Friction 8 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd

Friction88

Engineering Mechanics: Statics in SI Units, 12e


Chapter Objectives

• Introduce the concept of dry friction • To present specific applications of frictional force

analysis on wedges, screws, belts, and bearings• To investigate the concept of rolling resistance


Chapter Outline

1. Characteristics of Dry Friction

2. Problems Involving Dry Friction

3. Wedges

4. Frictional Forces on Screws

5. Frictional Forces on Flat Belts

6. Frictional Forces on Collar Bearings, Pivot Bearings, and Disks

7. Frictional Forces on Journal Bearings

8. Rolling Resistance


8.1 Characteristics of Dry Friction

Friction• Force that resists the movement of two contacting

surfaces that slide relative to one another• Acts tangent to the surfaces at points of contact with

other body• Opposing possible or existing motion of the body

relative to points of contact• Two types of friction – Fluid and Coulomb Friction



• Fluid friction exist when the contacting surface are separated by a film of fluid (gas or liquid)

• Depends on velocity of the fluid and its ability to resist shear force

• Coulomb friction occurs between contacting surfaces of bodies in the absence of a lubricating fluid



Theory of Dry Friction• Consider the effects caused by pulling horizontally on

a block of uniform weight W which is resting on a rough horizontal surface

• Consider the surfaces of contact to be nonrigid or deformable and other parts of the block to be rigid



Theory of Dry Friction

• Normal force ∆Nn and frictional force∆Fn act along the contact surface

• For equilibrium, normal forces act upward to balance the block’s weight W, frictional forces act to the left to prevent force P from moving the block to the right



Theory of Dry Friction• Many microscopic irregularities exist between the two

surfaces of floor and the block

• Reactive forces ∆Rn developed at each of the protuberances

• Each reactive force consist of both a frictional component ∆Fn and normal component ∆Nn




Equilibrium• Effect of normal and frictional loadings are indicated

by their resultant N and F

• Distribution of ∆Fn indicates that F is tangent to the contacting surface, opposite to the direction of P

• Normal force N is determined from the distribution of ∆Nn




Equilibrium• N is directed upward to balance W• N acts a distance x to the right of the line of action of

W• This location coincides with the centroid or the

geometric center of the loading diagram in order to balance the “tipping effect” caused by P




Impending Motion• As P is slowly increased, F correspondingly increase

until it attains a certain maximum value F, called the limiting static frictional force

• Limiting static frictional force Fs is directly proportional to the resultant normal force N

Fs = μsN




Impending Motion

• Constant of proportionality μs is known as the coefficient of static friction

• Angle Φs that Rs makes with N is called the angle of static friction

sss

s N

N

N

F 111 tantantan




Typical Values of μs

Contact Materials Coefficient of Static Friction μs

Metal on ice 0.03 – 0.05

Wood on wood 0.30 – 0.70

Leather on wood 0.20 – 0.50

Leather on metal 0.30 – 0.60

Aluminum on aluminum 1.10 – 1.70




Motion

• When P is greater than Fs, the frictional force is slightly smaller value than Fs, called kinetic frictional force

• The block will not be held in equilibrium (P > Fs) but slide with increasing speed




Motion

• The drop from Fs (static) to Fk (kinetic) can by explained by examining the contacting surfaces

• When P > Fs, P has the capacity to shear off the peaks at the contact surfaces




• Resultant frictional force Fk is directly proportional to the magnitude of the resultant normal force N

Fk = μkN

• Constant of proportionality μk is coefficient of kinetic friction

• μk are typically 25% smaller than μs

• Resultant Rk has a line of action defined by Φk, angle of kinetic friction

kkk

k N

N

N

F 111 tantantan



Theory of Dry Friction• F is a static frictional force if equilibrium is maintained• F is a limiting static frictional force when it reaches a

maximum value needed to maintain equilibrium• F is termed a kinetic frictional force when sliding

occurs at the contacting surface



Characteristics of Dry Friction• The frictional force acts tangent to the contacting

surfaces

• The max static frictional force Fs is independent of the area of contact

• The max static frictional force is greater than kinetic frictional force

• When slipping, the max static frictional force is proportional to the normal force and kinetic frictional force is proportional to the normal force


8.2 Problems Involving Dry Friction

Types of Friction Problems• In all cases, geometry and dimensions are assumed

to be known• 3 types of mechanics problem involving dry friction

- Equilibrium

- Impending motion at all points

- Impending motion at some points



Types of Friction Problems

Equilibrium• Total number of unknowns = Total number of available

equilibrium equations

• Frictional forces must satisfy F ≤ μsN; 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



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 8.1

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 μ = 0.3.


