Transcript Chapter 5 - SteadyServerPages

Chapter 5
• Applying
Newton’s Laws
Equilibrium
An object is in equilibrium when
the net force acting on it is zero.
In component form, this is
The net force on each
man in the tower is
zero.
Slide 5-13
Equilibrium
A hanging street sign with more than one force acting on it
𝐹1
𝐹2
𝐹3
𝐹𝑥1 + 𝐹𝑥2 + 𝐹𝑥3 =0
𝐹𝑦1 + 𝐹𝑦2 + 𝐹𝑦3 =0
Equilibrium
What are the components of the forces?
𝐹1
𝐹𝑦1
𝐹𝑥1
𝐹𝑦3
𝐹𝑥2
𝐹2
0
𝐹𝑥1 + 𝐹𝑥2 + 𝐹𝑥3 = 0
𝐹3
0
𝐹𝑦1 + 𝐹𝑦2 + 𝐹𝑦3 =0
Slide 5-14
Example Problem
A 100 kg block with a weight of 980 N hangs on a rope. Find
the tension in the rope if
A. the block is stationary.
B. it’s moving upward at a steady speed of 5 m/s.
C. it’s accelerating upward at 5 m/s2.
Slide 5-15
Example Problem
A 100 kg block with a weight of 980 N hangs on a rope. Find
the tension in the rope if
A.
the block is stationary.
𝐹𝑦 = 0
𝐹𝑦 = 𝑇𝑅 + 𝐹𝑔 = 0
𝑇𝑅
100kg
𝑇𝑅 + 𝑚𝑔 = 0
m
𝑇𝑅 = 100kg ∙ 9.8 2 = 980N
s
𝐹𝑔
Slide 5-15
A 100 kg block with a weight of 980 N hangs on a rope. Find the tension in the
rope if
A.
B.
C.
the block is stationary.
it’s moving upward at a steady speed of 5 m/s.
it’s accelerating upward at 5 m/s2.
𝐹𝑦 = 0
𝐹𝑦 = 𝑇𝑅 + 𝐹𝑔 = 0
𝑇𝑅
𝑣
100kg
𝑇𝑅 + 𝑚𝑔 = 0
m
𝑇𝑅 = 100kg ∙ 9.8 2 = 980N
s
𝐹𝑔
Slide 5-15
A 100 kg block with a weight of 980 N hangs on a rope. Find the tension in the
rope if
A.
B.
C.
the block is stationary.
it’s moving upward at a steady speed of 5 m/s.
it’s accelerating upward at 5 m/s2.
𝐹𝑦 = 𝑚 𝑎𝑦
𝑇𝑅
𝑎
𝑇𝑅 + 𝐹𝑔 = 𝑚𝑎𝑦
100kg
𝑇𝑅 = 𝑚𝑎𝑦 − 𝑚𝑔
𝑇𝑅 = 𝑚 𝑎𝑦 − 𝑔
m
m
𝑇𝑅 = 100kg 5 2 − −9.8 2
s
s
= 1480N
𝐹𝑔
Slide 5-15
Example Problem
A wooden box, with a mass of 22 kg, is pulled at a constant
speed with a rope that makes an angle of 25° with the wooden
floor. If the coefficient of kinetic friction is 𝜇𝑘 = .5, what is the
tension in the rope?
Slide 5-16
A wooden box, with a mass of 22 kg, is pulled at a constant
speed with a rope that makes an angle of 25° with the wooden
floor. If the coefficient of kinetic friction is 𝜇𝑘 = .5, what is the
tension in the rope?
𝐹𝑦 = 𝑇𝑅𝑦 + 𝐹𝑔 + 𝑁 = 0
𝐹𝑥 = 𝑇𝑅𝑥 + 𝐹𝑓 = 0
𝑇𝑅𝑦 = −𝑚𝑔 − 𝑁
𝑇𝑅𝑥 = −𝜇𝑘 𝑁
𝑇𝑅
𝑁
𝐹𝑓
25°
𝐹𝑔
𝑇𝑅𝑦
𝑇𝑅𝑥
Slide 5-16
A wooden box, with a mass of 22 kg, is pulled at a constant speed with a rope that makes
an angle of 25° with the wooden floor. If the coefficient of kinetic friction is 𝜇𝑘 = .5, what is
the tension in the rope?
