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Physics 207: Lecture 2 Notes

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Differentiate between Newton's 1st, 2nd and 3rd Laws. Use Newton's 3rd Law in problem solving ... The bird is initially sitting at rest on the perch. ... – PowerPoint PPT presentation

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Title: Physics 207: Lecture 2 Notes


1
Lecture 8
  • Goals
  • Differentiate between Newtons 1st, 2nd and 3rd
    Laws
  • Use Newtons 3rd Law in problem solving

Assignment HW4, (Chapters 6 7, due 10/1,
Wednesday) Finish Chapter 7 1st Exam Thursday,
Oct. 2nd from 715-845 PM Chapters 1-7
2
Inclined plane with Normal and Frictional
Forces
  • Static Equilibrium Case
  • Dynamic Equilibrium (see 1)
  • Dynamic case with non-zero acceleration

Normal means perpendicular
Normal Force
Friction Force
f
S F 0 Fx 0 mg sin q f Fy 0 mg
cos q N with mg sin q f mS N if mg sin q
gt mS N, must slide Critical angle mk tan q
mg sin q
q
y
mg cos q
q
q
x
Block weight is mg
3
Inclined plane with Normal and Frictional
Forces
  • Static Equilibrium Case
  • Dynamic Equilibrium
  • Friction opposite velocity
  • (down the incline)

Normal means perpendicular
Normal Force
v
Friction Force
fK
S F 0 Fx 0 mg sin q fk Fy 0 mg
cos q N fk mk N mk mg cos q Fx 0 mg
sin q mk mg cos q mk tan q (only
one angle)
mg sin q
q
y
mg cos q
q
q
x
mg
4
Inclined plane with Normal and Frictional
Forces
3. Dynamic case with non-zero acceleration Result
depends on direction of velocity
Fx max mg sin q fk Fy 0 mg cos q
N fk mk N mk mg cos q Fx max mg sin
q mk mg cos q ax g sin q mk g cos q
5
The inclined plane coming and going (not
static)the component of mg along the surface gt
kinetic friction
  • Fx max mg sin q uk N
  • Fy may 0 -mg cos q N

Putting it all together gives two different
accelerations, ax g sin q uk g cos q. A tidy
result but ultimately it is the process of
applying Newtons Laws that is key.
6
Velocity and acceleration plots
Notice that the acceleration is always down the
slide and that, even at the turnaround point, the
block is always motion although there is an
infinitesimal point at which the velocity of the
block passes through zero. At this moment,
depending on the static friction the block may
become stuck.
7
The flying bird in the cage
  • 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?
  • So, what is holding the airplane up in the sky? 

8
Friction in a viscous mediumDrag Force Quantified
  • With a cross sectional area, A (in m2),
    coefficient of drag of 1.0 (most objects), ?
    sea-level density of air, and velocity, v (m/s),
    the drag force is
  • D ½ C ? 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 30 Newtons
  • Minimizing drag is often important

9
Fish Schools
10
  • By swimming in synchrony in the correct
    formation, each fish can take advantage of moving
    water created by the fish in front to reduce
    drag.
  • Fish swimming in schools can swim 2 to 6 times as
    long as individual fish.

11
Free Fall
  • Terminal velocity reached when Fdrag Fgrav (
    mg)
  • For 75 kg person with a frontal area of 0.5 m2,
  • vterm ? 50 m/s, or 110 mph
  • which is reached in about 5 seconds, over 125 m
    of fall

12
Trajectories with Air Resistance
  • Baseball launched at 45 with v 50 m/s
  • Without air resistance, reaches about 63 m high,
    254 m range
  • With air resistance, about 31 m high, 122 m range

Vacuum trajectory vs. air trajectory for 45
launch angle.
13
Newtons Laws
  • Law 1 An object subject to no external forces is
    at rest or moves with a constant velocity if
    viewed from an inertial reference frame.
  • Law 2 For any object, FNET ??F ma
  • Law 3 Forces occur in pairs FA , B -
    FB , A
  • (For every action there is an equal and
    opposite reaction.)

14
Newtons 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 Newtons 3rd law concerns force
pairs which act on two different objects (not
on the same object) !
15
Gravity
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 Newtons law
of gravity, as follows
16
Cavendishs Experiment
F m1 g G m1 m2 / r2 g G m2 / r2 If we
know big G, little g and r then will can find m2
the mass of the Earth!!!
17
Example (non-contact)
Consider the forces on an object undergoing
projectile motion
Question By how much does g change at an
altitude of 40 miles? (Radius of the Earth 4000
mi)
18
Example
Consider the following two cases (a falling ball
and ball on table), Compare and contrast Free
Body Diagram and Action-Reaction Force Pair
sketch
19
Example
The Free Body Diagram
Ball Falls
For Static Situation N mg
20
Normal Forces
Certain forces act to keep an object in place.
These have what ever force needed to balance all
others (until a breaking point).
Main goal at this point Identify force pairs
and apply Newtons third law
21
Example
First Free-body diagram Second Action/reaction
pair forces
22
Exercise Newtons Third Law
A fly is deformed by hitting the windshield of a
speeding bus.
v
The force exerted by the bus on the fly is,
  1. greater than
  2. equal to
  3. less than

