12/1 do now – on a new sheet
1. A 10 kilogram block moves down a 30° incline at
constant velocity.
A. What is the magnitude of the component of the
gravity force that is parallel to the incline?
B. What is the friction force acting on the block?
2. A 10 kilogram block is pulled by a 60 N force at 60°
above the horizontal at constant velocity along the
horizontal surface.
A. What is the magnitude of the component of the 60 N
force that is parallel to the horizontal surface?
B. What is the friction force acting on the block?
assignment
• Castle learning – due tomorrow
• Due
– Force review packet – count as a quiz
– Egg drop project
• Progress report due 12/12 – make
corrections!!!
Circular Motion and Satellite Motion Chapter Outline
Lesson 1: Motion Characteristics for Circular Motion
Lesson 2: Applications of Circular Motion
Lesson 3: Universal Gravitation
Lesson 4: Satellite Motion
Chapter test is on Friday 12/12
Objective - Lesson 1: Motion
Characteristics for Circular Motion
1.
2.
3.
4.
Speed and Velocity
Acceleration
The Centripetal Force Requirement
Mathematics of Circular Motion
Uniform circular motion
• Uniform circular motion is the motion of an
object in a circle with a constant or uniform
speed. The velocity is changing because
the direction of motion is changing.
Speed is constant
v avg
d 2R
 
t
T
• R is radius of the circle
• 2πR is the circumference of the circle
• T is time to make one lap around the circle, also
known as period.
The average speed and the radius of the circle are
directly proportional.
Velocity is changing
The Direction of the Velocity Vector
The direction of the velocity
vector at every instant is in a
direction tangent to the
circle.
Acceleration
• An object moving in uniform circular motion is moving in a
circle with a uniform or constant speed. The velocity
vector is constant in magnitude but changing in direction,
which is tangent to the circle..
• Since the velocity is changing. The object is accelerating.
v v f v i
a

t
t
vi represents the initial velocity
vf represents the final velocity
t represents the time interval
Direction of the Acceleration Vector
• The velocity change is directed towards the center of the
circle.
• The acceleration is in the same direction as this velocity
change. The acceleration is directed towards the center of
the circle.
The Centripetal Force
• According to Newton's second law of motion, an
object which experiences an acceleration
requires a net force.
• The direction of the net force is in the same
direction as the acceleration. Since
acceleration is directed toward the center, the
net force must be toward the center. This net
force is referred to as the centripetal force. The
word centripetal means center seeking.
• The Centripetal Force is the NET force.
Inertia, Force and Acceleration
Centrifugal force is a fictitious force.
Centripetal force is the reason objects
moves in a circle
As a car makes
a turn in a
circle,
centripetal
force is
provided by
friction.
As a bucket of
water tied to a
string and spun in
a circle, centripetal
force is provided
by the tension
acting upon the
bucket.
As the moon
orbits the Earth
in a circle,
centripetal
force is
provided by
gravity acting
upon the moon.
Summary
• An object in uniform circular motion moves at
____________ speed. Its velocity is ___________
to the circle and its acceleration is directed toward
the ___________ of the circle. The object
experiences ____________ which is directed in
the same direction as the acceleration, toward
the _________ of the circle. This net force is
called ______________ force. The fact that the
centripetal force is directed perpendicular to the
tangential velocity means that the force can alter
the direction of the object's velocity vector without
altering its magnitude.
Check Your Understanding
• A tube is been placed upon the table and shaped into a
three-quarters circle. A golf ball is pushed into the tube
at one end at high speed. The ball rolls through the tube
and exits at the opposite end. Describe the path of the
golf ball as it exits the tube.
The ball will move along a path
which is tangent to the spiral at
the point where it exits the tube. At
that point, the ball will no longer
curve or spiral, but rather travel in
a straight line in the tangential
direction.
example
• The initial and final speed of a ball at two different points in
time is shown below. The direction of the ball is indicated by
the arrow. For each case, indicate if there is an acceleration.
Explain why or why not. Indicate the direction of the
acceleration.
No, no change
a.
in velocity
b.
yes, change in
velocity, right
c.
yes, change in
velocity, left
d.
yes, change in
velocity, left
example
• Identify the three controls on an
automobile which allow the car to be
accelerated.