Solution

Resultant normal force NC act a distance x from the crate’s center line in order to counteract the tipping effect caused by P.

3 unknowns to be determined by 3 equations of equilibrium.


SolvingmmmxNNF

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

Solution


Since x is negative, the resultant force acts (slightly) to the left of the crate’s center line.

No tipping will occur since x ≤ 0.4m

Max frictional force which can be developed at the surface of contact

Fmax = μsNC = 0.3(236N) = 70.8N

Since F = 69.3N < 70.8N, the crate will not slip thou it is close to doing so.

Solution


8.3 Wedges

• A simple machine used to transform an applied force into much larger forces, directed at approximately right angles to the applied force

• Used to give small displacements or adjustments to heavy load

• Consider the wedge used to lift a block of weight W by applying a force P to the wedge


8.3 Wedges

• FBD of the block and the wedge

• Exclude the weight of the wedge since it is small compared to weight of the block


Example 8.6

The uniform stone has a mass of 500kg and is held in place in the horizontal position using a wedge at B. if the coefficient of static friction μs = 0.3, at the surfaces of contact, determine the minimum force P needed to remove the wedge. Is the wedge self-locking? Assume that the stone does not slip at A.


Solution

Minimum force P requires F = μs NA at the surfaces of contact with the wedge.

FBD of the stone and the wedge as below.

On the wedge, friction force opposes the motion and on the stone at A, FA ≤ μsNA, slipping does not occur.


Solution

5 unknowns, 3 equilibrium equations for the stone and 2 for the wedge.

kNNP

NN

NN

F

NP

F

NN

mNNmNNmN

M

C

C

y

C

x

B

BB

A

15.19.1154

5.2452

0)7sin1.2383(3.07cos1.2383

;0

03.0)7cos1.2383(3.07sin1.2383

;0

1.2383

0)1)(7sin3.0()1)(7cos()5.0(4905

;0


Solution

Since P is positive, the wedge must be pulled out.

If P is zero, the wedge would remain in place (self-locking) and the frictional forces developed at B and C would satisfy

FB < μsNB

FC < μsNC


8.4 Frictional Forces on Screws

• 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 wedge wrapped around a cylinder



• 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 2πr, where r is the mean radius of the head

• Applying the force equations of equilibrium, we have srWM tan



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(θ – Φ)


Example 8.7

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 μs = 0.25, determine the moment M that must be applied to draw the end screws closer together. Is the turnbuckle self-locking?


Solution

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


8.5 Frictional Forces on Flat Belts

• 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



• 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 = μ dN• Applying equilibrium equations

02

sin2

sin)(

;0

02

cos)(2

cos

;0

dT

ddTTdN

F

ddTTdN

dT

F

y

x



• We have

eTT

T

TIn

dT

dT

TTTT

dT

dT

TddN

dTdN

T

T

12

1

2

0

21

2

1

,,0,


Example 8.8

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 μs = 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 8.8

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 B

180° = π rad, angle of contact between drum and cord

β = (135°/180°)π = 3/4π rad

NN

e

NT

eTN

eTT s

4.27780.1

500500

500

;

4/325.01

4/325.01

12


Example 8.8

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


8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks

• Pivot and collar bearings are used to support axial load on a rotating shaft

• Laws of dry friction is applied to determine the moment M needed to turn the shaft when it supports an axial force P


8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks

Frictional Analysis• The collar bearing on the shaft is subjected to an axial

force P and has a total contact area π(R22 – R1

2)

• Normal pressure p is considered to be uniformly distributed over this area – a reasonable assumption provided the bearing is new and evenly distributed

• Since ∑Fz = 0, p measured as a force per unit area

p = P/π(R22 – R1

2)


Example 8.9

The uniform bar has a total mass m. if it is assumed that the normal pressure acting at the contracting surface varies linearly along the length of the bar, determine the couple moment M required to rotate the bar. Assume that the bar’s width a is negligible in comparison to its length l. the coefficient of static friction is equal to μs.


Solution

FBD of the bar is as shown.

Bar has a total weight of W = mg.

