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Chapter 2
Review of Forces and Moments
2
...
Most of this material is identical
to material covered in EN030, and is provided here as a review
...
1
...
2
...
1 Definition of a force
Engineering design calculations nearly always use classical (Newtonian) mechanics
...
But we don’t really know why this is (except perhaps the last one)
...
If objects accelerate very quickly towards their preferred configuration, then we say that
there’s a big force acting on them
...
We can’t see a force; we can only deduce its existence by observing its effect
...
A `particle’ is a small mass at some position in space
...
When the sum of the forces acting on a particle is zero, its velocity is
constant;
2
...
The forces exerted by two particles on each other are equal in magnitude
and opposite in direction
...
But for
engineering calculations we can usually take the earth to be fixed, and happily apply Newton’s laws
...
2
...
2 Causes of force
Forces may arise from a number of different effects, including
(i) Gravity;
(ii) Electromagnetism or electrostatics;
(iii) Pressure exerted by fluid or gas on part of a structure
(v) Wind or fluid induced drag or lift forces;
(vi) Contact forces, which act wherever a structure or component touches anything;
(vii) Friction forces, which also act at contacts
...
For example, gravity forces can be calculated
using Newton’s law of gravitation; electrostatic forces acting between charged particles are governed by
Coulomb’s law; electromagnetic forces acting between current carrying wires are governed by Ampere’s
law; buoyancy forces are governed by laws describing hydrostatic forces in fluids
...
6
...
For example, to determine friction forces acting in a machine, you may
need to measure the coefficient of friction for the contacting surfaces
...
Lift and drag forces are described in Section 2
...
Friction forces are
discussed in Section 12
...
Contact forces can
either be measured, or they can be calculated by analyzing forces and deformation in the system of
interest
...
2
...
3 Units of force and typical magnitudes
In SI units, the standard unit of force is the Newton, given the symbol N
...
The fundamental unit of force in the SI convention is kg m/s2
In US units, the standard unit of force is the pound, given the symbol lb or lbf (the latter is an abbreviation
for pound force, to distinguish it from pounds weight)
A force of 1 lbf causes a mass of 1 slug to accelerate at 1 ft/s2
US units have a frightfully confusing way of representing mass – often the mass of an object is reported
as weight, in lb or lbm (the latter is an abbreviation for pound mass)
...
Therefore, the mass of an object in
slugs must be computed from its weight in pounds using the formula
m (slugs) =
W (lb)
g (ft/s 2 )
where g=32
...
A force of 1 lb(f) causes a mass of 1 lb(m) to accelerate at 32
...
448 N
0
...
onlineconversion
...
A few typical force
magnitudes (from `The Sizesaurus’, by Stephen Strauss, Avon Books, NY, 1997) are listed in the table
below
Force
Gravitational Pull of the Sun on Earth
Newtons
3
...
3 × 107
Thrust of a large jet engine
7
...
3 ×102
Maximum force exerted upwards by a forearm
2
...
1× 10−2
Force between an electron and the nucleus of a Hydrogen atom 8 × 10−6
22
Pounds Force
7
...
5 ×1019
7
...
7 ×105
1
...
6 ×102
60
1
...
8 × 10−8
2
...
4 Classification of forces: External forces, constraint forces and internal forces
...
External forces arise from interaction between the system of interest and its surroundings
...
Force laws governing these
effects are listed later in this section
...
Constraint forces are very complex, and will be discussed in detail in Section 8
...
For example, a stretched
rope has a tension force acting inside it, holding the rope together
...
These internal forces ultimately lead to structural failure, and also
cause the structure to deform
...
We will
not, unfortunately, be able to develop a full theory of internal forces in this course – a proper discussion
requires understanding of partial differential equations, as well as vector and tensor calculus
...
2
...
5 Mathematical representation of a force
...
k
Fz
j
Fy
A force is therefore always expressed mathematically as a vector
Fx
z
quantity
...
The procedure is
x
1
...
Using geometry or trigonometry, calculate the force component along each of the three reference
directions ( Fx , Fy , Fz ) or ( F1 , F2 , F3 ) ;
3
...
To do this, you need to report the position vector of the point where the force acts on the structure
...
To do so,
you need to:
1
...
Choose basis vectors {i, j, k} or {e1 , e 2 , e 2 } that establish three fixed directions in space (usually
we use the same basis for both force and position vectors)
3
...
The position vector is then reported as
r = rx i + ry j + rz k = r1e1 + r2e 2 + r3e3 (appropriate units)
2
...
6 Measuring forces
Engineers often need to measure forces
...
The magnitude of the acceleration tells us the magnitude of the
force; the direction of motion of the mass tells us the direction of the force
...
In addition to causing acceleration, forces cause objects to deform – for example, a force will
stretch or compress a spring; or bend a beam
...
The simplest application of this phenomenon is a spring scale
...
The spring stiffness can be measured experimentally to calibrate the spring
...
But the same principle is used in more
sophisticated instruments too
...
The simplest load cell works much like a spring scale – you can load it in one
direction, and it will provide an electrical signal proportional to the magnitude of the force
...
Really fancy load cells
measure both force vectors, and torque or moment vectors
...
The instrument on the right is
called a `proving ring’ – there’s a short article describing how it works at
http://www
...
nist
...
htm
A sophisticated force transducer produced by MTS systems, which is capable of measuring forces and
moments acting on a car’s wheel in-situ
...
mts
...
...
One example (from
http://www
...
gov/isrc/Load_Cell/load_cell
...
In this case the `spring’ is actually a tubular piece of high-strength
steel
...
The
deformation is detected using `strain gages’ attached to the cylinder
...
When the wire gets shorter, its electrical resistance decreases –
this resistance change can be measured, and can be used to work out the
force
...
The most sensitive load cell currently available is the atomic force
microscope (AFM) – which as the name suggests, is
intended to measure forces between small numbers
of atoms
...
When the tip is
brought near a sample, atomic interactions exert a
force on the tip and cause the cantilever to bend
...
The device is capable of measuring forces of about
1 pN (that’s 10−12 N!!), and is used to explore the
properties of surfaces, and biological materials such
as DNA strands and cell membranes
...
di
...
Here are a few considerations
that will guide your purchase
...
How many force (and maybe moment) components do you need to measure? Instruments that
measure several force components are more expensive…
2
...