The accelerator allows the car to increase
speed. The brake pedal allows the car to
decrease the speed. And the steering wheel
allows the car to change direction.
example
An object is moving in a clockwise direction around a
circle at constant speed.
1. Which vector below represents the direction of the
velocity vector when the object is located at point B on
the circle?
d
2. Which vector below represents the direction of the
acceleration vector when the object is located at point
C on the circle?
b
3. Which vector below represents the direction of the
force vector when the object is located at point A on
d
the circle?
Class work
• Packet pp. #1-8
• Packet is due Friday
12/2 do now
• The diagram shows a 5.00-kilogram block at rest on a
horizontal, frictionless table. Which diagram best
represents the force exerted on the block by the table?
Explain.
A.
B.
C.
D.
assignment
• Castle learning – due tomorrow
• Packet due Friday
• Progress report due 12/12 – make
corrections!!!
Uniform circular motion review
• What do we know about
1. Speed
2. Velocity
3. Acceleration
4. Centripetal force
5. Centrifugal force
questions
• The centripetal force acting on the space
shuttle as it orbits Earth is equal to the
shuttle’s
A.inertia
B.momentum
C.velocity
D.weight
Mathematics of Circular Motion
vavg
2   R

T
Fnet  m  a
m: mass
v: speed
R: radius
T: period
Fnet
v2
a
R
v2
 m
R
4  2  R
Fnet  m 
T2
Relationships between quantities
vavg
2   R

T
v2
ac 
R
v2
Fc  m 
R
vavg and R are direct: R double, v double
vavg and T are inverse: T double, v halves
ac and v are direct squared: v double, ac quadruple.
Fc and v are direct squared: v double, Fc quadruple.
ac and R are inverse: R double, ac halves.
Fc and R are inverse: R double, Fc halves.
Fc and m are direct: m double, Fc doubles.
ac and m are not related: m double, ac unchanged.
example
•
1.
2.
3.
4.
A car going around a curve is acted upon
by a centripetal force, F. If the speed of
the car were twice as great, the centripetal
force necessary to keep it moving in the
same path would be
F
F = m v2 / r
2F
F in directly proportional to v2
F/2
4F
2
F in increased by 2
example
• Anna Litical is practicing a centripetal force
demonstration at home. She fills a bucket with water, ties
it to a strong rope, and spins it in a circle. Anna spins the
bucket when it is half-full of water and when it is quarterfull of water. In which case is more force required to spin
the bucket in a circle?
It will require more force to accelerate a half-full bucket
of water compared to a quarter-full bucket. According to
the equation Fnet = m•v2 / R, force and mass are
directly proportional. So the greater the mass, the
greater the force.
Example
•
1.
2.
3.
4.
The diagram shows a 5.0-kilogram cart
traveling clockwise in a horizontal circle of
radius 2.0 meters at a constant speed of 4.0
meters per second. If the mass of the cart was
doubled, the magnitude of the centripetal
acceleration of the cart would be
doubled
halved
unchanged
quadrupled
ac = v2 / R
example
•
1.
2.
3.
4.
A 60.-kilogram adult and 30.-kilogram child are passengers on
a rotor ride at an amusement park. When the rotating hollow
cylinder reaches a certain constant speed, v, the floor moves
downward. Both passengers stay "pinned" against the wall of
the rotor, as shown in the diagram. The magnitude of the
frictional force between the adult and the wall of the spinning
rotor is F. What is the magnitude of the frictional force between
the child and the wall of the spinning rotor?
F
Ff = µFNorm
2F
½F
FNorm = Fnet = mv2/R
¼F
The child is half as massive as the adult, therefore FNorm of
the child is half of that of the adult and its Ff is half of that
of friction of the adult also.
example
•
1.
2.
3.
4.
Two masses, A and B, move in circular paths as
shown in the diagram. The centripetal acceleration of
mass A, compared to that of mass B, is
the same
twice as great
one-half as great
four times as great
F = m v2 / r
Class work
• Work on packet
12/3 do now
• In the diagram, as angle θ between the lawnmower
handle and the horizontal increases, the horizontal
component of F
A. decreases
B. increases
C. remains the same
assignment
• Castle learning – due tomorrow
• Packet due Friday
• Progress report due 12/12 – make
corrections!!!
Solving problems using
Equations
•
•
•
•
Identify the given and unknown.
