Download Ch33 - Wells College

Chapter 5: The Laws of Motion
Chapter 5 Goals:
• to introduce the concept of inertia and its relationship
to changes in motion
• to show how the net force is the physical cause of
changes in motion
• to present and understand Newton’s First and Second
laws of motion
• to delve into the weight force
• to discuss Newton’s Third law
• to build intuition about two more important force
examples: the tension force and the normal force
• to work with both kinds of the tricky friction forces
(kinetic and static friction)
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Newton’s First Law
Also called the law of inertia
“An object in motion remains in motion, and an object at
rest remains at rest, unless a net force acts on the object”
Not a quantitatively useful result, but it captures an
incredibly insightful observation about motion
Until Isaac Newton, most people assumed that motion
would cease unless the force continued to act, but this
was because ‘friction’ is ubiquitous
If an observer sees motion that obeys N1 (again, with
no forces present), the observer is said to be in an
‘inertial frame of reference’
Example of non-inertial reference frame: the bed of an
accelerating pickup truck
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Some of the Forces in Nature
The Spring Force?? The Tension Force!!
The Normal Force!!
The Weight Force!!
The Electric Force!!
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The Impulse Force!!
The Magnetic Force!!
Relationship of Mass to Weight
weight force Fg is the gravity force on a body of mass m
mass m is the amount of matter in a body [m] = kg, slug
W  mg and the direction is toward the center of the Earth
g = 9.81 m/s2, so a 1 kg mass has a weight of 9.8 N
g = 32.2 ft/s2, so a 1 slug mass has a weight of 32.2 lb
• if an object ONLY feels the gravity force, then it is in
free-fall, which means that it is accelerating downward at
g (whether moving up or down!).
•If the object is YOU, you would say you were
‘weightLESS’ but in fact what you lack is a force to
counteract W.
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Weight Force and Tension Force
• Lamp is object
• There are two forces acting on
the object, which are depicted as
vectors here:
• Tension force (from chain) is
UP and therefore POSITIVE
• Gravity force (from Earth) is
DOWN and therefore
NEGATIVE
• By N1, if the two forces add to
zero, the lamp is either
motionless or moving at
constant velocity (its
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acceleration is zero!)
Newton’s Second Law
Serves to define the mass of the object
“When a net force acts on a body, the body accelerates in
the direction of the net force. The acceleration is directly
proportional to the force, and inversely proportional to
the body’s mass”
Fnet
a
or
Fnet  ma
Fnet : F
m
SI: [a] = m/s2 [m] = kg  [F] = kg-m/s2 := N(ewton)

A 1 N force causes a 1 kg mass to accelerate at 1 m/s2
USA: [a] = ft/s2 [m] = slug  [F] = slug-ft/s2 := lb (‘pound’)
A 1 lb force causes a 1 slug mass to accelerate at 1 ft/s2
All of this so far is essentially in one dimension…
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What is a force?
“When a net force acts on a body, the body accelerates in
the direction of the net force. The acceleration is directly
proportional to the force, and inversely proportional to
the body’s mass”
Often you literally feel them: when an object presses on
you, your biology can sense it
Usually: but from the moment you came to life you have
felt gravity, and it acts at all points in your body, so you
are NOT aware of it until it goes away
What is a force? A force is a push or a pull.
Enuf said? NO, but that definition has survived and
works really well. It is a tautology: anything that is a
push or a pull must be a force
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Weight Force and Normal Force
• Monitor is object
• There are two forces acting on
the object:
• Normal force (from desk) is
UP and therefore POSITIVE
• Gravity force (from Earth) is
DOWN and therefore
NEGATIVE
• By N1, if the two forces add to
• Book calls the normal force Ftm, zero, the monitor is either
for force of table on monitor
motionless or moving at
•Book calls the weight force FEm,
constant velocity (its
for force of Earth on monitor
acceleration is zero!)
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The Third law: involving multiple players
• all forces are ultimately caused by bodies: the
source of the force is some body
• N3: “If the first body exerts a force on a second
body, the second body exerts an equal (in magnitude)
but opposite (in direction ) force on the first body”
• the two forces do not act on the same body!!