𝑇𝑅𝑥 = −𝜇𝑘 𝑁
𝑇𝑅𝑦 = −𝑚𝑔 − 𝑁
𝑇𝑅 cos 𝜃 = −𝜇𝑘 𝑁
𝑇𝑅 sin 𝜃 = −𝑚𝑔 − 𝑁
𝑇𝑅 cos 𝜃
𝑇𝑅 sin 𝜃 = −𝑚𝑔 +
𝜇𝑘
𝑇𝑅 cos 𝜃
= −𝑁
𝜇𝑘
𝑇𝑅
𝑁
𝐹𝑓
25°
𝐹𝑔
𝑇𝑅𝑦
𝑇𝑅𝑥
Slide 5-16
A wooden box, with a mass of 22 kg, is pulled at a constant speed with a rope that makes
an angle of 25° with the wooden floor. If the coefficient of kinetic friction is 𝜇𝑘 = .5, what is
the tension in the rope?
𝑇𝑅 cos 𝜃
𝑇𝑅 sin 𝜃 = −𝑚𝑔 +
𝜇𝑘
𝑇𝑅 cos 𝜃
𝑇𝑅 sin 𝜃 −
= −𝑚𝑔
𝜇𝑘
𝑇𝑅
cos 𝜃
sin 𝜃 −
= −𝑚𝑔
𝜇𝑘
−𝑚𝑔
𝑇𝑅 =
cos 𝜃
sin 𝜃 −
𝜇𝑘
Example Problem
A ball weighing 50 N is pulled back by a rope to an angle of 20°.
What is the tension in the pulling rope?
Slide 5-19
Example Problem
A ball weighing 50 N is pulled back by a rope to an angle of 20°.
What is the tension in the pulling rope?
𝑇𝑅
𝑇𝑅𝑦
𝑇
𝐹𝑔
𝑇𝑅𝑥
Slide 5-19
A ball weighing 50 N is pulled back by a rope to an angle of 20°.
What is the tension in the pulling rope?
𝐹𝑥 = 𝑇𝑅𝑥 + 𝑇 = 0
𝑇𝑅𝑥 = −𝑇
𝑇
𝑅𝑦
°
cos 20 =
𝑇𝑅𝑥
𝐹𝑦 = 𝑇𝑅𝑦 + 𝐹𝑔 = 0
𝑇𝑅𝑦 = −𝑚𝑔 = 50N
𝑇𝑅𝑦
𝑇𝑅
𝑇
𝐹𝑔
𝑇𝑅𝑥
Slide 5-19
Using Newton’s Second Law
Slide 5-20
Dynamics with the 2nd law
𝑎
𝑎𝑦
𝐹s
𝑎𝑥
Σ𝐹𝑥 = 𝐹s𝑥 + 𝑤𝑥 = 𝑚𝑎𝑥
𝑤
Σ𝐹𝑦 = 𝐹s𝑦 + 𝑤𝑦 = 𝑚𝑎𝑦
Example Problem
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s.
It then crosses a rough patch of snow which exerts a friction force
of -12 N. How far does it slide on the snow before coming to rest?
Slide 5-21
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s.
It then crosses a rough patch of snow which exerts a friction force
of -12 N. How far does it slide on the snow before coming to rest?
𝑣𝑥
Slide 5-21
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s.
It then crosses a rough patch of snow which exerts a friction force
of -12 N. How far does it slide on the snow before coming to rest?
𝐹𝑓
Slide 5-21
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s.
It then crosses a rough patch of snow which exerts a friction force
of -12 N. How far does it slide on the snow before coming to rest?