Choose an equation
Substitute number with units
Answer with units
vavg
2   R

T
v2
ac 
R
v2
Fc  m 
R
Example
• A 900-kg car moving at 10 m/s takes a turn
around a circle with a radius of 25.0 m.
Determine the acceleration and the net force
acting upon the car.
Given: = 900 kg; v = 10.0 m/s R = 25.0 m
Find: a = ? Fnet = ?
a = v2 / R
a = (10.0 m/s)2 / (25.0 m)
a = (100 m2/s2) / (25.0 m)
a = 4 m/s2
Fnet = m • a
Fnet = (900 kg) • (4 m/s2)
Fnet = 3600 N
example
• A 95-kg halfback makes a turn on the football field. The
halfback sweeps out a path which is a portion of a circle with
a radius of 12-meters. The halfback makes a quarter of a
turn around the circle in 2.1 seconds. Determine the speed,
acceleration and net force acting upon the halfback.
Given: m = 95.0 kg; R = 12.0 m; Traveled 1/4 of the circumference in 2.1 s
Find: v = ?
a = ? Fnet = ?
a = v2 / R
a = (8.97 m/s)2/(12.0 m)
v=d/t
v = (1/4•2•π•12.0m) /(2.1s)
a = 6.71 m/s2
v = 8.97 m/s
Fnet = m*a
Fnet = (95.0 kg)*(6.71 m/s2)
Fnet = 637 N
example
• Determine the centripetal force acting upon a 40-kg child
who makes 10 revolutions around the Cliffhanger in 29.3
seconds. The radius of the barrel is 2.90 meters.
Given: m = 40 kg; R = 2.90 m; T = 2.93 s (since 10 cycles takes 29.3 s).
Find: Fnet
Step 1: find speed: v = (2 • π • R) / T = 6.22 m/s.
Step 2: find the acceleration: a = v2 / R = (6.22 m/s)2/(2.90 m) = 13.3 m/s2
Step 3: find net force: Fnet = m • a = (40 kg)(13.3 m/s/s) = 532 N.
12/4 do now
• The diagram shows a sled and rider sliding down a
snow-covered hill that makes an angle of 30.° with the
horizontal. Which vector best represents the direction of
the normal force, FN, exerted by the hill on the sled?
A
B
C
D
assignment
• Castle learning – due tomorrow
• Packet due tomorrow
• Progress report due 12/12 – make
corrections!!!
Question
• A go-cart travels around a flat, horizontal, circular track
with a radius of 25 meters. The mass of the go-cart with
the rider is 200. kilograms. The magnitude of the
maximum centripetal force exerted by the track on the
go-cart is 1200. newtons. Which change would increase
the maximum speed at which the go-cart could travel
without sliding off this track?
A. Decrease the coefficient of friction between the go-cart
and the track.
2
v2 a  v
Fc  m 
c
B. Decrease the radius of the track.
R
R
C. Increase the radius of the track.
2   R
v

D. Increase the mass of the go-cart.
avg
T
Lesson 2: Applications of
Circular Motion
1. Newton's Second law - Revisited
2. Amusement Park Physics
Newton's Second Law - Revisited
Where Fnet is the sum (the resultant) of all forces acting
on the object.
Newton's second law is used in combination of circular
motion equations to analyze a variety of physical
situations.
Note: centripetal force is the net force!
Steps in solving problems involving forces
1. Drawing Free-Body Diagrams
2. Determining the Net Force from
Knowledge of Individual Force Values
3. Determining Acceleration from Net
Force Or Determining Individual Force
Values from Knowledge of the
Acceleration
Case 1: car driven in a circle
When a car is moving in a horizontal circle on a
level surface, there is a net force (centripetal
force) acting on it, the net force is FRICTION, which
is directed toward the center of the circle.
example
• A 945-kg car makes a 180-degree turn with a speed of 10.0 m/s.
The radius of the circle through which the car is turning is 25.0 m.
Determine the force of friction and the coefficient of friction acting
upon the car.
Given: m = 945 kg; v = 10.0 m/s; R = 25.0 m
Find: Ffrict = ? μ = ?
m*v2/R
Ff = Fnet =
Ff = (945 kg)*(10m/s)2/25m
Ff = 3780 N
FNorm
Ff
Ff = μFNorm; FNorm = mg;
Ff = μmg
3780 N = μ(9270 N);
μ = 0.41
Fgrav
example
• The coefficient of friction acting upon a 945-kg car is 0.850.