Load range – what is the minimum force you need to measure?
4
...
Temperature stability – how much will the reading on the cell change if the temperature changes?
6
...
Frequency response – how rapidly will the cell respond to time varying loads? What is the
maximum frequency of loading that can be measured?
8
...
Cost
2
...
7 Force Laws
In this section, we list equations that can be used to calculate forces associated with
(i)
Gravity
(ii)
Forces exerted by linear springs
(iii)
Electrostatic forces
(iv)
Electromagnetic forces
(v)
Hydrostatic forces and buoyancy
(vi)
Aero- and hydro-dynamic lift and drag forces
Gravitation
Gravity forces acting on masses that are a large distance apart
Consider two masses m1 and m2 that are a distance d
apart
...
Mass
m2 will experience a force of equal magnitude, acting in the opposite direction
...
673 × 10−11 m3 kg -1s -2
The law is strictly only valid if the masses are very small (infinitely small, in fact) compared with d – so
the formula works best for calculating the force exerted by one planet or another; or the force exerted by
the earth on a satellite
...
The earth is spherical, with a radius R
2
...
The object’s height h above the earths
surface is small compared to R
If the first two assumptions are valid, then one can
show that Newton’s law of gravitation implies that a
mass m at a height h above the earth’s surface experiences a force
F=−
GMm
er
( R + h) 2
where M is the mass of the earth; m is the mass of the object; R is the earth’s radius, G is the gravitation
constant and e r is a unit vector pointing from the center of the earth to the mass m
...
One might guess that points close to the earth’s surface under the mass would attract the mass
more than those far away, so the earth would exert a larger gravitational force than a very small object
with the same mass located at the earth’s center
...
If the third assumption (h<
GM
( R + h)
2
≈ g ⇒ F = − mgj
where g is a constant, and j is a `vertical’ unit vector (i
...
perpendicular to the earth’s surface)
...
81ms −2
...
For most engineering calculations the
center of gravity of an object can be assumed to be the same as its center of mass
...
The location of the center of mass
for several other common shapes is shown below
...
3
...
Surveyors
know perfectly well that the earth is no-where near spherical; its density is also not uniform
...
So using the simple
gravitational formulas in surveying applications (e
...
to find the `vertical’ direction) can lead to large
errors
...
Gravity is actually a
distributed force
...
The distributed force can be replaced by a
single, statically equivalent force, but the point where the equivalent force acts depends on the relative
positions of the two objects, and is not generally a fixed point in either solid
...
For example,
the resultant force of gravity exerted on the earth by the sun and moon does not act at the center of mass
of the earth
...
Forces exerted by springs
A solid object (e
...
a rubber band) can be made to exert forces by stretching it
...
Solid objects can also exert moments, or torques
– we will define these shortly
...
Here, we restrict attention to the simplest case:
forces exerted by linear springs
...
You can attach it
to something at both ends
...
The forces exerted by the ends of the spring always act along the line
of the spring
...
In the SI system, k has units of N/m
...
If you don’t do this,
your sign convention will be inconsistent with the formula
F = k ( L − a) , which assumes that a compressed spring (L
...
The device is extremely useful for damping vibrations
...
For an example of a precision dashpot see
http://www
...
com/beta/html/dashpot_defined
...
The magnitude of the force exerted by a fluid filled dashpot
is given by the formula
F =η
dL
dt
where L is the length, and η is the rate constant of the dashpot
...
In the SI system, η has units of Ns/m
...
Forces exerted by an ‘Inerter’
The ‘Inerter’ is a device that exerts forces proportional to the relative
acceleration of its two ends
...
admin
...
ac
...
The device works
by spinning a flywheel between two moving rods, as sketched in the figure
...
The magnitude of the force exerted by an inerter is given by the formula
F =μ
d 2L
dt 2
where L is the length, and μ is the inertia constant of the dashpot
...
Electrostatic forces
As an engineer, you will need to be able to design structures and machines that manage forces
...
It’s also difficult (but not
impossible) to design a spring with a variable stiffness or unstretched length
...
Electrostatic and electromagnetic forces are among the most important ones
...
The concepts of electrical potential, current
and charge are based on experiments
...
Electrostatic forces acting on two small charged objects that are a large distance apart
Coulomb’s Law states that if like charges Q1 and Q2
are induced on two particles that are a distance d
apart, then particle 1 will experience a force
F=−
Q1Q2
e12
4πε d 2
(acting away from particle 2), where ε is a
fundamental physical constant known as the
Permittivity of the medium surrounding the particles
(like the Gravitational constant, its value must be
determined by experiment)
...
Permittivity is more usually specified using derived units, in
Farads per meter
...
The value of ε for air is very close to that of a vacuum
...
In SI units its value is approximately 8
...
It’s worth giving results for two cases that
arise frequently in engineering designs:
Forces acting between charged flat parallel plates
F
Two parallel plates, which have equal and opposite charges Q and
are separated by a distance d , experience an attractive force with
magnitude
F = Q 2 /(2ε )
The force can be thought of as acting at the center of gravity of the
plates
...
F
Applications of electrostatic forces:
Electrostatic forces are small, and don’t have many applications in conventional mechanical systems
...
The basic idea is to construct a parallel-plate capacitor, and then to apply force to the machine by
connecting the plates to a power-supply
...
An experimental comb drive MEMS
actuator developed at Sandia National
Labs,
http://mems
...
gov/scripts/index
...
Its purpose is to rotate the mirror at the center, which
acts as an optical switch
...
A detailed
discussion of forces in these systems will be deferred to future courses
...
These systems generate charged particles (electrons), for example by heating a tungsten
wire
...
The
force then causes the particles to accelerate – but we can’t talk about accelerations in this course so you’ll
have to take EN4 to find out what happens next…
Electromagnetic forces
Electromagnetic forces are exploited more widely than electrostatic
forces, in the design of electric motors, generators, and electromagnets
...
In SI units, μ0 has fundamental units of kg m 2sec-1 A -2 , but is usually
specified in derived units of Henry per meter
...
d
F
I2
F
L
+
V2
-
The value of μ0 is exactly 4π × 10−7 H/m
Electromagnetic forces between more generally shaped current carrying wires and magnets are governed
by a complex set of equations
...
Applications of electromagnetic forces
Electromagnetic forces are widely exploited in the design of electric motors; force actuators; solenoids;
and electromagnets
...
