to N

Fm = µs N; µs – static friction co-efficient

Similarly, Fk = µk N; µk – kinetic friction co-efficient

µs and µk depends on the natureof surface; not on contact areaof surface

µk = 0.75 µs



P

W

N

F

A B

Fm

Fk

F

p

Equilibrium Motion

More irregularitiesinteraction

Less irregularitiesinteraction

• ‘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 about ‘B’




• Four situations can occur when a rigid body is in contact with

a horizontal surface:

• No friction,

(P x = 0)

• No motion,

(P x < F m)

• Motion impending,

(P x = F m)

• Motion,

(P x > F m)



Angles of Friction

• It is sometimes convenient to replace normal force N and friction force F by

their resultant R:

• No friction • Motion impending• No motion

ss

sms

N

N

N

F

µ φ

φ

=

==

tan

tan

• Motion

k k

k k k

N

N

N

F

µ φ

φ

=

==

tan

tan



Angles of Friction

• Consider block of weight W resting on board with

variable inclination angle q.

• No friction • No motion • Motion

impending

• Motion



Problems Involving Dry Friction

• 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



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 µs N)

First category: to know a body slips or not

R. Ganesh Narayanan 14

• Determine ‘F’ required for equilibrium, by solving equilibrium equns;

Also find ‘N’

• Compare ‘F’ obtained with maximum value ‘Fm’ i.e., from Fm = µs N• ‘F’ is smaller or equal to ‘Fm’, then body is at rest

• Otherwise body starts moving

• Actual friction force magnitude = Fk = µk N

Solution



SOLUTION:

• 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

Sample Problem 14.1

A 500-N force acts as shown on a 1.5 kN

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.

greater, block will not slide.

• If maximum friction force is less

than friction force required for

equilibrium, block will slide.

Calculate kinetic-friction force.



SOLUTION:

• Determine values of friction force and normal

reaction force from plane required to maintain

equilibrium.

:0=∑ F ( ) 0N1500-N)500(5

3

54 =− F

N500−=F

0N500N500- 34 =− N

Sample Problem 14.1

• Calculate maximum friction force and compare

with friction force required for equilibrium. If it is

greater, block will not slide.( ) N375N150025.0 === msm F N F µ

The block will slide down the plane.

y

N1500= N



• If maximum friction force is less than friction

force required for equilibrium, block will slide.

Calculate kinetic-friction force.

( )N150020.0=

==N F F k k actual

N300=actualF

Sample Problem 14.1

Fm

Fk

F

p

Equilibrium Motion



Meriam/Kraige; 6/8

M

30°

Cylinder weight: 30 kg; Dia: 400 mm

Static friction co-efft: 0.30 between cylinder and surface

Calculate the applied CW couple M which cause thecylinder to slip

30 x 9.81

ΣFx = 0 = -NA+0.3NB Cos 30-NB Sin 30 = 0

ΣFy = 0 =>-294.3+0.3NA+NBCos 30-0.3NB Sin 30 = 0

NA

FA = 0.3 NANB

FB = 0.3 NB

M

C

Find NA & NB by solving these two equns.

ΣMC = 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 N-m



Meriam/Kraige; 6/5

Wooden block: 1.2 kg; Paint: 9 kg

Determine the magnitude and direction of (1) thefriction force exerted by roof surface on the woodenblock, (2) total force exerted by roof surface on the

Paint

Woodenblock

12

4

Roofsurface

Θ

Θ = tan-1 (4/12) = 18.43°

(2) Total force = 10.2 x 9.81 = 100.06 N UP

N

F

10.2x 9.81

X

Y

(1)ΣFx = 0 => -F+100.06 sin 18.43 => F = 31.6 N

ΣFy = 0 => N = 95 N



Beer/Johnston

20 x 9.81 = 196.2 N30 x 9.81 = 294.3 N

For 20 kg block For 30 kg block

F1 N1

(a)

N1

F1F2

N2



(B)490.5 N

P

N



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 µs is the sameat both surfaces of contact, determine the smallest value of µs

for which equilibrium can be maintained.

Slip impends at both A and B, FA= µsNA, FB= µsNB

ΣFx=0=> FA−NB=0, NB=FA=µsNA

A

B

B

NB

FANA

W

1.25 1.25

O

ΣFy=0=> NA−W+FB=0, NA+FB=W

NA+µsNB=W; W = NA(1+µs2)

ΣMo = 0 => (6) NB - (2.5) (NA) +(W) (1.25) = 0

6µsNA - 2.5 NA + NA(1+µs2) 1.25 = 0

µs = -2.4 ± 2.6 = > Min µs = 0.2



ME101: Engineering Mechanics

Lecture 15

01st February 2011

Friction



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.

