Chapter 4: Dynamics - Weebly
Transcript of Chapter 4: Dynamics - Weebly
Chapter 4:
Dynamics Newton’s Laws
What if we all jumped at once?
Newton’s 1st Law
• Objects with mass have Inertia: the
tendency to stay at rest (or moving!)
• The more mass an object has, the more
difficult it is to move it (or stop it!)
• Activity: penny/cardboard
LAB P. 76-79
Lab p. 76-79
Write out headings, purpose
procedure (choose one of a-h)
Observations: Copy out data
tables
Lab p. 76-79: choose on of A-H
A: Caitlyn/Susan
B: Carson
C: Lindsay,Miranda
D: Hailey/Brett
E: Steven^2,Kev,Zach
F: Elisabeth/Lalia
G: Noah
H: Liam/Captain Proton!
Email logger pro file to yourself - lab
Newton’s 2nd Law
• The force necessary to move (or stop)
objects depends on:
– mass
– acceleration
amFnet
• Ex 1: how much force is necessary to
accelerate a 80kg student at 10 m/s2?
280 10 800net
mF ma kg Ns
What do we mean “net” force?
• Net force is zero if there are no
unbalanced forces
• We usually do not notice forces until
they become unbalanced
• Ex. What are the forces acting on a
suction cup?
Free Body Diagrams
• The point of a FBD is to simplify the
dynamics involved
• We only point out the forces acting on the
body in question
• To get to the point, we draw the body as a…
point!
• The forces are drawn
pointing away from the
body
F1 F2
F3
More than one force?
• Ex 1: Cody applies a 50 N force to a
2.5kg book to slide it across the table.
Find the acceleration if there is a 45 N
friction force resisting this motion
netF ma
50 ( 45 )
2.5
netF N Na
m kg
20.2
5.2
5
sm
kg
N
Fa Ff
• Ex 2: Find the force necessary to
accelerate a 1200 kg rocket at 3.4 m/s2,
if there is a 250 N force of air
resistance. Fa
Fr
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝐹𝑎+𝐹𝑟= 𝑚𝑎
Exercises p. 81 #1-6
Weight
• Weight is the force of gravity acting on
an object
• Where g is the gravitational field strength at
some location in space
• Near the surface of the earth, each kg of mass
has a weight of 9.8 N
• Ex 1: what is the weight of a 86 kg student?
843gF N
m
Fg
mgF
Ex 2
• What is the weight of the same student,
while carrying a 55N bag of tricks?
843gF N
Ex 3
• What is the mass of a student with a
weight of 700N?
m
Fg
g
Fm
kgN
Nm
/8.9
700 kgm 71
Ex 3
• What is the weight of the same student
in orbit around the earth, where
g=9.2m/s2?
• 792N
Ex 4:
• What is the
weight of the
same student
on the moon,
whose
gravitational
field strength is
1/6 earth’s?
• 140N
Exercises p. 82-3 #1-5
Prep Lab p. 85
Percent Difference
• Comparing two numbers: subtract, then
divide
9.782 − 9.832
9.832
𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 − 𝐴𝑐𝑐𝑒𝑝𝑡𝑒𝑑
𝐴𝑐𝑐𝑒𝑝𝑡𝑒𝑑
Lab p. 76-79: choose on of A-H
A: Caitlyn/Susan
B: Carson
C: Lindsay,Miranda
D: Hailey/Brett
E: Steven^2,Kev,Zach
F: Elisabeth/Lalia
G: Noah
H: Liam/Captain Proton!
Email logger pro file to yourself - lab
Newton’s 3rd Law
• For every action there is an equal and
opposite reaction
• When you hit something, it hits back!
Normal Force
• When an object is in contact with a supporting
surface, it pushes down on that surface
• Newton’s 3rd Law states the surface pushes back
with an equal and opposite force
• This is normally (but not always!) equal to the
object’s weight
• We sometimes refer to Normal force as the
“apparent weight”
Simple case: object at rest
• Ex 3: What is the normal force acting on
the 2.5 kg book resting on your desk?
– What forces act on the book?
• Gravity and Normal force
– Free body diagram
– Apply 2nd law
FN
Fg
0netF ma
0g NF F
N gF F mg FN
Fg
2.5 9.8 24.5NNF kg Nkg
Extended object at rest
• Ex 3: what is the normal force acting on
your book as you lean on it with a 35 N
force?
– What forces act on the book?
• Gravity, Applied and Normal force
– Free body diagram
– Apply 2nd law
0netF ma
FN
Fa Fg
0netF ma
0g a NF F F
N g aF F F
FN
Fa Fg ( 24.5 ) ( 35 )NF N N
60NF N
Accelerating object
• Ex 4: find the apparent weight of a 50 kg
student accelerating upwards at 3.4 m/s2
– What forces act on the student?
