9789810682446_slides_Chapter 08

Download Report

Transcript 9789810682446_slides_Chapter 08

Engineering Mechanics:
Statics in SI Units, 12e
8
Friction
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1.
2.
3.
4.
5.
6.
Characteristics of Dry Friction
Problems Involving Dry Friction
Wedges
Frictional Forces on Screws
Frictional Forces on Flat Belts
Frictional Forces on Collar Bearings, Pivot Bearings,
and Disks
7. Frictional Forces on Journal Bearings
8. Rolling Resistance
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry 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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
Theory of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
Theory of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
Theory of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
Theory of Dry Friction
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
 Fs 
1   s N 
1
s  tan    tan 

tan
s

N
 N 
1
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
Theory of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
Theory of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
Theory of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
Theory of Dry Friction
• 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
 Fk
N
k  tan 1 

1   k N 
1

tan


  tan  k

 N 
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.1 Characteristics of Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.2 Problems Involving Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.2 Problems Involving Dry Friction
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
   Fx  0;
80 cos 30 N  F  0
   Fy  0;
 80 sin 30 N  N C  196.2 N  0
 M O  0;
80 sin 30 N (0.4m)  80 cos 30 N (0.2m)  N C ( x)  0
Solving
F  69.3N , N C  236N
x  0.00908m  9.08mm
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
5 unknowns, 3 equilibrium equations for the stone and 2
for the wedge.
 M A  0;
 4905 N (0.5m)  ( N B cos 7  N )(1m)  (0.3N B sin 7  N )(1m)  0
N B  2383.1N
   Fx  0;
2383.1sin 7   0.3(2383.1 cos 7  )  P  0.3 N C  0
   Fy  0;
N C  2383.1 cos 7  N  0.3(2383.1sin 7  )  0
N C  2452.5 N
P  1154.9 N  1.15kN
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.4 Frictional Forces on Screws
• 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
M  rW tan s   
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.4 Frictional Forces on Screws
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)
Copyright © 2010 Pearson Education South Asia Pte Ltd
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?
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Since friction at two screws must be overcome, this
requires
M  2Wr tan    
W  2000 N , r  5mm, s  tan 1  s  tan 1 0.25  14.04
  tan 1  / 2r   tan 1 2mm / 2 5mm  3.64
Solving


M  2 2000 N 5mm tan 14.04  3.64 
 6374.7 N .mm  6.37 N .m
When the moment is removed, the turnbuckle will be selflocking
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.5 Frictional Forces on Flat Belts
• 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
 Fx  0;
 d 
 d 
T cos


dN

(
T

dT
)
cos


0
 2 
 2 
 Fy  0;
 d 
 d 
dN  (T  dT ) sin 

T
sin


0
 2 
 2 
Copyright © 2010 Pearson Education South Asia Pte Ltd
8.5 Frictional Forces on Flat Belts
• We have
dN  dT
dN  Td
dT
 d
T
T  T1 ,   0, T  T2 ,   

dT
T1 T   0 d
T
In 2  
T1
T2
T2  T1e 
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
T2  T1e  s  ;
500 N  T1e 0.253 / 4  
T1 
500 N
e 0.253 / 4  
500 N

 277.4 N
1.80
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 8.8
For section of the cord passing over the drum at C
W < 277.4N
T2  T1e  s  ;
277.4  We 0.253 / 4  
W  153.9 N
W
153.9 N
m

 15.7 kg
2
g 9.81m / s
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
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 – R12)
• 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 – R12)
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
   Fz  0;
1    
 mg  2   wo   0
2  2  
2mg
wo 

Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Since w = 0 at x = l/2, for distributed load function,
 2 x  2mg  2 x 
w  wo 1   
1  
 
 
 

For magnitude of normal force acting on a segment of
area having length dx,
2mg  2 x 
dN  wdx 
1  dx
 
 
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
For magnitude of the frictional force acting on the same
element of area,
2 s mg  2 x 
dF   s dN 
1  dx
 
 
For moment created by this force about z axis,
2  s mg

dM  xdF 
 2x 
x1  dx
 

Summation of moments by integration,
 M z  0; M  2
/2
0
2 s mg  2 x 
x1  dx  0

 

/2
4 s mg  x 2 2 x 3 
 mg
 
  s
M
  2
3  0
6
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
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,
 M z  0;
M  ( R sin k )r  0
M  Rr sin k
• If the bearing is partially lubricated, μk is small,
μk = tanΦk ≈ sinΦk ≈ Φk
• Frictional resistance
M ≈ Rrμk
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
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,
sin s  (tan s  s )
rf  r s  (5mm)(0.4)  2mm
 M P2  0;981N (52mm)  T (48mm)  0
T  1063N  1.06kN and s  tan 1 0.4  20.8
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 8.10
Part (a)
For radius of friction circle,
rf  r sin s  5 sin 21.8  1.86mm
Therefore,
 M P2  0;
981N (50mm  1.86mm)  T (50mm  1.86mm)  0
T  1057 N  1.06kN
Copyright © 2010 Pearson Education South Asia Pte Ltd
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,
 M P3  0;
981N (48mm)  T (52mm)  0
T  906 N
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
Copyright © 2010 Pearson Education South Asia Pte Ltd
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,
 M A  0;
9.81 cos 1.2 N (a )  9.81sin 1.2 N (100mm)  0
Solving
a  2.09mm
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
C)  s N
B) = s N
D) = k N
Copyright © 2010 Pearson Education South Asia Pte Ltd
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)
100 lb
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.
B
A) 
B) 
C)
D)
A
Copyright © 2010 Pearson Education South Asia Pte Ltd
P(A)
P(B)
P(C)
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
Copyright © 2010 Pearson Education South Asia Pte Ltd
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
Copyright © 2010 Pearson Education South Asia Pte Ltd