The magnetic field can either be induced by a permanent magnet (as
in a DC motor); or can be induced by passing a current through a second wire (used in some DC motors,
and all AC motors)
...
Two examples of DC motors – the picture on the right is cut open to show the windings
...
execpc
...
htm
Hydrostatic and buoyancy forces
When an object is immersed in a stationary fluid, its surface
is subjected to a pressure
...
pa
HMS Bounty
d
A pressure is a distributed force
...
The total force on a surface must be calculated by
integration
...
The pressure in a stationary fluid varies linearly with depth below the fluid surface
p = pa + ρ gd
where pa is atmospheric pressure (often neglected as it’s generally small compared with the second
term); ρ is the fluid density; g is the acceleration due to gravity; and d is depth below the fluid surface
...
The magnitude of the resultant force is equal to
the weight of water displaced by the object
...
Thus, if the fluid has mass density ρ , and a
Center of mass of
submerged portion VI
Volume VI lies below
fluid surface
F
volume VI of the object lies below the surface
of the fluid, the resultant force due to fluid
pressure is
F = ρ gVI j
The force acts at the center of buoyancy of the immersed object
...
e
...
The buoyancy force acts in addition to gravity loading
...
The force of gravity acts (as usual) at the center of mass of the entire
object
...
Hydrodynamic forces are also of
great interest to engineers who
design bearings and car tires, since
hydrodynamic forces can cause one
surface to float above another, so
reducing friction to very low
levels
...
A Drag force, which acts parallel to the direction of air or fluid flow
2
...
The forces act at a point known as the center of lift of the object – but there’s no simple way to predict
where this point is
...
e
...
This asymmetry can result from the shape of the object itself (this effect is exploited in the
design of airplane wings); or because the object is spinning (this effect is exploited by people who throw,
kick, or hit balls for a living)
...
The friction force depends on the
object’s shape and size; on the speed of the flow; and on the viscosity of the fluid, which is a
measure of the shear resistance of the fluid
...
Viscosity is often given the symbol η , and has the rather strange units in the SI
system of Nsm −2
...
The conversion factor is 1P = 0
...
(Just to be
confusing, there’s another measure of viscosity, called kinematic viscosity, or specific
viscosity, which is ς = η / ρ , where ρ is the mass density of the material
...
73 ×10−5 Nsm −2 for a standard atmosphere (see
http://users
...
edu/~ierardi/PDF/air_nu_plot
...
001Nsm −2 ; SAE40 motor oil η ≈ 0
...
See if you can find out
what this cool word means – it’s a handy thing to bring up if you work in a fast food
restaurant
...
The pressure arises because the air accelerates as it
flows around the object
...
The pressure drag force depends on
the objects shape and size, the speed of the flow, and the fluid’s mass density ρ
...
If you want to
watch a fight, ask two airplane pilots to discuss the origin of lift in your presence
...
If this is the case, and you still want to watch a fight, you could try
http://www
...
com/, or go to a British soccer match)
...
A correct explanation of the origin of lift forces can be found at
http://www
...
nasa
...
html (this site has some neat Java applets that
calculate pressure and flow past airfoils)
...
Various measures of area are used in practice – when you look up values for drag
coefficients you have to check what’s been used
...
Vehicle
manufacturers usually use the projected frontal area (equal to car height x car width for practical
purposes) when reporting drag coefficient
...
The drag and lift coefficients are not constant, but depend on a number of factors, including:
1
...
The object’s orientation relative to the flow (aerodynamicists refer to this as the `angle of
attack’)
3
...
Size can be
quantified by
AL or
AD ; other numbers are often used too
...
Dimensional analysis shows that CD
and CL can only depend on these factors through a dimensionless constant known as
`Reynold’s number’, defined as
ρV A
η
Re =
For example, the graph on the right shows the
variation of drag coefficient with Reynolds
number for a smooth sphere, with diameter D
...
01 - 0
...
Lift
coefficients for most airfoils are of order 1 or 2,
but can be raised as high as 10 by special
techniques such as blowing air over the wing)
0
...
1
1
10
100
104
1000
Re = ρ VD / η
105
The variation of drag coefficient with Reynolds
number Re for a smooth sphere
...
g
...
hanleyinnovations
...
html
...
desktopaero
...
They usually have to be measured to get really accurate numbers
...
org/Resources/database
...
Lift and drag forces on an airfoil are
computed using the usual formula
⎛1
⎞
FL = ⎜ ρV 2 ⎟ CL AW
⎝2
⎠
⎛1
⎞
FD = ⎜ ρV 2 ⎟ CD AW
⎝2
⎠
FL
c
FD
V
α
The wing area AW = cL where c is the chord of the wing (see the picture) and L is its length, is used in
defining both the lift and drag coefficient
...
For reasonable
values of α (below stall - say less than 10 degrees) the behavior can be approximated by
CL = k Lα
CD = k Dp + k DI α 2
where k L , k Dp and k DI are more or less constant for any given airfoil shape, for practical ranges of
Reynolds number
...
The second term k DI α 2 is called induced drag, and is an undesirable byproduct of lift
...
H
...
S
...
That’s NF)
...
1deg −1 , and in fact a simple model known as
`thin airfoil theory’ predicts that lift coefficient
should vary by 2π per radian (that works out as
0
...
1deg −1 , L is the length of the wing and c is its width; while e is a constant known as the
`Oswald efficiency factor
...
9 for a high performance
wing (eg a jet aircraft or glider) and of order 0
...
The parasite drag coefficient k Dp is of order 0
...
005 or lower for a commercial airliner
...
They do this using `Molecular Dynamics,’ which is a computer method for integrating the equations of
motion for every atom in the solid
...
Specifying these
forces is usually the most difficult part of the calculation
...
In the simplest models, the atoms are assumed to
interact through pair forces
...
a
• The magnitude of the force is a function of the distance between
them
...
Various functions are used to specify the detailed shape of the force-separation law
...
This function was originally intended to model the atoms in a Noble gas – like He or Ar, etc
...
It would not be a good model of a metal, or
covalently bonded solids
...
More complicated functions exist that can account for this
kind of behavior, but there is still a great deal of uncertainty in the choice of function for a particular
material
...
2 Moments
The moment of a force is a measure of its tendency to rotate an object about some point
...