2
𝑉𝑓
=
2
−𝑉𝑖
2
𝑉𝑖
𝐹 = 𝑚𝑎
+ 2𝑎∆𝑥
𝐹
=2
∆𝑥
𝑚
−𝑉𝑖2 𝑚
∆𝑥 =
2𝐹
∆𝑥
𝐹
𝑎=
𝑚
Slide 5-21
Mass and Weight
• Mass and weight are not the same
𝑤 = 𝑚𝑔
m
The moon’s gravity
• The moon has about 1/6 of the
gravity of earth
𝑤m =
1
𝑚 6𝑔
m
Weightlessness
• Falling doesn’t mean you have no
weight
• But this is the term we use anyway
𝑣
𝑎
Mass and Weight
–w = may = m(–g)
w = mg
Slide 5-23
Which of the following statements
about mass and weight is correct?
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A. Your mass is a measure of the
force gravity exerts on you
B. Your mass is same everywhere
in the universe
C. Your weight is the same
everywhere in the universe
D. Your weight is a measure of
your resistance of being
accelerated
30
Apparent Weight
Slide 5-24
Example Problem
A 50 kg student gets in a 1000 kg elevator at rest. As the elevator
begins to move, she has an apparent weight of 600 N for the first 3
s. How far has the elevator moved, and in which direction, at the
end of 3 s?
Slide 5-25
A 50 kg student gets in an elevator at rest. As the elevator begins to
move, she has an apparent weight of 600 N for the first 3 s. How far
has the elevator moved, and in which direction, at the end of 3 s?
𝐹𝑦 = 𝑤𝑎 + 𝑤𝑠 = 𝑚𝑎
𝑤𝑎 = 600N
𝑤𝑎
𝑎
𝑤𝑎 + 𝑤𝑠
𝑎=
𝑚
𝑤𝑠 = 𝑚𝑔
1 2
1 2
∆𝑦 = 𝑣𝑖𝑦 𝑡 + 𝑎𝑡 → ∆𝑦 = 𝑎𝑡
2
2
0
1 𝑤𝑎 + 𝑤𝑠
∆𝑦 =
2
𝑚
𝑡2
𝑤𝑠
Slide 5-25
Normal Forces
𝐹N
m
𝑤
𝐹N = 𝑚𝑔
Normal Forces
+𝑦
𝐹N
𝑎
+𝑥
− 𝑤 cos 𝜃
𝑤
𝜃
− 𝑤 sin 𝜃
Normal Forces
𝑤 = 𝑚𝑔
𝐹N
+𝑦
+𝑥
Σ𝐹𝑥 = −𝑚𝑔 sin 𝜃 = 𝑚𝑎𝑥
Σ𝐹𝑦 = −𝑚𝑔 cos 𝜃 + 𝐹N = 𝑚𝑎𝑦
−𝑚𝑔 cos 𝜃 + 𝐹N = 0
𝐹N = 𝑚𝑔 cos 𝜃
𝑎
− 𝑤 cos 𝜃
𝑤
𝜃
− 𝑤 sin 𝜃
Static Friction
fs max = µsn
Slide 5-30
Kinetic Friction
fk = µkn
Slide 5-31
Coefficients of static friction are typically greater
than coefficients of kinetic friction.
50%
50%
Fa
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Tr
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A. True
B. False
Include static friction
𝐹f = 𝜇𝑠 𝐹N
𝐹N
𝑎
+𝑦
+𝑥
− 𝑤 cos 𝜃
𝑤
𝜃
− 𝑤 sin 𝜃
Kinetic and Rolling friction
𝐹f = 𝜇𝑘 𝐹N
𝐹f = 𝜇𝑟 𝐹N
𝑣
𝑣
𝐹f
𝐹f
Working with Friction Forces
Slide 5-32
Drag
An object moving in a gas or liquid experiences a drag force
Drag coefficient. Depends on
details of the object’s shape.
“Streamlining” reduces drag by
making CD smaller. For a typical
object, CD 0.5.
Density of gas or liquid. Air has
a density of 1.29 kg/m3.
A is the object’s cross section area
when facing into the wind.
Drag depends on the square of the speed.
This is a really important factor that limits the
top speed of cars and bicycles. Going twice
as fast requires 4 times as much force and,
as we’ll see later, 8 times as much power.