Intensity wo of the distributed lead at the center (x = 0) is determined from vertical force equilibrium.

mgw

wmg

F

o

o

z

2

022

12

;0


Solution

Since w = 0 at x = l/2, for distributed load function,

For magnitude of normal force acting on a segment of area having length dx,

xmgx

ww o

21

221

dxxmg

wdxdN

2

12


Solution

For magnitude of the frictional force acting on the same element of area,

For moment created by this force about z axis,

Summation of moments by integration,

dxxmg

dNdF ss

2

12

dxx

xmg

xdFdM s

2

12

63

2

2

4

02

12

2;0

2/

0

32

2/

0

mgxxmgM

dxx

xmg

MM

ss

sz


8.7 Frictional Forces on Journal Bearings

• When a shaft or axle is subjected to lateral loads, a journal bearing is used for support

• Well-lubricated journal bearings are subjected to the laws of fluid mechanisms

• When the bearing is not lubricated, analysis of the frictional resistance can be based on the laws of dry friction

• If the lateral load is P, the bearing reactive force R acting at A is equal and opposite to P


8.7 Frictional Forces on Journal Bearings

• Moment needed to maintain constant rotation of the shaft can be found by the summation of moments about the z axis of the shaft,

• If the bearing is partially lubricated, μk is small,μk = tanΦk ≈ sinΦk ≈ Φk

• Frictional resistance

M ≈ Rrμk

k

k

z

RrM

rRM

M

sin

0)sin(

;0


Example 8.10

The 100mm diameter pulley fits loosely on a 10mm diameter shaft for which the coefficient of static friction is μs = 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 belt and the pulley and neglect the weight of the pulley.


Example 8.10

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 circle’s radius, rf = r sinΦs.

Using the simplification,

8.204.0tan and 06.11063

0)48()52(981;0

2)4.0)(5(

)(tansin

1

2

s

P

sf

sss

kNNT

mmTmmNM

mmmmrr


Example 8.10

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


Example 8.10

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,

NT

mmTmmN

M P

906

0)52()48(981

;03


8.8 Rolling Resistance

• A rigid cylinder of weight W rolls at constant velocity along a rigid surface, the normal force is at tangent point of contact

• Hard material cylinder will compresses the soft surface underneath it


8.8 Rolling Resistance

• We consider the resultant of the entire normal pressure acting on the cylinder

N = Nd + Nr

• To keep the cylinder in equilibrium, rolling at constant rate, N must be concurrent with the driving force P and the weight W

• Summation of moment about A,

Wa = P (r cosθ)

Wa ≈ Pr

P ≈ (Wa)/r


Example 8.11

A 10kg steel wheel has a radius of 100mm and rest on an inclined plans made of wood. If θ is increased so that the wheel begins to roll down the incline with constant velocity when θ = 1.2°, determine the coefficient of rolling resistance.


Solution

FBD of the wheel is as shown.

Wheel has impending motion.

Normal reaction N acts at point A defined by dimension a.

Summing moments about point A,

Solving

mma

mmNaN

M A

09.2

0)100(2.1sin81.9)(2.1cos81.9

;0


QUIZ

1. A friction force always acts _____ to the contact surface.

A) Normal B) At 45°

C) Parallel D) At the angle of static friction

2. If a block is stationary, then the friction force acting on it is ________ .A) s N B) = s N

C) s N D) = k N


QUIZ

3. A 100 lb box with a wide base is pulled by a force P and s = 0.4. Which force orientation requires the least force to begin sliding?

A) P(A) B) P(B)

C) P(C) D) Not determined

4. A ladder is positioned as shown. Please indicate the direction of the friction force on the ladder at B.

A) B) C) D)

P(A)

P(B)P(C)

100 lb

A

B


QUIZ

5. A wedge allows a ______ force P to lift a _________ weight W.

A) (large, large) B) (small, small)

C) (small, large) D) (large, small)

6. Considering friction forces and the indicated motion of the belt, how are belt tensions T1 and T2 related?

A) T1 > T2 B) T1 = T2

C) T1 < T2 D) T1 = T2 e


QUIZ

7. When determining the force P needed to lift the block of weight W, it is easier to draw a FBD of ______ first.

A) The wedge B) The block

C) The horizontal ground D) The vertical wall

8. In the analysis of frictional forces on a flat belt, T2 = T1 e . In this equation, equals ______ .

A) Angle of contact in deg B) Angle of contact in rad

C) Coefficient of static friction D) Coefficient of kinetic friction

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Friction 8 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson Education South Asia Pte Ltd