We begin by stating the mathematical definition of the
moment of a force about a point
...
2
...
To calculate the moment of a force about some point, we need to know
three things:
1
...
The position vector (relative to some convenient origin) of the
point where the force is acting ( x, y, z ) or better
r = xi + yj + zk
r-rA
F
rA
r
i
3
...
2
...
Suppose that N forces F (1) , F (2)
...
r ( N )
...
You can calculate the resultant
moment by first calculating the moment of each force, and then adding all the moments together (using
vector sums)
...
Taking moments about a different point for each force and adding the result is meaningless!
2
...
3 Examples of moment calculations using the vector formulas
We work through a few examples of moment calculations
j
Example 1: The beam shown below is uniform and has weight
W
...
L
i
A
B
W
We know (from the table provided earlier) that the center of
gravity is half-way along the beam
...
The position vector of the force relative to A is
r = ( L / 2)i
The moment about A therefore
M A = r × F = ( L / 2)i × (−W ) j = −(WL / 2)k
To calculate the moment about B, we take B as the origin
...
Member AB of a roof-truss is subjected to a vertical gravitational force W and a horizontal
wind load P
...
P
Both the wind load and weight act at the center of
gravity
...
The structure shown is subjected to
a force T acting at E along the line EF
...
This example requires a lot more work
...
We
know the magnitude of the force is T, so we only
need to work out its direction
...
It’s not hard to see that the
vector EF is
EF = ai − 3aj + 2ak
We can divide by the length of EF ( a 14 ) to
find a unit vector pointing in the correct
direction
e EF = (i − 3 j + 2k ) / 14
The force vector is
F = T (i − 3j + 2k ) / 14
Next, we need to write down the necessary position vectors
Force: r = 2ai + 3aj
Point A: rA = −2ai
Point D: rD = 4aj
Finally, we can work through the necessary cross products
M D = (r − rD ) × F
M A = (r − rA ) × F
= (4ai + 3aj) × T (i − 3 j + 2k ) / 14
(
)
i
j
= (2ai − aj) × T (i − 3 j + 2k ) / 14
k
(
)
i
j
k
= Ta / 14 4 3 0
1 −3 2
= Ta / 14 2 −1 0
1 −3 2
= Ta ( 6i − 8 j − 15k ) / 14
= Ta ( −2i − 4 j − 5k ) / 14
Clearly, vector notation is very helpful when solving 3D problems!
Example 4
...
Calculate expressions for the
moments exerted by the pressure acting on the beam about
points A and B
...
2
...
1
...
2
...
3
...
Let’s explore these statements in more detail
...
Consider a beam that’s pivoted about
some point (e
...
a see-saw)
...
We could
(a) Hang a weight W a distance d to the right of the pivot
(b) Hang a weight 2W a distance d/2 to the right of the pivot
A force applied to a pivoted beam
causes the beam to rotate
(c) Hang a weight W/2 a distance 2d to the right of the pivot
(d) Hang a weight αW a distance d / α to the right of the pivot
...
This is certainly consistent with M O = r × F
To see where the cross product in the definition comes
from, we need to do a rather more sophisticated
experiment
...
Does this have a turning tendency Fd?
d
θ
F
Fcosθ
j
k
d
i
Fsinθ
A little reflection shows that this cannot be the case
...
But the component parallel to the beam will not tend to turn the beam
...
Let’s compare this with M O = r × F
...
That’s why the definition of a moment needs a cross product
...
We can get some insight
by calculating M O = r × F for forces acting on our beam to
the right and left of the pivot
d
j
For the force acting on the left of the pivot, we just found
r = −di
F = − F cosθ i − F sin θ j
⇒ r × F = dF sin θ k
θ
k
F
d
θ
For the force acting on the right of the pivot
r = di
F = − F cos θ i − F sin θ j
⇒ r × F = − dF sin θ k
F
i
Thus, the force on the left exerts a moment along the +k direction, while the force on the right exerts a
moment in the –k direction
...
Clearly, the direction of the moment has something to do with the direction of the
turning tendency
...
It’s best to use the screw rule to visualize the effect of a moment – hold
your right hand as shown, with the thumb pointing along the direction
of the moment
...
Try this for the beam problem
...
With
your thumb pointing along –k, your fingers curl clockwise
...
2
...
As long as you can write down
A
r-rA
position vectors and force vectors correctly, and can do a cross
F
product, it is totally fool-proof
...
i
1
...
For 2D problems,
(r − rA ) and F lie in the same plane, so the direction of the moment must be perpendicular to this plane
...
F
r
A
Place your pencil over
the position vector r
Push the pencil
in the direction
of F here
j
k
i
A
What is the direction of
the moment of F about A?
Pinch the pencil lightly
here so point A is fixed
k
The pencil rotates counterclockwise
so (by right hand rule) the moment
must be towards us in +k direction
You can do a quick experiment to see whether the direction is +k or –k
...
To do so,
(i)
(ii)
(iii)
Place your pencil on the page so that it lies on the line connecting A to the force
...
Push on the pencil in the direction of the force at B
...
If it rotates
clockwise, the direction of the moment is into the picture (–k)
...
In practice you will soon find that you can very quickly tell the direction of a moment (in 2D, anyway)
just by looking at the picture, but the experiment might help until you develop this intuition
...
The magnitude of a moment about some point is equal to the
perpendicular distance from that point to the line of action of the force,
multiplied by the magnitude of the force
...
Calculate the moment exerted by the gravitational force about
points A and B
...
The pencil trick shows that W exerts a clockwise moment about A
...
Its use is best illustrated by
example
...
r-rA
Similarly, the perpendicular distance to B is L/2, and W exerts a counterclockwise moment about B
...
Member AB of a roof-truss is subjected to a
vertical gravitational force W and a horizontal wind load
P
...
L/2
B
j
(L/2)tanθ
P
i
W
The perpendicular distance from the line of action of W
θ
to B is L/2
...
Therefore W exerts a moment M B = ( L / 2)Wk
2L
The perpendicular distance from the line of action of P
to B is ( L / 2) tan θ
...
Therefore
M B = ( L tan θ / 2) Pk
The total moment is
M B = ( L / 2) {W + P tan θ } k
Example 3: It is traditional in elementary
statics courses to solve lots of problems
involving ladders (oh boy! Aren’t you glad
you signed up for engineering?)