• Example: a baby of mass m sits in a chair of mass
M, which is sitting on the floor
• find all N3 pairs and express things precisely as
vectors!! Use the unit vector j
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Remarks on T and n
• the direction of n is by definition always ‘normal’
to the surface
• if a surface exerts a non-normally directed force on
a body, it is usually treated as some kind of friction—
or GLUE!
• often, the normal force magically adjusts itself to
suit what the kinematics is doing!
• if a weight is supported by a surface, the normal
force is equal to the weight in magnitude—but ONLY
if the weight is not accelerating!
• the contact force between two objects is partly a
normal force, and also may include friction
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Remarks on T and n
• the direction of T is always ‘along’ the string or rod
• T lives without change in all parts of the string or cable
that is ‘under’ tension
• T can go around corners by the use of pulleys
• if a weight is tied to a string and it hangs, the tension is
the weight—but ONLY if the weight is not accelerating!
• often, the tension magically adjusts itself to suit what
the kinematics is doing!
Example: a 2 kg hockey skate is accelerating vertically
upward at 4 m/s2 because a child is pulling upward on
its lace. Find the tension and the weight.
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Some Scenarios Combining T, n and Fg
Let the weight Fg of the body be 12 N (so what is mass?)
At the beginning, the body is on a horizontal surface
Tension T is provided by a hand pulling up on the rope
 The body may or may not be in equilibrium (that is, a
may or may not be zero), and there may or may not be a
third normal force n
T=10 N
T=12 N
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T=15 N
Newton’s Second Law ‘higher dimensions’:
Forces are now vectors
ma  Fnet  the net force is a vector
Fnet 
 Force  every force is a vector
• Fnet is also called the resultant force
• to grapple with solving N2, we will use the component
method of processing vectors: the Free Body Diagram
Do we have explicit expressions for a force?
• weight force: Fg = mg (g is down and |g|=9.81 m/s2)
• spring force: usually it is along one of the coordinate
directions, so for example Fspr = –i k x
• friction force: two types; both proportional to |n| but
direction is perpendicular to n: tangent to the surface
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Elementary example
A 4 kg box accelerates at 2
m/s2 to the right on a
frictionless floor, because a
child pulls on the rope with
tension T. The rope makes an
angle 60° with the horizontal.
4 kg
What do we know about forces on the body,
and its acceleration?
• we know a, in both magnitude and direction
• we know Fg , in both magnitude and direction
• we know T, but in direction only
• we know n, but in direction only
• are there any other forces? NO FRICTION!
• the problems are to find magnitudes: |T| = T & |n| = n
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The Free Body Diagram Method
1. draw the body as a simple shape: dot, circle, rectangle
2. draw force arrows, length to scale if you can, either tip or
tail on the body, in accurate directions [At this stage in
the course all forces act at a single point].
3. choose a 2d Cartesian coordinate system with origin at
that point, oriented with one + coordinate along the
acceleration direction if you know that. Draw + and –
axes for both coordinates.
4. off to the side, I draw a double arrow for the acceleration
if anything is known about it.
5. working component by component, invoke Fnet = ma
6. Now let’s apply the FBD method to that example!
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Another example
A box of mass m, on a plane
at angle q, is tied by a
parallel rope to the wall.
a) Find |T| and |n|
b) If the rope is cut, find a.
q
What do we know?
• for (a) we know a = 0.
• we know Fg , in both magnitude and direction
• we know T and n, but in direction only
• the problems are to find magnitudes: |T| = T & |n| = n
• for (b) we know T = 0 and we know all about Fg
• we know the direction of a
• the problems are to find magnitudes: |a| = a & |n| = n
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Another example: the Atwood’s machine
• two masses m1 and m2
• assume m2 > m1
• massless perfectly limp string
• massless pulley, no friction
• pulley is supported by force S
What do we know?
m1
m2
• There are in fact 3 bodies to deal with
• we know Fg1 and Fg2, in both magnitude and direction
• we know T and S, but in direction only
• we know a1 & a2 but in direction only
• kinematical constraint: we know a2 = – a1 so let them
be a1 = a and a2 = – a, respectively
• What do we seek to find, therefore??