The car is making a 180-degree turn around a curve with a
radius of 35.0 m. Determine the maximum speed with which
the car can make the turn.
Given: m = 945 kg; μ = 0.85; R = 35.0 m
Find: v = ? (the minimum speed would be the speed achieved with
the given friction coefficient)
Ff = Fnet
FNorm
Ff = μFN = μFN ;
Ff = (0.85)(9270N) = 7880 N
Fnet = m*v2/R
7880 N = (945 kg)(v2) / (35.0 m);
Ff
Fgrav
v = 17.1 m/s
Case 2: a swing bucket
In a swing bucket, the net force (Fc) is the
combinations of gravity and tension.
At top:
Fnet = Ftens + Fg
Fnet = mv2/R
At bottom:
Fnet = Ftens - Fg
Fnet = mv2/R
Example
• A 1.50-kg bucket of water is tied by a rope and whirled in a circle with
a radius of 1.00 m. At the top of the circular loop, the speed of the
bucket is 4.00 m/s. Determine the acceleration, the net force and the
individual force values when the bucket is at the top of the circular
loop.
Fgrav = mg = 14.7 N
a = v2 / R = 16 m/s2
Fnet = ma = 24 N, down
Fnet = Fgrav + Ftens
24 N = 14.7 N + Ften
Ftens = 24 N - 14.7 N = 9.3 N
m = 1.5 kg
a = ________ m/s/s
Fnet = _________ N
Example
• A 1.50-kg bucket of water is tied by a rope and whirled in a
circle with a radius of 1.00 m. At the bottom of the circular
loop, the speed of the bucket is 6.00 m/s. Determine the
acceleration, the net force and the individual force values
when the bucket is at the bottom of the circular loop.
Fgrav = m • g = 14.7 N
a = v2 / R = 36 m/s2
Fnet = ma = 54 N, up
Fnet = Ften - Fgrav
54 N = Ften – 14.7 N
Ften = 54 N +14.7 N = 68.7 N
m = 1.5 kg
a = ________ m/s/s
Fnet = _________ N
Case 3: Roller Coasters and
Amusement Park Physics
In a roller coaster, the centripetal force is provided by the
combination of normal force and gravity.
example
• Anna Litical is riding on The Demon at Great America. Anna
experiences a downwards acceleration of 15.6 m/s2 at the top of the
loop and an upwards acceleration of 26.3 m/s2 at the bottom of the
loop. Use Newton's second law to determine the normal force acting
upon Anna's 864 kg roller coaster car at top and at bottom of the
loop.
Given:
m = 864 kg
atop = 15.6 m/s2 , down
abottom = 26.3 m/s2 , up
Fgrav = m•g = 8476 N.
Find:
Fnorm at top and bottom
top of Loop
• Fnet = m * a
Fnet = (864 kg)∙(15.6 m/s2 ) = 13 478 N, downward
Fnet = Fnorm + Fgrav
13478 N = Fnorm + 8475 N
Fnorm = 5002 N
Bottom of Loop
• Fnet = m * a
Fnet = (864 kg) * (26.3 m/s2 ) = 22 723 N, upward
Fnet = Fnorm + (-Fgrav)
22723 N = Fnorm + (-8476 N)
Fnorm = 31199 N
Class work
• Work on packet
• lab
12/5 do now
• A 10.-kilogram box, sliding to the right across a rough horizontal
floor, accelerates at -2.0 meters per second2 due to the force of
friction.
A. Calculate the magnitude of the net force acting on the box.
B. Calculate the coefficient of kinetic friction between the box and the
floor.
assignment
• Castle learning – due tomorrow
• Packet due Monday
• Progress report due 12/12 – make
corrections!!!
Lesson 3: Universal Gravitation
1. What is Gravity?
2. The Apple, the Moon, and the Inverse
Square Law
3. Newton's Law of Universal Gravitation
4. Cavendish and the Value of G
5. The Value of g
What is Gravity?