...
Forces acting on the ladder are shown
as well
...
N θ
F
j
k
i
B
L
B
C
V
θ
A
d
C
W
H
A
(L/2) cos θ
The perpendicular distance from point A to the line along which N acts is d / cos θ
...
Therefore the trick (perpendicular distance times force) gives
M A = (d / cos θ ) N k
(for the force acting at B)
The perpendicular distance from point A to the line along which W is acting is ( L / 2) cos θ
...
Therefore
M A = −( L / 2) cos θ W k
(for the weight force)
Let’s compare these with the answer we get using M A = (r − rA ) × F
...
Then, for the force at B
(rB − rA ) = di + d tan θ j
FB = − N sin θ i + N cos θ j
⇒ M = (rB − rA ) × FB = ( dN cos θ + d tan θ N sin θ ) k
= ( dN (cos 2 θ + sin 2 θ ) / cos θ ) k = (dN / cos θ )k
giving the same answer as before, but with a whole lot more effort!
Similarly, for the weight force
(rC − rA ) = ( L / 2) cos θ i + ( L / 2) sin θ j
FC = −Wj
⇒ M = (rC − rA ) × FB = −( L / 2) cos θ W k
A
3
...
j
r-rA
A
F
rA
r-rA
rA
j
r
r
This is rather obvious in light
of trick (2), but it’s worth
stating anyway
...
The component of moment exerted by a force about an axis
through a point can be calculated by (i) finding the two force
components perpendicular to the axis; then (ii) multiplying each
force component by its perpendicular distance from the axis; and
(iii) adding the contributions of each force component following the
right-hand screw convention
...
i
k
Fz
First, let’s review what we mean by the component of a moment
about some axis
...
z
O
Fx
i
y
Fy
j
The trick gives you a quick way to calculate one of the components
...
The rule says
(i)
Identify the force components perpendicular to the i axis – that’s Fz and Fy in this case;
(ii)
Multiply each force component by its perpendicular distance from the axis
...
From the picture, we can see that Fz is a distance y from the axis, and
Fy is a distance z from the axis
...
(iii)
Add the two contributions according to the right hand screw rule
...
We can
do the pencil experiment to figure this out – the answer is that Fz exerts a moment along +i,
while Fy causes a moment along –i
...
Example: The structure shown is subjected to a vertical force
V and horizontal force H acting at E
...
Our trick gives the answer immediately
...
The
total k component of moment is
M Az = a (3H − 4V )
This trick clearly can save a great deal of time
...
2
...
A natural question arises – is there a way to rotate a body without moving it? And is there a kind of force
that causes only rotation without translation?
The answer to both questions is yes
...
3
...
The simplest example of a force couple consists of two equal and opposite
forces +F and −F acting some distance apart
...
This gives a quick way to calculate the moment
induced by a force couple:
-F
r+ - r_
r_
j
r+
F
i
Two equal and opposite
forces exert a purely
rotational force
...
It doesn’t matter which force you use to do this calculation
...
Its effect is to induce rotation without translation
...
The physical
significance of M is equivalent to the physical significance of the moment of a force about some point
...
The magnitude of M specifies
the intensity of the rotational force
...
They include
1
...
3
...
The forces exerted by your hand on a screw-driver
The forces exerted by the tip of a screw-driver on the head of a screw
The forces exerted by one part of a constant velocity joint on another
Drag forces acting on a spinning propeller
2
...
2 Pure moments, couples and torques
...
Its effect is to induce rotation, without translation – just like a force
couple
...
A pure moment is a vector quantity – it has magnitude and direction
...
The direction of a
moment indicates the axis associated with its rotational force (following the right
hand screw convention); the magnitude represents the intensity of the force
...
The concept of a pure moment takes some getting used to
...
A 3D moment
A 2D moment
d
j
k
W
i
The picture above shows the un-balanced beam
...
We can also balance the beam by applying a pure moment to it
...
d
d
W
M= - dWk
W
d
M= - dWk
W
j
M= - dWk
d
k
i
W
You could even apply the moment to the left of the beam – even right on top of the force W if you like!
2
...
3 Units and typical magnitudes of moments
In the SI system, moments have units of Nm (Newton-meters)
...
738 ft lb; or 1 ft-lb = 1
...
Typical magnitudes are:
•
•
•
Max torque exerted by a small Lego motor: 0
...
3
...
We
showed an example of a forcetransducer attached to the wheel of a car
during our earlier discussion of force
transducers
...
(So then is
Oprah a talk wench?) When you tighten
the bolts on a precision machine, it’s
important to torque them correctly
...
If you don’t apply
enough, the bolt will work itself loose
during service
...
The device may be
mechanical, or electronic
...
mac
...
asp ) is shown below
...
3
...
For example,
(i)
(i)
(ii)
(iii)
(iv)
The driving axle on your car turns the wheels by exerting a moment on them
...
In fact, motors are
usually rated by their torque capacity
...
You apply a torque to the input shaft,
and get a bigger or smaller torque from the output shaft
...
There are also some clever gearboxes that allow you to add
torques together – they are used in split-power variable speed transmissions, for example
...
Unlike a gearbox, however, the input
and output shafts don’t rotate at the same speed
...
It is used as part of an
automatic transmission system in a car
...
For example, the resistance you feel to turning the
steering wheel of your car is caused by moments acting on the wheels where they touch the
ground
...
(v)
Moments appear as internal forces in structural members or components
...
A shaft will twist because of an internal moment whose direction is parallel to the shaft
...
2
...
For example, a building
consists of a steel frame that is responsible for carrying most of the weight of the building and its
contents
...
Similarly, an automobile’s engine and transmission system contain hundreds of parts, all designed
to transmit forces exerted on the engine’s cylinder heads to the ground
...
We do this by developing a set of rules that specify the forces associated with various types of joints and
connections
...
For all our preceding examples, (e
...
gravity, lift and drag forces, and so on) we always knew everything
about the forces – magnitude, direction, and where the force acts
...
Usually (but not always), they will specify where the forces act; and they will specify that
the forces and moments can only act along certain directions
...
2
...
1 Constraint forces: overview of general nature of constraint forces
The general nature of a contact force is nicely illustrated by a
familiar example – a person, standing on a floor (a Sumo
wrestler was selected as a model, since they are particularly
interested in making sure they remain in contact with a floor!)