Slide 5-34
Drag
𝑣
𝐹𝐷
𝐹𝐷 =
1
𝐶𝑑 𝜌𝐴𝑣 2
2
Direction opposite of motion
Drag
𝐹𝐷 =
1
𝜌𝐴𝑣 2
4
𝑣
𝜌 is the fluid density
A is the cross-sectional area
𝐶𝑑 is the drag coefficient(most objects
have 𝐶𝑑 = 1/2)
Terminal Speed
Sky divers go no faster
than this
It’s about 150 mph for
average objects
At this speed…
Σ𝐹𝑦 = 0
𝐹𝐷
𝑣𝑡
Example Problem
A car traveling at 20 m/s stops in a distance of 50 m. Assume that
the deceleration is constant. The coefficients of friction between a
passenger and the seat are μs = 0.5 and μk = 0.3. Could a 70 kg
passenger slide off the seat if not wearing a seat belt?
Slide 5-33
Example Problem
A car traveling at 20 m/s stops in a distance of 50 m. Assume that
the deceleration is constant. The coefficients of friction between a
passenger and the seat are μs = 0.5 and μk = 0.3. Could a 70 kg
passenger slide off the seat if not wearing a seat belt?
m
𝑣 = 20
s
∆𝑥 = 50m
Slide 5-33
Example Problem
A car traveling at 20 m/s stops in a distance of 50 m. Assume that
the deceleration is constant. The coefficients of friction between a
passenger and the seat are μs = 0.5 and μk = 0.3. Could a 70 kg
passenger slide off the seat if not wearing a seat belt?
m
𝑣 = 20
s
∆𝑥 = 50m
0
𝑉𝑓2 = 𝑉𝑖2 + 2𝑎∆𝑥
𝑎=
2
−𝑉𝑖
2∆𝑥
𝐹𝑓 = 𝜇𝑠 𝑁 = 𝜇𝑠 𝑚𝑔
𝑚𝑎 ≷ 𝜇𝑠 𝑚𝑔
Slide 5-33
Cross-Section Area
Slide 5-35
Terminal Speed
A falling object speeds up
until reaching terminal speed,
then falls at that speed
without further change.
If two objects have the same
size and shape, the more
massive object has a larger
terminal speed.
At terminal speed, the net
force is zero and the object
falls at constant speed with
zero acceleration.
Slide 5-36
Summer 2014 Beginning of lecture
Applying Newton’s Third Law: Interacting Objects
Slide 5-37
Example Problem
Block A has a mass of 1 kg; block B’s mass is 4 kg. They are
pushed with a force of magnitude 10 N across a frictionless
surface.
a. What is the acceleration of the blocks?
b. With what force does A push on B? B push on A?
Slide 5-39
Block A has a mass of 1 kg; block B’s mass is 4 kg. They are
pushed with a force of magnitude 10 N across a frictionless
surface.
a. What is the acceleration of the blocks?
b. With what force does A push on B? B push on A?
𝐹𝑥 = 10N = 1kg + 4kg 𝑎𝑥
10N
m
𝑎𝑥 =
=2 2
5kg
s
10N
Slide 5-39
Block A has a mass of 1 kg; block B’s mass is 4 kg. They are
pushed with a force of magnitude 10 N across a frictionless
surface.
a. What is the acceleration of the blocks?
b. With what force does B push on A? A push on B?
Force from block A
on hand
Force from hand on
block A
Slide 5-39
Block A has a mass of 1 kg; block B’s mass is 4 kg. They are
pushed with a force of magnitude 10 N across a frictionless
surface.
a. What is the acceleration of the blocks?
b. With what force does A push on B? B push on A?
𝐹𝑥𝐴 = 𝐹ℎ𝐴 + 𝐹𝐵𝐴 = 1kg
m
2 2
s
10N + 𝐹𝐵𝐴 = 2N
Force from hand on
block A
𝐹𝐵𝐴 = −8N
Force from block B
on block A
Slide 5-39
Block A has a mass of 1 kg; block B’s mass is 4 kg. They are pushed with
a force of magnitude 10 N across a frictionless surface.
a. What is the acceleration of the blocks?
b. With what force does A push on B? B push on A?
