Physics 207 – Lecture 9
Lecture 9
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Goals
v Describe Friction in Air (Ch. 6), (not on 1st Exam)
v Differentiate between Newton’s 1st, 2nd and 3rd Laws
v Use Newton’s 3rd Law in problem solving
1st Exam Thurs., Oct. 6th from 7:15-8:45 PM Chapters 1-6 & 7
(“light”, direct applications of the third law)
Rooms: 2103 (302, 303, 306, 309, 310, 313)
2141 (304, 307, 308, 312) ,
2223 (311) Chamberlin Hall (plus quiet room)
Physics 207: Lecture 9, Pg 1
Friction in a viscous medium
Drag Force Quantified
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With a cross sectional area, A (in m2), coefficient of
drag of 1.0 (most objects), ρ sea-level density of air,
and velocity, v (m/s), the drag force is:
D = ½C ρ A v2 ≅ c A v2 in Newtons
c = ¼ kg/m3
In falling, when D = mg, then at terminal velocity
Example: Bicycling at 10 m/s (22 m.p.h.), with projected
area of 0.5 m2 exerts a force of ~30 Newtons
v At low speeds air drag is proportional to v but at high
speeds it is v2
v Minimizing drag is often important
Physics 207: Lecture 9, Pg 2
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Physics 207 – Lecture 9
Newton’s Third Law:
If object 1 exerts a force on object 2 (F2,1 ) then object 2
exerts an equal and opposite force on object 1 (F1,2)
F1,2 = -F2,1
For every “action” there is an equal and opposite “reaction”
IMPORTANT:
Newton’s 3rd law concerns force pairs which
act on two different objects (not on the same object) !
Physics 207: Lecture 9, Pg 3
Force Pairs vs. Free Body Diagrams
Consider the following two cases (a falling ball
and ball on table),
Compare and contrast Free Body Diagram
and
Action-Reaction Force Pair sketch
Physics 207: Lecture 9, Pg 4
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Physics 207 – Lecture 9
Forces just on a single body (1st and 2nd Laws only)
mg
FB,T= N
mg
Ball Falls
For Static Situation
N = mg
Physics 207: Lecture 9, Pg 5
Force Pairs (3rd Law)
1st and 2nd Laws
Free-body diagram
Relates force to acceleration
rd
3 Law
Action/reaction pairs
Shows how forces act between objects
FB,E = -mg
FB,T= N
FT,B= -N
FE,B = mg
FB,E = -mg
FE,B = mg
Physics 207: Lecture 9, Pg 6
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Physics 207 – Lecture 9
Example (non-contact)
Consider the forces on an object undergoing projectile motion
FB,E = - mB g
FB,E = - mB g
FE,B = mB g
FE,B = mB g
EARTH
Question: By how much does g change at an altitude of
40 miles? (Radius of the Earth ~4000 mi)
Physics 207: Lecture 9, Pg 7
Note on Gravitational Forces
Newton also recognized that gravity is an
attractive, long-range force between any two
objects.
When two objects with masses m1 and m2 are
separated by distance r, each object “pulls” on the
other with a force given by Newton’s law of gravity,
as follows:
Physics 207: Lecture 9, Pg 8
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Physics 207 – Lecture 9
Example (non-contact)
Consider the force on a satellite undergoing projectile motion
40 km above the surface of the earth:
FB,E = - mB g
FB,E = - mB g
FE,B = mB g
FE,B = mB g
EARTH
Compare: g = G m2 / 40002
g’ = G m2 / (4000+40)2
g’ / g = 40002 / (4000+40)2 = 0.98
Physics 207: Lecture 9, Pg 9
A conceptual question: A flying bird in a cage
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You have a bird in a cage that is resting on your
upward turned palm. The cage is completely
sealed to the outside (at least while we run the
experiment!). The bird is initially sitting at rest on
the perch. It decides it needs a bit of exercise and
starts to fly.
Question: How does the weight of the cage plus
bird vary when the bird is flying up, when the bird is
flying sideways, when the bird is flying down?
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Follow up question:
So, what is holding the airplane up in the sky?
Physics 207: Lecture 9, Pg 10
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Physics 207 – Lecture 9
3rd Law : Static Friction with a bicycle wheel
You are pedaling hard
and the bicycle is
speeding up.