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Three FBDs
• Take + = up
• Same |T|
m1
• We use Newton’s
Third Law here: to
a1
every action there
is an equal and
opposite reaction
T
S
T
m2
a2
Fg 1
Fg 2
T
T
m1  T  m1 g  m1a  T  m1 a  g 
• now to insert
accelerations
• a1 is +, a2 is –
m2  T  m2 g   m2 a  T  m2  a  g 
equate T  T  a m1  m2   g  m1  m2 
 m2  m1 
 2m1m2 
 T  g 

results : a  g 
 m2  m1 
 m2  m1 
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Implications of the Atwood’s machine
• a is proportional to g but
reduced to nearly zero if masses
are close: nice for slowing down
accelerated motion!!
• T is NOT equal to either of the
weights unless the two masses
are equal: then it is mg (in
which case a = 0 anyway)
• what about the third FBD?
S
T
T
• it says S – 2T = 0 or S = 2T
• interesting consequence of pulleys is how, if stacked,
they can have 2 or 3 or 4 or more tensions on them
• block and tackle for lifting heavy objects!
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The force(s) of friction: kinetic (sliding)
• kinetic friction: when a body is in contact with a
surface and the surfaces are in relative motion in their
mutual plane, there will be kinetic friction fk on the
body due to the surface
• the force’s direction is in the plane and opposite to
the motion
• the force’s magnitude is expressed by The Law of
Kinetic Friction: fk = |fk | = mk n
• roughly speaking, kinetic friction occurs because of
the roughness of the surface
• the force depends on both materials via the
coefficient of kinetic friction mk (a pure number,
typically of the order of .1, .2, up to 1.0 or more
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The force(s) of friction: static (no relative
motion))
• static friction: when a body is in contact with a
surface and the surfaces are attempting to have relative
motion in their mutual plane, but fail to move
relatively, there is static friction fs on the body
• it is as if the body is stuck to the surface!
• the force’s direction is in the plane and opposite to
the attempted but failed motion
• the force’s magnitude can be understood, depending
whether the body accelerates or not… subtle!
• this force is somewhat magical, too: it adjusts itself to
suit the situation… but only up to a point
• its maximum magnitude is ms n so fs = |fs | ≤ ms n
• this is The Law of Static Friction
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The force(s) of friction: breakaway (static to the
max)
• breakaway: when there is no relative motion (due to
static friction) but the body is on the verge of ‘breaking
away’ into relative motion, the resistive friction has
reached the limiting friction fs,max
• its direction is in the plane and opposite to the motion
• its magnitude is expressed by fs,max = ms n
• coefficient of static friction depends on both materials
•ms is a pure number, also typically of the order of .1, .2,
up to 1.0 or more
• Usually, ms > mk ; otherwise things get weird!
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Another example: static friction
A box of mass m, on a plane
at angle q, is prevented from
accelerating by the resistive
force. Find fs and n.
• we know a = 0; we know Fg
• we know fs and n, but in
direction only
FBD
x :  mg sin q  f s  0  f s  mg sin q
y : mg cos q  n  0  n  mg cos q
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q
+y
+x
fs
q F
g
n
Another example: limiting friction
A box of mass m, on a plane
at angle qB, is on the verge of
breakaway. Find ms and n
q
• we know a = 0; we know Fg
• we know fs,max and n, but in
direction only
+y
+x F
s,max
FBD
x :  mg sin q B  f s,max  0  f s ,max  mg sin q B
y : mg cos q B  n  0  n  mg cos q B
but Fs ,max  m s n  mg sin q B  m s mg cos q B
dividing, we get that m s  tan q B
nice!!
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qB F
g
n
Another example
• box of mass m, on a plane
at angle q, is tied by a
horizontal rope to the wall.
• find T and n .
q
The wet diaper problem
q
q
• wet diaper has weight Fg
• clothesline droops at angle q
• the body is the junction point

show that tens ion is T 
Fg
2 sin q
 you cannot prevent droop without
breaking the clotheslin e
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Fg