...
If the floor is slippery, you know
that the force on you acts perpendicular to the floor, but you
can’t make any measurements on the properties of the floor or
your feet to determine what the force will be
...
(This is
generally considered to be a good thing, although there are
occasions when it would be helpful to be able to break this law)
...
Let’s say you weight 300lb (if you don’t, a visit to Dunkin Donuts will help you reach this
weight)
...
The
magnitude of the total contact force is therefore 300lb
...
From this specific example, we can draw the following general rules regarding contact and joint forces
(1) All contacts and joints impose constraints on the relative motion of the touching or connected
components – that is to say, they allow only certain types of relative motion at the joint
...
g
...
This
means that for all intents and purposes, a constraint force acts in more than one direction at the
same time
...
(3) The direction of the forces and moments acting on the connected objects must be consistent with
the allowable relative motion at the joint (detailed explanation below)
(4) The magnitude of the forces acting at a joint or contact is always unknown
...
Because forces acting at joints impose constraints on motion, they are often called constraint forces
...
2
...
2 How to determine directions of reaction forces and moments at a joint
Let’s explore the meaning of statement (3) above in more detail, with some specific examples
...
How do we know this?
Because, according to (3) above, forces at the contact have to be consistent with the nature of relative
motion at the contact or joint
...
That means there can’t be a force
acting parallel to the floor
...
Consequently, there can’t be any moment
acting on you
...
A force must act to prevent this
...
Consequently, the floor can
only exert a repulsive force on you, it can’t attract you
...
Each time you
meet a new kind of joint, you should ask
(1) Does the connection allow the two connected solids move relative to each other? If so, what
is the direction of motion? There can be no component of reaction force along the direction of
relative motion
...
(3) For certain types of joint, a more appropriate question may be ‘Is it really healthy/legal for
me to smoke this?’
2
...
3 Drawing free body diagrams with constraint forces
When we solve problems with constraints, we are nearly always interested in analyzing forces in a
structure containing many parts, or the motion of a machine with a number of separate moving
components
...
To avoid making mistakes, it is critical to use a systematic, and logical, procedure for drawing free body
diagrams and labeling forces
...
When drawing free
body diagrams yourself, you will find it helpful to consult Section 4
...
4 for the nature of reaction forces
associated with various constraints
...
The figure shows Mickey
Mouse standing on a beam supported by a pin joint at one end
and a slider joint at the other
...
We draw a picture of the system, isolated from its
surroundings (disconnect all the joints, remove contacts, etc)
...
2L/3
j
i
A
B
Pin joint
L/3
Slider joint
2L/3
RAx
Notice how we’ve introduced variables to denote the unknown
force components
...
It is a good idea to use double subscripts –
the first subscript shows where the force acts, the second shows its direction
...
We’ve used the fact that A is a pin joint, and therefore exerts both vertical and horizontal forces; while B
is a roller joint, and exerts only a vertical force
...
For example, it’s
fairly clear that RAx = 0 in this example, but it would be incorrect to leave off this force
...
2D Mickey Mouse problem 2 Mickey mouse of weight WM stands on
L/3
2L/3
a balcony of weight WB as shown
...
j
Romeo, Romeo,
wherefore art thou
Romeo?
i
A free body diagram for the balcony and strut together is shown on
the right
...
Both A and C are pin joints, and therefore exert both
horizontal and vertical forces
...
In cases like this you have a choice of (a)
treating the two parts together as a single system; or (b) considering
the strut and floor as two separate systems
...
Pin joint
60o
C
Pin joint
L/3
2L/3
RAx
A
Rcx
RAy
WB WM
B
L/ 3
C
Rcy
The picture shows free body diagrams for both components
...
The recommended procedure is
1
...
Denote reaction forces acting between components with
(1/
the following convention
...
The subscript Bx
denotes that the force acts at B, and it acts in the positive
x direction
...
The forces RBx 2) , RBy 2) exerted by component (1) on
L/3
2L/3
RAx
(1)
A
RAy
B
WB W
M
R(1/2)Bx
(2)
Rcx
R(1/2)Bx
R(1/2)By
R(1/2)By
L/ 3
C
Rcy
component (2) are drawn in the positive x and y directions on the free body diagram for
component (2)
...
The forces exerted by component (2) on component (1) are equal and opposite to RBx 2) , RBy 2)
...
You need to think of the reaction force components as acting in two directions at
the same time
...
2
...
4 Reaction Forces and Moments associated with various types of joint
Clamped, or welded joints
No relative motion or rotation is possible
...
Three
components of reaction force must be present to prevent motion in all three
directions
...
Three components of moment must be
present to prevent relative rotation
...
The forces and moments are
labeled according to the conventions described in the
preceding section
...
A pinned joint is like a door hinge, or the joint of your elbow
...
Reaction forces: No relative motion is possible at the joint
...
Reaction moments: Relative rotation is possible about one axis
(perpendicular to the hinge) but is prevented about axes perpendicular
to the hinge
...
R(1/2)Ay
(2)
2D pinned joints are often represented as shown in the picture below
A
A
RAx
R(1/2)Ay
RAy
(1)
(1/2)
A R Ax
R(1/2)Ax A
(2)
R(1/2)Ay
Roller and journal bearings
Bearings are used to support rotating shafts
...
We’ll look at a couple of different ones
...
Reaction forces: No relative motion is possible at this kind of
bearing
...
Reaction moments: Relative rotation is allowed about one axis
(parallel to the shaft), but prevented about the other two
...
Example2: Some types of bearing allow the shaft both to rotate, and
to slide through the bearing as shown below
Reaction forces: No relative motion is possible transverse to the shaft,
but the shaft can slide freely through the bearing
...
Reaction moments: Relative rotation is allowed about one axis
(parallel to the shaft), but prevented about the other two
...
Roller bearings don’t often appear in 2D problems
...
Swivel joint: Like a pinned joint, but allows rotation about two axes
...
Reaction forces: All relative motion is prevented by the joint
...
Reaction moments: rotation is permitted about two axes, but prevented
about the third
...
Swivel joints don’t often appear in 2D problems
...
Ball and socket joint Your hip joint is a good example of a ball and
socket joint
...
Reaction forces: Prevents any relative motion
...
Reaction moments
...