• Gravity and Normal force
– Free body diagram
– Apply 2nd law
netF ma
FN
Fg
FN
Fg
maFF gN
maFF gN
mamgFN
4.3508.950 NF
NFN 660
Accelerating object
• Ex 4: find the acceleration of a 55 kg
student with a normal force of 1300N
– What forces act on the student?
• Gravity and Normal force
– Free body diagram
– Apply 2nd law
netF ma
FN
Fg
FN
Fg
maFF gN
m
FFa
gN
55
)8.9(551300
Na
28.13s
ma
Do Lab 4-3
• P. 85-86
Friction
• Once we know normal
force, we can calculate
friction
• Friction force can be
calculated as normal force
times “mu”
• Ff=μFN
• But how do we find mu?
• Lab 3-3 p. 55
FN
Fg
Ff Fa
Friction lab 3-3
• Conclusion: find the
average value of μ from
part 1
• Make a statement about
what factors affect friction
force
• As always, state sources of
error and possible
improvements
FN
Fg
Ff Fa
ReTest procedure
• Come in at lunch, do
corrections, get help
• Then book date to come in
another lunch to do retest
Friction
• We find friction force is proportional to the Normal force and a “stickiness factor” (AKA coefficient of friction)
Nf FF
• Ex: find the friction force acting on your
25N textbook as it slides across the table if
=0.55
• Applying 2nd law to find Normal force:
x y
Nf FF maFnet
maFF Ng
0 Ng FF
NFF gN 25
NFf 2555.0
NFf 14
Nf FF
• Ex: find the coefficient of friction if it takes
a 13 N force to slide your 2.2 kg book along
the table at a constant speed.
N
f
F
F
)8.9(2.2
13 60.0
Exercises
• P. 60 #1-4
• Lab p. 55 done!!
Friction on “static” objects
• If an object is not moving, forces must
be balanced
– friction force must be equal to the applied
horizontal force
• The maximum static friction force is
given by the previous formula, so we
have:
Nsf FF
• Where s is the coefficient of static friction, and
tends to be larger than k for kinetic friction
When does static become
kinetic? • Static friction increases
with applied force until it “breaks free”
– at this point, it starts moving and kinetic or “sliding” friction takes over
• Ex 6: for what applied force if s=0.65 does your textbook start moving?
– What will the acceleration be?
• We already have Normal force:
x y
Nf FF max NFN 25
NFf 2565.0max
NFf 16max
• Once it breaks free we have kinetic friction again:
x y
NFa 16
maFnet
m
FFa
fa
298.0s
ma
kg
NN
55.2
75.1325.16
Friction Lab
• Add a column to table 1 for coefficient of
friction
• Do questions #1, 3 (average) & 4 p. 56
• Do question #1 p. 58 (#2: sources of
error should be done in your conclusion)
Universal
Gravitation
• Newton’s most original contribution!
• All objects in the universe exert a gravitational pull on each other
• But why doesn’t the moon fall?
• Newton realized
objects don’t fall
to the Earth’s
surface if they
have a high
enough tangential
velocity
Universal gravitation:
• Fg=GMm/r2
– Where G is the
universal gravitational
constant
How do we find G?
Weigh the Earth!?
Mm
rFG
g
2
• Weigh the Earth!
Pendulum Method
• If only we had a large enough mass to get a
measurable force…
• Ex 1: Calculate G if you get a 4.6×10-6 N
force on a 1.5kg pendulum 150m away from
a mountain with a mass of 6.4×108 kg
Mm
rFG
g
2
Mm
rFG
g
2
26
8
4.6 10 150
1.5 6.4 10
N mG
kg kg
2
2101008.1kg
NmG
2
21010kg
NmG
Cavendish’s experiment
• To get a more
precise
measurement of G,
he used a torsional
pendulum
• Cavendish was able
to get a value of:
2
2111067.6kg
NmG
• Ex 2: What is the
Earth’s mass if 1kg
of mass has a
weight of 9.8N?
2r
GMmFg
mG
FrM
g
2
6 2
211
2
9.8 (6.38 10 )
1 6.67 10
N mM
Nmkgkg
kgM 241098.5
• Ex 3: Find the force
of attraction
between Kat (54 kg)
and Mikael (95 kg) if
they sit 4.0 m apart.
2r
GMmFg
2r
GMmFg
2
112
2
6.67 10 95 54
(4.0 )g
Nm kg kgkg
Fm
NFg8101.2
Start Exercises
• P. 67 question 6 a) through c)
• *part (a) consider what happens when we replace r with 2r?
21r
GMmF
222r
GMmF
What about g?
• How does gravitational field strength “g”
fit into all of this?
• Ex 4: find g at an
altitude of 130 km.
Careful!
2r
GMmmg
2
11 242
6 5 2
6.67 10 5.98 10
(6.38 10 1.3 10 )
Nm kgkg
gm m
kgNg 41.9
mgFg 2r
GMg
Ex. 3: find g on the moon.