Sample Problem 14.2

The moveable bracket shown may be

placed at any height on the 75 mm.

diameter pipe. If the coefficient of

friction between the pipe and bracket is0.25, determine the minimum distance

x at which the load can be supported.

Neglect the weight of the bracket.



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.

B Bs B

A As A

N N F

N N F

25.0

25.0

==

==

µ

• Apply conditions for static equilibrium to find minimum x.

Sample Problem 14.2

:0=∑ F 0=− A B N N A B N N =

:0=∑ yF

W N

W N N

W F F

A

B A

B A

=

=−+

=−+

5.0

025.025.0

0

W N N B A 2==

:0=∑ B M ( ) ( ) ( )

( )

( ) ( ) 05.37275.182150

05.3725.075150

0mm5.35mm75mm150

=+−−

=+−−

=−−−

W WxW W

W Wx N N

xW F N

A A

A A

mm300= x



• Wedges - simple • Block as free-body • Wedge as free-body

Wedges

heavy loads.

• Force required to lift

block is significantly

less than block weight.

• Friction preventswedge from sliding

out.

• Want to find minimum

force P to raise block.

0

:0

0

:0

21

21

=+−−

=

=+−

=

∑

∑

N N W

F

N N F

s

y

s

x

µ

µ

or

021 =++ W R R

( )

( ) 06sin6cos

:0

0

6sin6cos

32

32

=°−°+−

=

=+

°−°−−

=

∑

s

y

ss

x

N N

F

P

N N

µ

µ µ

or

032 =+− R RP



• 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.

Square-Threaded Screws

r PaQ =

• Impending motion

upwards. Solve

for Q.

• Self-locking,

solve for Q to lower

load.

,θ φ >s • Non-locking,

solve for Q to hold load.

,θ φ >s



-

the force P applied to a cord firmly attached to its periphery. The force P is

slowly increased and kept normal to the flat surface of the semi-cylinder. If

slipping is observed just as θ reaches 400, determine the coefficient of static

friction µ s and the value of P when slipping occurs.



06_06



06_07



A clamp is used to hold two pieces of

wood together as shown. The clamp

SOLUTION

• Calculate lead angle and pitch angle.

• Using block and plane analogy with

impending motion up the plane, calculatethe clamping force with a force triangle.

• With impending motion down the plane,

Sample Problem 14.5

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.

ca cu a e e orce an orque requ re o

loosen the clamp.



SOLUTION

• Calculate lead angle and pitch angle. For the double

threaded screw, the lead L is equal to twice the pitch.

( )

30.0tan

1273.0mm10mm22

2tan

==

===

ss

r L

µ φ

π π θ °= 3.7θ

°= 7.16sφ

Sample Problem 14.5

• Using block and plane analogy with impending

motion up the plane, calculate clamping force with

force triangle.

kN8mm5

mN40mN40 =

⋅=⋅= Qr Q

( )°

==+24tan

kN8tan W

W

Qsφ θ

kN97.17=W



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

R. Ganesh Narayanan 37

FBD joint D :

ΣFy = 0 => 2FADsin25°−4 kN=0

FAD = FCD = 4.73 kN

By symmetry:

4 kN

FAD FCD

25° 25°D



FBD joint A:

4.73 kN

FAC

FAE = 4.73

25°

25°A

ΣFx = 0 => FAC

−2(4.73) cos25°=0

FAC = 8.57 kN

W = FAC = 8.57

Joint A

38

L = Pitch = 2 mmφ

θ

RP = M/r

Π (7.5)

θ

FBD j i t A



FBD joint A:

4.73 kN

FAC

FAE = 4.73

25°

25°A

ΣFx = 0 => FAC

−2(4.73) cos25°=0

FAC = 8.57 kN

W = FAC = 8.57

Joint A

Here θ is used instead of α used earlier

39

L = Pitch = 2 mmφ

θ

RP = M/r

Π (7.5)

θ

MA = rW tan (Φ+α) = (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



A novel Electrical machine



Trent 1000



06_08



06_09



Journal Bearings Axle Friction



Journal Bearings. Axle Friction

• Angle between R and

normal to bearing

surface is the angle of kinetic friction jk .

k

k

Rr

Rr M

µ

φ

≈

= sin

• May treat bearing

reaction as force-

couple system.