No reaction
moments can be present
...
When they do, they look just like a pinned joint
...
There must be two components of reaction force, acting
along directions of constrained motion
...
There must be two components of reaction
moment
...
Reaction forces: Relative motion is prevented in two directions,
but allowed in the third
...
Reaction moments: Rotation is permitted around two axes, but
prevented around the third
...
In 2D, a slider with swivel looks identical to a slider with a pin joint
...
4
...
The nature of the forces acting at a contact depends on three things:
(1) Whether the contact is lubricated, i
...
whether friction acts at the contact
(2) Whether there is significant rolling resistance at the contact
(3) Whether the contact is conformal, or nonconformal
...
For now, we will consider only two
limiting cases (a) fully lubricated (frictionless) contacts; and (b) ideally rough (infinite friction) contacts
...
Forces acting at frictionless nonconformal contacts
A contact is said to be nonconformal if the two objects initially touch at
a point
...
Examples include contact between two balls, a ball and a
flat surface, or contact between two non-parallel cylinders
...
Three rules that help to establish the direction of frictionless
contact forces are:
(1) When one of the two contacting surfaces is flat, the
force must act perpendicular to the flat surface;
(2) When two solids contact along sharp edges, the
contact force must be perpendicular to both edges
...
Forces acting at rough (infinite friction) nonconformal contacts
A rough nonconformal contact behaves somewhat like a pinned
joint
...
Unlike a pin joint, however, the contact can
only sustain a repulsive normal force
...
If
the normal force is zero (eg when the two surfaces are about to
separate), the tangential forces TA1 = TA 2 = 0 as well
...
Therefore there must be no moment acting
on the contacting solids at the point of contact
...
If you do this, it’s easy to enforce the
N A ≥ 0 constraint
...
There’s a minor disadvantage to doing this – it’s not easy to check whether the normal force between the
surface is repulsive
...
Examples include contact between the face of a cube and a
flat surface; contact between the flat end of a cylinder and a flat surface;
or a circular pin inside a matching circular hole
...
It’s really hard to calculate the pressure
distribution (you have to model the deformation of the two contacting
solids), so instead we replace the pressure by a statically equivalent force
...
You can of course make really weird conformal contacts – like a jigsaw connection – that can completely
prevent both relative translation and rotation of the contacting solids
...
Forces acting at ideally rough (infinite friction) conformal contacts
No relative motion can occur at the contact
...
The forces can act anywhere within the area of contact – (its actual
position is determined by force and moment balance)
The component of force acting normal to the surface must be
repulsive
...
A moment must act
about an axis perpendicular to the contact to prevent relative rotation
about this axis
...
4
...
There are two particularly common structural
or machine elements that can be treated using short-cuts
...
A Two-force member is a component or structural member that
1
...
2
...
Member BC is a two-force member,
because its weight is negligible, and it has only two
pin joints connecting it to other members
...
2L/3
L/3
Romeo, Romeo,
wherefore art thou
Romeo?
i
B
A Pin joint
Pin joint
60o
This is not a 2-force
member
C
Pin joint
This is a 2-force member
The following rules are very helpful
•
•
Only one component of reaction force acts at the joints on a 2-force member
The reaction force component acts along a line connecting the two joints
...
(3)
A generic 2 force member is shown in the
figure
...
By convention, a positive reaction force is
normally taken to pull at each end of the
member, as shown
...
(3)
B
FAB
A
FAB
(2)
B
FAB
(1)
A
A
(1)
FAB
(2)
Forces on a freely rotating wheel with negligible weight: Wheels are so ubiquitous that it’s worth
developing a short-cut to deal with them
...
The contact between wheel and ground is assumed to be ideally rough (infinite
friction)
...
The picture shows a free body diagram for
a 2D wheel mounted on a frictionless
bearing
...
The two forces must be equal
and opposite, and must act along the same
line
...
R(2/1)By
R(2/1)By
B
(1)
(2)
A
RAy
j
RAy
i
For a freely rotating 3D wheel, there are 2 components of reaction force acting at the contact
between the wheel and ground
...
e
...
The picture below shows all the forces and moments acting on a freely rotating 3D wheel
...
A view from in front of the wheel shows the directions of the forces and moments more clearly
The forces and moments shown are the only nonzero components of reaction force
...
The details are left as an exercise
...
You can only use these shortcuts if:
1
...
The wheel rotates freely (no bearing friction, and nothing driving the wheel);
3
...
If any of these conditions are violated you must solve the problem by applying all the proper reaction
forces at contacts and bearings, and drawing a separate free body diagram for the wheel
...
5 Friction Forces
Friction forces act wherever two solids touch
...
It’s worth reviewing our earlier discussion of contact forces
...
e
...
We have so far only discussed two types of contact (a) fully lubricated (frictionless) contacts; and (b)
ideally rough (infinite friction) contacts
...
In contrast, for an ideally rough (infinite friction) contact, three
components of force are present as indicated on the figure on the right
...
The contacting surfaces will experience
both a normal and tangential force
...
The tangential forces can act in any direction, but their magnitude is limited
...
This is why it’s easy to walk up a dry, rough slope, but very difficult to
walk up an icy slope
...
The picture shows the big MM walking up a slope with
angle θ , and shows the forces acting on M and the slope
...
If
the tangential force gets too large, then Mickey will start to slip down
the slope
...
Sometimes (e
...
when we design moving machinery)
we are trying to calculate the forces that are needed to overcome friction and keep the parts moving
...
g
...
2
...
1 Experimental measurement of friction forces
To do both these calculations, we need to know how to determine the critical tangential forces that cause
contacting surfaces to slip
...
Leonardo da Vinci was
apparently the first person to do this – his experiments were repeated by Amontons and Coulomb about
100 years later
...
The experiment is conceptually very
simple – it’s illustrated in the figure
...
We
then try to slide the two solids relative to
each other by applying a tangential force
T
...
A simple
equilibrium calculation shows that, as long
as the weight of the components can be
neglected, the contacting surfaces must be
subject to a normal force N and a tangential force T
...
We could measure the critical tangential force as a
function of N, the area of contact A, the materials and lubricants involved, the surface finish, and other
variables such as temperature
...
plinttribology
...
co
...
htm ) is shown below
...
There are many other techniques for measuring friction
...