• We know that gravity is a force and we represent it by
the symbol Fg. It causes an acceleration of all objects
around it. The acceleration is referred as the
acceleration of gravity. On and near Earth's surface,
the value for the acceleration of gravity (g) is
approximately 9.81 m/s/s. It is the same acceleration
value for all objects, regardless of their mass (and
assuming that the only significant force is gravity).
The Apple, the Moon, and the
Inverse Square Law
• It was known that the moon orbits Earth and Earth orbits
the sun. Since both the moon and Earth moves in a
circular path, they are accelerating. But what is the force
that causes these accelerations?
• According to legend, a breakthrough came to Newton at
age 24 in an apple orchard in England.
•
Newton's was able to relate the cause
for heavenly motion (the orbit of the
moon about the earth) to the cause for
Earthly motion (the falling of an apple
to the Earth). This connection led him
to his notion of universal gravitation.
• It was known that earthbound objects (such as falling
apples) accelerate towards the earth at a rate of 9.81 m/s2.
And it was also known that the moon accelerated towards
the earth at a rate of 0.00272 m/s2.
The difference in the
accelerations between the moon
and the apple is due to the
difference in distance from the
center of the earth to the two
objects.
The force of gravity
follows an inverse
square law.
Do now
• On assigned sheet: page 77 #163-166
• Homework – packet pp. #13-15
Newton’s law of Universal Gravitation
• Any two bodies in the universe attract each other with a
force that is directly proportional to the product of their
masses and inversely proportional to the square of the
distance between them.
F1 and F2 are
action/reaction forces.
G is gravitational constant. G = 6.67x10-11N∙m2/kg2
m1 is the mass of object 1
m2 is the mass of object 2
r is the distance between the objects’ centers
Gravity and distance - Inverse Square Law
• The relationship between the force of gravity (Fg)
between any two objects and the distance (d) that
separates their centers is inverse squared.
• If the separation distance is increased by a factor of 2,
then the force of gravity is decreased by a factor of four
(22). And if the separation distance is increased by a
factor of 3, then the force of gravity is decreased by a
factor of nine (32).
Fg
r
Check Your Understanding
1 . Suppose that two objects attract each other with a gravitational force of
16 units. If the distance between the two objects is doubled, what is the
new force of attraction between the two objects?
16/22 = 4 units
2. Suppose that two objects attract each other with a gravitational force of
16 units. If the distance between the two objects is tripled, then what is
the new force of attraction between the two objects?
16/32 = 1.8 units
3. Suppose that two objects attract each other with a gravitational force of
16 units. If the distance between the two objects is reduced in half, then
what is the new force of attraction between the two objects?
16 x 22 = 64 units
4. Suppose that two objects attract each other with a gravitational force of
16 units. If the distance between the two objects is reduced by a factor
of 5, then what is the new force of attraction between the two objects?
16 x 52 = 400 units
Relationship between quantities
Example
• Determine the force of gravitational attraction between
the earth (m = 5.98 x 1024 kg) and a 70-kg physics
student if the student is standing at sea level, a distance
of 6.38 x 106 m from earth's center.
Fgrav = mg = (70 kg)(9.8 m/s2) = 686 N
The Universality of Gravity
Mass of
Object 1
(kg)
Mass of Object 2
(kg)
Separation
Distance
(m)
Force of
Gravity
(N)
Student
70 kg
Earth
5.98 x1024 kg
6.60 x 106 m
(low-height orbit)
686 N
Student
70 kg
Physics Student
70 kg
1m
3.27 • 10-7 N
Student
70 kg
Physics Book
1 kg
1m
4.67 • 10-9 N
Student
70 kg
Jupiter
1.901 x 1027 kg
6.98 x 107 m
(on surface)
1822 N
Gravity is a force pulling together all matter. The more matter,
the more gravity.
Example
• The weight of an object was determined at five different
distances from the center of Earth. The results are shown in the
table below. Position A represents results for the object at the
surface of Earth. What is the approximate mass of the object?
100 kg
The Value of g
Fg  m2  g
G  mEarth  m2
Fg 
r2
G  mEarth  m2
m2  g 
r2
G  mEarth
g
r2
The acceleration of gravity is dependent upon the mass of
the earth (approx. 5.98x1024 kg) and the distance (r) from
the center of the earth.