• For graphical solution,

R must be tangent to

circle of friction.

k

k f

r

r r

µ

φ

≈

= sin

Th B i Di k F i i



Consider rotating hollow shaft:

( )21

22 R R

APr

A A

Pr N r F r M

k

k k

−

∆

=

∆=∆=∆=∆

π

µ

µ µ

22

2PR

k µ π

Thrust Bearings. Disk Friction

For full circle of radius R,

PR M k µ 32=

( )21

22

31

32

32

0

2

1

2

2 1

R R

R RP

R R

k

R

−

−=

−

µ

π



Wheel Friction. Rolling Resistance

• 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 andground cause resultant of

ground reaction to be

applied at B. P is required

to balance moment of W

about B.

Pr = Wb

b = coef of rolling

resistance

Sample Problem 14 6



A pulley of diameter 100 mm can

rotate about a fixed shaft of

diameter 50 mm. The coefficient

of static friction between the

pulley and shaft is 0.20.

Determine:

•

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-

Sample Problem 14.6

required to start raising a2.5 kN load,

• the smallest vertical force P

required to hold the load,

and• the smallest horizontal force

P required to start raising

the same load.

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.

Sample Problem 14 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

Sample Problem 14.6

( ) mm520.0mm25sin =≈≈= f ss f r r r r ϕ

Summing moments about B,

( )( ) ( ) 0mm45N2500mm55:0 =−=∑ P M B

N3060=P

Sample Problem 14 6



The perpendicular distance from center O of pulley to

line of action of R is again 0.20 in. Summing

• Impending motion is counter-clockwise as load is held

stationary with smallest force P. Sum moments about

C to find P.

Sample Problem 14.6

moments about C,

( )( ) ( ) 0mm55N2500mm45:0 =−=∑ P M C

N2050=P

Sample Problem 14.6



• With the load on the left and force P actinghorizontally 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 beconcurrent. Line of action of R must pass through

intersection of W and P and be tangent to circle of

=

Sample Problem 14.6

( )

°=

===

1.4

0707.02mm50

mm5sin

θ

θ ODOE

From the force triangle,( ) ( ) °=−°= 9.40cotN250045cot θ W P

N2890=P

Belt Friction



Belt Friction

• Relate T 1 and T 2 when belt is about to slide to right.

• Draw free-body diagram for element of belt

( ) 02

cos2

cos:0 =∆−∆

−∆

∆+=∑ N T T T F s x µ θ θ

( ) 02

sin2

sin:0 =∆

−∆

∆+−∆=∑θ θ

T T T N F y

• Combine to eliminate ∆ N divide throu h b ∆θ

( )22sin

22cos

θ θ µ θ

θ ∆∆

∆+−∆

∆∆ T T T s

• In the limit as ∆θ goes to zero,

0=−

T d

dT

s µ θ

• Separate variables and integrate from β θ θ == to0

β µ β µ seT

T

T

T s ==

1

2

1

2 orln

Sample Problem 14.7



SOLUTION:

• Since angle of contact is smaller,

slippage will occur on pulley B first.

Determine belt tensions based onpulley B.

• Taking pulley A as a free-body, sum

Sa p e ob e .

A flat belt connects pulley A to pulley B.The coefficients of friction are ms = 0.25

and mk = 0.20 between both pulleys and

the belt.

Knowing that the maximum allowabletension in the belt is 3000-N, determine

the largest torque which can be exerted

by the belt on pulley A.

determine torque.

Sample Problem 14.7



SOLUTION:

• Since angle of contact is smaller, slippage will

occur on pulley B first. Determine belt tensions

based on pulley B.

( )688.1

N3000 3ρ225.0

11

2=== e

T e

T

T s β µ

p

• Taking pulley A as free-body, sum moments about

pulley center to determine torque.

( )( ) 0N3000N3.1777mm200:0=−+=

∑A A M M

mN245 ⋅= A M

N3.1777

1.688

1 ==T

R f b k



1. Vector Mechanics for Engineers – Statics & Dynamics, Beer &

Johnston; 7th edition

2. Engineering Mechanics Statics & Dynamics, Shames; 4th

edition3. Engineering Mechanics Statics Vol. 1, Engineering Mechanics

Dynamics Vol. 2, Meriam & Kraige; 5th edition

Reference books

57

STATICS – MID SEMESTER – DYNAMICS

Tutorial: Thursday 8 am to 8.55 am

4. Schaum’s solved problems series Vol. 1: Statics; Vol. 2:Dynamics, Joseph F. Shelley

lecture slides statics 2011 lectures 14 15 [compatibility mode]

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