Two examples
are shown below
...
ist
...
de/leistung/gf4/ qualitaet/bildgro4
...
The picture on the right, from www
...
ac
...
htm
shows a pin on disk experiment inside an environmental chamber
...
The force required to hold the pin stationary is
measured
...
The friction force can be deduced by measuring the torque required to keep the disks
moving
...
ms
...
gov/htmlhome/mituc/te53
...
A friction experiment must answer two questions:
(i)
(ii)
What is the critical tangential force that will cause the surfaces to start to slide? The force
required to initiate sliding is known as the static friction force
...
We might guess that the critical force required to cause sliding could depend on
(i)
The area of contact between the two surfaces
(ii)
The magnitude of the normal force acting at the contact
(iii)
Surface roughness
(iv)
The nature of the crud on the two surfaces
(v)
What the surfaces are made from
We might also guess that once the surfaces start to slide, the tangential force needed to maintain sliding
will depend on the sliding velocity, in addition to the variables listed
...
This is very weird
...
We’ll discuss why it doesn’t below
...
If the normal force is zero, the contact can’t support any tangential force
...
(iii) Surface roughness has a very modest effect on friction
...
(iv) The crud on the two surfaces has a big effect on friction
...
If there’s a thin layer of grease on the surfaces it can cut friction by a factor of
10
...
(v) Friction forces depend quite strongly on what the two surfaces are made from
...
Some materials (e
...
Teflon) don’t bond well to other materials
...
(v) If the surfaces start to slide, the tangential force often (but not always) drops slightly
...
Otherwise, kinetic friction forces behave
just like static friction – they are independent of contact area, are proportional to the normal force, etc
...
Increasing sliding speed by a factor of 10 might drop the friction force by a few percent
...
For example, friction forces acting on the tip of an
atomic force microscope probe will behave completely differently (but you’ll have to read the scientific
literature to find out how and why!)
...
Friction
forces between rubber and other materials don’t obey all the rules listed above
...
5
...
(2)
contact, area A
T
N
Friction forces at 2D contacts
N
(1)
Friction forces at a 2D contact are
described by the following laws:
(i) If the two contacting surfaces do
not slide, then
T < μN
T
(ii) The two surfaces will start to slip if
T = μN
(iii) If the two surfaces are sliding, then
T = ±μ N
The sign in this formula must be selected so that T opposes the direction of slip
...
For most
engineering contacts, 0 < μ < 1
...
Probably we need to explain statement (iii) in more detail
...
If (1) is stationary and
(2) moves to the right, then this is the correct direction for the force and we’d use T = + μ N
...
Friction forces at 3D contacts
3D contacts are the same, but more complicated
...
To describe this mathematically, we
introduce a basis {e1 , e 2 , e3 } with e1 , e 2 in the plane of the contact,
(1/ 2)
exerted by
and e3 normal to the contact
...
The relative velocity can be computed from
the velocities v1 and v 2 of the two contacting solids, using the equation v12 = v1 − v 2
...
5
...
Material
Approx friction coefficient
Clean metals in air
0
...
5-1
...
1-0
...
1-0
...
05-0
...
05-1
...
04-0
...
05-0
...
05-0
...
0001-0
...
For example, Lim and Ashby (Cambridge
University Internal Report CUED/C-mat
...
123 January 1986) have
catalogued a large number of experimental measurements of friction
coefficient for steel on steel, and present the data graphically as shown
below
...
0001 to 3
...
For example, the picture below (from Lim and Ashby, Acta Met 37 3
(1989) p 767) shows the time variation of friction coefficient during a
pin-on-disk experiment
...
5
...
One value, known as the
coefficient of static friction and denoted by μ s , is used to model static friction in the equation giving the
condition necessary to initiate slip at a contact
T < μs N
A second value, known as the coefficient of kinetic friction, and denoted by μ k , is used in the equation
for the force required to maintain steady sliding between two surfaces
T = ± μk N
I don’t like to do this (I’m such a rebel)
...
05)
...
The real reason to distinguish between static and kinetic friction coefficient is to provide a simple
explanation for slip-stick oscillations between two contacting surfaces
...
If the end of the spring is moved steadily to the right, the block sticks for a while until the force in the
spring gets large enough to overcome friction
...
If μ were constant, then this behavior would be
impossible
...
But if we’re not trying to model slip-stick oscillations,
it’s much more sensible to work with just one value of μ
...
Most sophisticated models of slip-stick oscillations (e
...
models of earthquakes at
faults) do this
...
6 The microscopic origin of friction forces
Friction is weird
...
Coulomb grappled with these problems and came up with an incorrect explanation
...
To understand friction, we must take a close look at the nature of surfaces
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Surface roughness can be controlled to some extent – a cast
surface is usually very rough; if the surface is machined the
roughness is somewhat less; roughness can be reduced further
by grinding, lapping or polishing the surfaces
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Many surfaces can be thought of as having a
fractal geometry
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When you zoom in, it looks like the
surface got stretched vertically – surfaces are rougher at short
wavelengths than at long ones)
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Various statistical
measures are used to quantify surface roughness, but a discussion of these parameters is beyond the scope
of this course
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The surfaces will only touch at
high spots (these are known in the trade as `asperities’) on the two surfaces
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The asperity tips are squashed flat where they contact, so that there is a finite
total area of contact between the two surfaces
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N
Nominal contact area Anom
True contact
area Atrue
The true contact area can be estimated by measuring the surface roughness, and then calculating how the
surfaces deform when brought into contact
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The best estimates we have
today all agree that:
The true area of contact between two rough surfaces is proportional to the normal force pressing them
together
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This is true for all materials (except for rubbers, which are so compliant that the true contact area is close
to the nominal contact area), and is just a consequence of the statistical properties of surface roughness
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Finally, to understand the cause of the Coulomb/Amonton friction law, we need to visualize what happens
when two rough surfaces slide against each other
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It’s possible to remove
this film in a lab experiment – in which case friction behavior changes dramatically and no longer follows
Coulomb/Amonton law – but for real engineering surfaces it’s always present
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It will start to deform, and so allow the two asperities to
slide past each other, when the tangential force per unit area acting on the film reaches the shear strength
of the film τ 0
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This simple argument also explains why
friction force is independent of contact area; why it is so sensitive to surface films, and why it can be
influenced (albeit only slightly) by surface roughness