2r
GMg
2
11 222
6 2
6.67 10 7.35 10
(1.74 10 )
Nm kgkg
gm
kgNg 62.1
Ex 4: Find the Sun’s
gravitational field here
2r
GMg
2
11 302
11 2
6.67 10 1.98 10
(1.50 10 )
Nm kgkg
gm
kgNg 0059.0
Do you feel lighter? • How much difference would a 85kg
student feel from noon to midnight?
mgFg
NFg 5.0
85 0.0059gNF kgkg
Solve for Mass
• Ex 1: Find the mass of the Earth
required to exert a force of 2.0x1020 N
on the Moon
2r
GMmFg
Gm
rFM
g
2
kgM 24100.6
Gravitational Definitions • This is where most people mix up
problems
• Mass, Force or Field?
– 25N
– Weight
– 5kg
– 9.8N/kg
– Apparent Weight
– Gravitational acceleration
– 25lb!?
– 1.63m/s2
Did it really
happen like that?
• Mostly true story but...
• 2nd law actually written in terms of "impulse":
Impulse
defined as a
change in
momentum
We use momentum to analyze collisions…
MOMENTUM IS THE PRODUCT
OF MASS AND VELOCITY
p=m·v
What is the momentum of a 120 kg
rugby player running at 11 m/s?
p=mv
p=120kg · 11 m/s
p=1320 kg ·m/s
More angular momentum
• http://www.youtube.com/watch?v=vj5XFK5p38c&fe
ature=youtube_gdata_player
http://www.youtube.com/watch?v=
M1fTD69CYtA&feature=related
Second Law Revisited?
Newton originally wrote his second law in this
form:
Rearrange by
multiplying both
sides by Δt:
Impulse
If t is small, F can be very large
Can we find the average
force Venus applies to this
57g tennis ball?
She returns a 46m/s serve
at 35m/s and the ball only
contacts the racquet for
6.5ms
#1-7 p. 89-90
Start with #1,2,5,7
FIAQD
Ex: #2 find Impulse
tF
sN 001.055
#1-7 p. 89-90
Start with #1,2,5,7
FIAQD
Ex: #5 find Impulse
vm
s
ms
mkg 6.125.2510
Principle of Momentum Conservation:
Total momentum
before is equal to
total momentum
after the collision
This is true for all
closed, isolated
systems
No
one allowed
in or out
No net
external
force
Chabal: timing is everything
Can we use momentum to analyze
this collision?
Momentum is conserved so p before
must equal p after
Rokocoko has a
mass of 105 kg and
runs at 7.5 m/s into
92 kg Betson
How fast are they
moving after the
collision?
Tm
umv 11
Most common physics 11 collision: moving object
collides with stationary one; they move off
together
Ex: a student with a mass of 105 kg runs at 7.5
m/s into a 92 kg classmate, find v
Same steps every time!
Variables
m1=
u1=
m2=
u2=
v'=
pi=pf
m1u1= (m1 + m2)v?
0 = m1v1+ m2v2?
• Explosions
For an explosion? Ex: bottle rocket
Variables
m1=0.320 kg
v1=?
m2=0.850 kg
v2=-25 m/s
pi=pf
0 = m1v1+ m2v2
m1v1 =- m2v2
v1=- m2v2/m1
v1=-0.850kg(-25m/s)/0.320kg
v1= 66 m/s
Practice Problems
• Finish q's 1-7p.89-90
• Start Ch 4 review p. 90
• Force quiz tomorrow
• Momentum Quiz Friday
• Test next week
• Start bottle rocket design
Ex: question 3
Variables
m1=0.060 kg
v1=600 m/s
m2=3.0 kg
V2=? m/s
pi=pf
0 = m1v1+ m2v2
m2v2 =- m1v1
v2=- m1v1/m2
v2=-0.060kg(600m/s)/3.0kg
v2= -12 m/s
Ex: question 6
Variables
m1=1500 kg
u1=44 m/s
m2=1000 kg
u2=-22 m/s
mT=2500 kg
v=? m/s
pi=pf
m1u1+ m2u2 = mTv
(m1u1+ m2u2)/mT= v
v=18 m/s
Ex: question 13
Variables
m1=0.250 kg
u1=5.0 m/s
m2=0.300 kg
u2=2.0m/s
mT=0.550 kg
v=? m/s
p=mv
m1u1+ m2u2 = mTv
(m1u1+ m2u2)/mT= v
v=3.4 m/s
Elastic vs. Inelastic?
• An elastic collision is
one where
mechanical energy is
conserved
• This can be
represented by a
bouncing ball
elastic inelastic completely
inelastic
Elasticity when bouncing
Chapter Review
• Finish p. 89-90 #1-7 Finish Chapter Review p. 90 q's 1-18
Momentum Quiz Friday
Chapter 4 Test next week
Bonus! Find the velocity of the
wreckage after this horrible
collision :(
Chapter Review
Finish p. 91 #10-18
Start “Test Yourself” p. 92
Momentum Quiz Today