𝐹𝑥𝐵 = 𝐹𝐴𝐵 = 𝑚𝐵 𝑎
From Newton’s 3rd law
𝐹𝐴𝐵 = −𝐹𝐵𝐴
−𝐹𝐵𝐴 = 𝑚𝐵 𝑎
− −8N = 𝑚𝐵 𝑎
8N = 𝑚𝐵 𝑎 = 4kg
m
2 2
s
Force from block B
on block A
Force from block A
on block B
Slide 5-39
Which pair of forces is a “third law pair”?
The string tension and the friction force acting on A
The normal force on A due to B and the weight of A
The normal force on A due to B and the weight of B
The friction force acting on A and the friction force acting on B
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B.
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Example Problem
What is the acceleration of block B?
Slide 5-42
Sum the forces on block B?
𝐹𝑥𝐵 = 20N + 𝐹𝐴𝐵 = 𝑚𝐵 𝑎𝐵
𝐹𝐴𝐵 = 𝐹𝑓 = 𝜇𝑘 𝑁𝐵𝐴 = 𝜇𝑘 𝑚𝐴 𝑔
20N + 𝜇𝑘 𝑚𝐴 𝑔
m
𝑎𝐵 =
= 5.764 2
𝑚𝐵
s
Slide 5-42
Ropes and Pulleys
All ropes are assumed massless with
uniform tension
All pulleys are assumed frictionless
𝑇
𝑇
Example Problem
Block A, with mass 4.0 kg, sits on a frictionless table. Block B, with
mass 2.0 kg, hangs from a rope connected through a pulley to
block A. What is the acceleration of block A?
Slide 5-44
Block A, with mass 4.0 kg, sits on a frictionless table. Block B, with
mass 2.0 kg, hangs from a rope connected through a pulley to
block A. What is the acceleration of block A?
𝑁
Same string, same tension??
𝑇𝐴
𝐹𝑔𝐴
𝐹𝑓 = 𝜇𝑘 𝑁 = 0 ∙ 𝑁
𝑇𝐵
𝐹𝑔𝐵
Slide 5-44
Block A, with mass 4.0 kg, sits on a frictionless table. Block B, with mass 2.0 kg,
hangs from a rope connected through a pulley to block A. What is the acceleration
of block A? SUM THE FORCES on each mass separately
𝑥
𝐹𝑥𝐴 = 𝑇𝐴 = 𝑚𝐴 𝑎𝐴
𝐹𝑦𝐵 = 𝑇𝐵 + 𝐹𝑔𝐵 = 𝑚𝐵 𝑎𝐵
Note that the motions of these two masses are coupled
𝑦
𝑇𝐴
𝑇𝐵
𝐹𝑔𝐵
𝑥
𝑇𝐴
𝐹𝑔𝐴
𝐹𝑔𝐴
𝑇𝐵
𝐹𝑔𝐵
𝑦
𝑥
Slide 5-44
Block A, with mass 4.0 kg, sits on a frictionless table. Block B, with
mass 2.0 kg, hangs from a rope connected through a pulley to
block A. What is the acceleration of block A?
𝑇𝐴 = 𝑚𝐴 𝑎𝐴
𝑇𝐵 + 𝐹𝑔𝐵 = 𝑚𝐵 𝑎𝐵
−𝑚𝐴 𝑎𝐴 + 𝐹𝑔𝐵 = 𝑚𝐵 𝑎𝐵
𝑇𝐴 = −𝑇𝐵
Newton’s Third
But… 𝑎𝐴 = 𝑎𝐵
𝑦
𝑥
So just call them both 𝑎
−𝑚𝐴 𝑎 + 𝐹𝑔𝐵 = 𝑚𝐵 𝑎
Now solve for 𝑎
𝑇
𝐹𝑔𝐴
𝑇
𝐹𝑔𝐵 𝑥
𝑦
𝐹𝑔𝐵
2.0kg ∙ 𝑔 1
𝑎=
=
= 𝑔
𝑚𝐴 + 𝑚𝑏
6.0kg
3
Slide 5-44
Summary
Slide 5-45
Summary
Slide 5-46