G and g have different values!!!
g is inversely proportional to the distance
squared – inverse square law.
g is different at different location
Location
Distance from
Earth's center (m)
Value of g
m/s2
Earth's surface
6.38 x 106 m
9.8
1000 km above
surface
7.38 x 106 m
7.33
5000 km above
surface
1.14 x 107 m
3.08
10000 km
above surface
1.64 x 107 m
1.49
50000 km
above surface
5.64 x 107 m
0.13
Gravity is a field force
• Gravitational field – a region in space where
an object experiences a gravitational force.
Every mass is surrounded by a gravitational
field.
The strength of gravitational
field is inversely related to the
distance from Earth. It is
stronger near Earth and
decreases as the distance
from the Earth increases.
Gravitational field strength is a
vector quantity (g)
• At any point in a gravitational field, gravitational field strength,
g, equals the force per unit mass at that point.
g
Fg
m
or
G  mEarth
g
2
r
– g (gravitational field strength) = 9.81 N/kg (near earth)
– g (gravitational acceleration) = 9.81 m/s2
The gravitational field lines are directed toward the center
of Earth.
12/10 do now
• On assigned sheet: #171-178
• Homework questions?
• Homework: packet pp. 17, 19-20
Questions
• Earth’s mass is approximately 81 times the
mass of the Moon. If Earth exerts a
gravitational force of magnitude F on the
Moon, the magnitude of the gravitational
force of the Moon on Earth is
A.F
B.F/81
C.9F
D.81F
Lesson 4: Satellite Motion
– Circular Motion Principles for Satellites
– Mathematics of Satellite Motion
– Weightlessness in Orbit
Circular Motion Principles for Satellites
• A satellite is any object that is orbiting the earth, sun or
other massive body. Every satellite's motion is governed
by the same physics principles and described by the
same mathematical equations.
Velocity, Acceleration and Force Vectors
• The motion of an orbiting satellite can be described by
the same motion characteristics as any object in circular
motion.
– The velocity of the satellite would be directed
tangent to the circle at every point along its path.
– The acceleration of the satellite would be directed
towards the center of the circle - towards the central
body that it is orbiting.
– And this acceleration is caused by a net force that is
directed inwards in the same direction as the
acceleration. This centripetal force is supplied by
gravity - the force that universally acts at a distance
between any two objects that have mass.
Mathematics of Satellite Motion
• If the satellite moves in circular motion, then the net
centripetal force is provided by the gravitational force acting
upon this orbiting satellite.
msat  v 2
Fc 
r
G  mEarth  msat
Fg 
r2
Fc  Fg
msat  v 2 G  mEarth  msat

r
r2
GM Earth
v
r
G  M central
a
2
r
The speed and acceleration of satellite
GM central
v
r
G  M central
a
2
r
Both the speed and acceleration of satellite depends only
on the mass of the central body about which the satellite
orbits, and the distance between the satellite and the center
of the central body. It has nothing to do with mass of the
satellite.
If a base ball and an big spaceship were sent to the same
distance r from the center of Earth, both will orbit Earth with
same speed and have the same acceleration.
Check your understanding
• A satellite is orbiting the earth. Which of the
following variables will affect the speed of the
satellite?
a. mass of the satellite
b. height above the earth's surface
c. mass of the earth
Weightlessness in Orbit
• Astronauts who are orbiting the Earth often experience
sensations of weightlessness. These sensations
experienced by orbiting astronauts are the same
sensations experienced by anyone who has been
temporarily suspended above the seat on an amusement
park ride.
• You can only feel the force of gravity pulling upon your
body through contact force, such as standing on a
floor, sitting in a chair, or lying on a bed. Without the
contact force (the normal force), there is no means of
feeling the non-contact force (the force of gravity).
• When you are falling on an amusement park ride, there
is still gravity act on you even thought you can not feel
it.
Test your preconceived notions
about weightlessness:
•
a.
b.
c.
d.
Astronauts on the orbiting space station are
weightless because...
there is no gravity in space and they do not
weigh anything.
space is a vacuum and there is no gravity in a
vacuum.
space is a vacuum and there is no air
resistance in a vacuum.
the astronauts are far from Earth's surface at a
location where gravitation has a minimal affect.
Scale Readings and Weight
If you want to know your weight,
you may stand on a scale to find it
out. However, the scale does not
always measure your weight. When
you stand on a scale in an elevator
going up and down or when you
standing on a scale on an inclined
plane, the scale reading is different
than when you are at rest or you
are traveling at constant speed.