What is the direction of
the frictional force?
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You are breaking and
the bicycle is slowing
down
What is the direction of
the frictional force?
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Physics 207: Lecture 9, Pg 11
Static Friction with a bicycle wheel
You are pedaling hard and
the bicycle is speeding up.
What is the direction of the
frictional force?
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Hint…you are accelerating
to the right
a=F/m
Ffriction, on B from E is to the right
Ffriction, on E from,B is to the left
Physics 207: Lecture 9, Pg 12
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Physics 207 – Lecture 9
Exercise
Newton’s Third Law
A fly is deformed by hitting the windshield of a speeding bus.
✿
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The force exerted by the bus on the fly is,
A.
B.
C.
greater than
equal to
less than
that exerted by the fly on the bus.
Physics 207: Lecture 9, Pg 13
Exercise
Newton’s 3rd Law
Same scenario but now we examine the accelerations
A fly is deformed by hitting the windshield of a speeding bus.
✿
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The magnitude of the acceleration, due to this collision, of
the bus is
A. greater than
B. equal to
C. less than
that of the fly.
Physics 207: Lecture 9, Pg 14
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Physics 207 – Lecture 9
Exercise
Newton’s 3rd Law
Solution
By Newton’s third law these two forces form an interaction
pair which are equal (but in opposing directions).
✿
Thus the forces are the same
However, by Newton’s second law Fnet = ma or a = Fnet/m.
So Fb, f = -Ff, b = F0
but |abus | = |F0 / mbus | << | afly | = | F0/mfly |
Answer for acceleration is (C)
Physics 207: Lecture 9, Pg 15
Exercise
Newton’s 3rd Law
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Two blocks are being pushed by a finger on a horizontal
frictionless floor.
How many action-reaction force pairs are present in this
exercise?
a
A.
B.
C.
D.
b
2
4
6
Something else
Physics 207: Lecture 9, Pg 16
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Physics 207 – Lecture 9
Force pairs on an Inclined plane
Forces on the block (static case)
Normal
Force
Friction
Force
f= µN
Forces on the plane by block
y
θ
x
Physics 207: Lecture 9, Pg 17
Force pairs on an Inclined plane
Forces on the block (sliding case, no friction)
Just one force on the plane by block
so if plane is to remain
stationary these two
components must be offset
by other force pairs
(N cos θ and N sin θ along
vertical and horizontal)
Normal
Force
y
θ
Physics 207: Lecture 9, Pg 18
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Physics 207 – Lecture 9
Example: Friction and Motion
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A box of mass m1 = 1 kg is being pulled by a horizontal
string having tension T = 40 N. It slides with friction
(µk= 0.5) on top of a second box having mass m2 = 2 kg,
which in turn slides on a smooth (frictionless) surface.
(g is said to be 10 m/s2)
v What is the acceleration of the second box ?
1st Question: What is the force on mass 2 from mass 1?
v
T
slides with friction (µk=0.5 )
m1
a=?
m2
slides without friction
Physics 207: Lecture 9, Pg 19
Example
Solution
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First draw FBD of the top box:
v
N1
T
fk = µKN1 = µKm1g
m1
m1g
Physics 207: Lecture 9, Pg 20
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Physics 207 – Lecture 9
Example
Solution
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Newtons 3rd law says the force box 2 exerts on
box 1 is equal and opposite to the force box 1
exerts on box 2.
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As we just saw, this force is due to friction:
Reaction
f2,1 = -ff1,2
Action
m1 f1,2 = µKm1g = 5 N
m2
f=5N
Physics 207: Lecture 9, Pg 21
Example
Solution
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Now consider the FBD of box 2:
N2
f2,1 = µkm1g= 5 N
m2
m1g
m2g
Physics 207: Lecture 9, Pg 22
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Physics 207 – Lecture 9
Example
Solution
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Finally, solve Fx = ma in the horizontal direction:
µK m1g = m2a
a =
m1µ k g
5N
=
m2
2 kg
= 2.5 m/s2
f2,1 = µKm1g
m2
Physics 207: Lecture 9, Pg 23
Recap
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Wednesday: Review for exam
Physics 207: Lecture 9, Pg 24
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