that exerted by the fly on the bus.
23
Exercise 2Newtons Third Law
Same scenario but now we examine the
accelerations
A fly is deformed by hitting the windshield of a
speeding bus.
v
The magnitude of the acceleration, due to this
collision, of the bus is
  1. greater than
  2. equal to
  3. less than

that of the fly.
24
Exercise 2Newtons Third LawSolution
By Newtons third law these two forces form an
interaction pair which are equal (but in opposing
directions).
?
Thus the forces are the same
However, by Newtons second law Fnet ma or a
Fnet/m. So Fb, f -Ff, b F0 but abus F0
/ mbus ltlt afly F0/mfly
Answer for acceleration is (C)
25
Exercise 3Newtons 3rd Law
  • Two blocks are being pushed by a finger on a
    horizontal frictionless floor.
  • How many action-reaction force pairs are present
    in this exercise?
  1. 2
  2. 4
  3. 6
  4. Something else

26
Exercise 3Solution
a
b
6
27
Example Friction and Motion
  • A box of mass m1 1 kg is being pulled by a
    horizontal string having tension T 40 N. It
    slides with friction
  • (mk 0.5) on top of a second box having mass
    m2 2 kg, which in turn slides on a smooth
    (frictionless) surface.
  • What is the acceleration of the second box ?
  • (This is what I solved for in class!)
  • But first, what is force on mass 2?
  • a 0 N (B) a 5 N (C) a 20 N (D)
    cant tell

slides with friction (mk0.5 )
T
m1

a ?
m2
slides without friction
28
ExampleSolution
  • First draw FBD of the top box

N1
m1
fk mKN1 mKm1g
T
m1g
29
ExampleSolution
  • 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.

Action
Reaction
f1,2 mKm1g 5 N
f2,1 -f1,2
m1
m2
  • a 0 N (B) a 5 N (C) a 20 N (D)
    cant tell

30
ExampleSolution
  • Now consider the FBD of box 2

N2
f2,1 mkm1g
m2
m1g
m2g
31
ExampleSolution
  • Finally, solve Fx ma in the horizontal
    direction

mK m1g m2a
2.5 m/s2
f2,1 mKm1g
m2
32
Example Friction and Motion, Replay
  • A box of mass m1 1 kg, initially at rest, is
    now pulled by a horizontal string having tension
    T 10 N. This box (1) is on top of a second box
    of mass m2 2 kg. The static and kinetic
    coefficients of friction between the 2 boxes are
    ?s1.5 and mk 0.5. The second box can slide
    freely (frictionless) on an smooth surface.
  • Compare the acceleration of box 1 to the
    acceleration of box 2 ?

a1
friction coefficients ms1.5 and mk0.5
T
m1

a2
slides without friction
m2
33
ExampleFriction and Motion, Replay in the static
case
  • A box of mass m1 1 kg, initially at rest, is
    now pulled by a horizontal string having tension
    T 10 N. This box (1) is on top of a second box
    of mass m2 2 kg. The static and kinetic
    coefficients of friction between the 2 boxes are
    ?s1.5 and mk 0.5. The second box can slide
    freely on an smooth surface (frictionless).
  • If there is no slippage then maximum frictional
    force between 1 2 is
  • (A) 20 N
  • (B) 15 N
  • (C) 5 N
  • (D) depends on T

34
Exercise 4Friction and Motion, Replay in the
static case
  • A box of mass m1 1 kg, initially at rest, is
    now pulled by a horizontal string having tension
    T 10 N. This box (1) is on top of a second box
    of mass m2 2 kg. The static and kinetic
    coefficients of friction between the 2 boxes are
    ?s1.5 and mk 0.5. The second box can slide
    freely on an smooth surface (frictionless).
  • If there is no slippage, what is the maximum
    frictional force between 1 2 is
  1. 20 N
  2. 15 N
  3. 5 N
  4. depends on T

35
Exercise 4Friction and Motion
N
fS ? ?S N ?S m1 g 1.5 x 1 kg x 10 m/s2
which is 15 N (so m2 cant break free)
fS
T
m1 g
  • fs 10 N and the acceleration of box 1 is
  • Acceleration of box 2 equals that of box 1, with
    a T / (m1m2) and the frictional force f is
    m2a
  • (Notice that if T were raised to 15 N then it
    would break free)

a1
friction coefficients ms1.5 and mk0.5
T
m1

a2
slides without friction
m2
36
Exercise Tension example
Compare the strings below in settings (a) and (b)
and their tensions.
  1. Ta ½ Tb
  2. Ta 2 Tb
  3. Ta Tb
  4. Correct answer is not given

37
Lecture 8
  • Goals
  • Differentiate between Newtons 1st, 2nd and 3rd
    Laws
  • Use Newtons 3rd Law in problem solving

Assignment HW4, (Chapters 6 7, due 10/1,
Wednesday) Finish Chapter 7 1st Exam Thursday,
Oct. 2nd from 715-845 PM Chapters 1-7
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