What does the scale measure?
Fnet = m*a
Fnet = 0 N
Fnorm equals
Fgrav
Fnorm = 784 N
Fnet = m*a
Fnet = 400 N, up
Fnorm > Fgrav by
400 N
Fnorm = 1184 N
Fnet = m*a
Fnet = 400 N, down
Fnorm < Fgrav by
400 N
Fnorm = 384 N
Fnet = m*a
Fnet = 784 N, down
Fnorm < Fgrav by
784 N
Fnorm = 0 N
•
Otis stands on a bathroom scale and reads the scale while ascending
and descending the John Hancock building. Otis' mass is 80 kg. Use a
free-body diagram and Newton's second law of motion to solve the
following problems.
a. What is the scale reading when Otis accelerates upward at 0.40
FN = 816 N
m/s2?
b. What is the scale reading when Otis is traveling upward at a
constant velocity of at 2.0 m/s?
FN = 784 N
c. As Otis approaches the top of the building, the elevator slows down
at a rate of 0.40 m/s2. Be cautious of the direction of the
acceleration. What does the scale read?
FN = 752 N
d. Otis stops at the top floor and then accelerates downward at a rate
of 0.40 m/s2. What does the scale read?
FN = 752 N
e. As Otis approaches the ground floor, the elevator slows down (an
upward acceleration) at a rate of 0.40 m/s2. Be cautious of the
direction of the acceleration. What does the scale read? FN = 816 N
f. Use the results of your calculations above to explain why Otis feels
less weighty when accelerating downward on the elevator and why
he feels heavy when accelerating upward on the elevator.
12/11 do now
• Packet page 76 #154-157
• Homework – castle learning
• Review sheets
• test is on castle learning
• Food drive – 10 items for 10 points – add points
to any assignment, your choice – last day to
bring food is Tuesday, 12/16/14
• 12/19 – physics shirt day
12/12 do now
• #179-180
• All castle learning assigned for this chapter are
opened for review – close tonight
• Class work - Review sheets,
• Take home quiz – due Monday
• test is on castle learning
• Project is due Thursday
• Food drive – 10 items for 10 points – add points
to any assignment, your choice – last day to bring
food is Tuesday, 12/16/14
• 12/19 – physics shirt day
Do now
• The path of a stunt car driven horizontally off a cliff is
represented in the diagram below. After leaving the cliff, the
car falls freely to point A in 0.50 second and to point B in
1.00 second. [Neglect friction.]
1. Determine the magnitude of the
horizontal component of the
velocity of the car at point B.
2. Determine the magnitude of the
vertical velocity of the car at
point A.
3. Calculate the magnitude of the
vertical displacement, dy, of the
car from point A to point B.
• A skier on waxed skis is pulled at constant speed across
level snow by a horizontal force of 39 newtons. Calculate
the normal force exerted on the skier. Round your
answer to the nearest whole number.
• A 10.-kilogram rubber block is pulled horizontally at
constant velocity across a sheet of ice. Calculate the
magnitude of the force of friction acting on the block.
Round your answer to the nearest tenth of a newton.
• A student and the waxed skis he is wearing have a
combined weight of 850 newtons. The skier travels down
a snow-covered hill and then glides to the east across a
snow-covered, horizontal surface. Calculate the
magnitude of the force of friction acting on the skis as
the skier glides across the snow-covered, horizontal
surface.
• An ice skater applies a horizontal force to a 20.-kilogram
block on frictionless, level ice, causing the block to
accelerate uniformly at 1.4 meters per second2 to the
right. After the skater stops pushing the block, it slides
onto a region of ice that is covered with a thin layer of
sand. The coefficient of kinetic friction between the block
and the sand-covered ice is 0.28. Calculate the
magnitude of the force of friction acting on the block as it
slides over the sand-covered ice.
• 747 jet, traveling at a velocity of 70. meters per second
north, touches down on a runway. The jet slows to rest
at the rate of 2.0 meters per second2. Calculate the total
distance the jet travels on the runway as it is brought to
rest.
• A 60-kilogram skydiver is falling at a constant speed
near the surface of Earth. What is the magnitude of the
force of air friction acting on the skydiver?