Transcript Newton’s First Law - Miss Gray's Superb Science Site

Warm-Up: Solve for F1 and F2, then
calculate the Fnet and any resulting
acceleration of this crate.
Types of Forces
A force is a push or pull acting upon an object as a
result of its interaction with another object.
Contact Forces
Frictional Force
Tension Force
Normal Force
Air Resistance Force
Applied Force
Spring Force
Non-Contact Forces
Gravitational Force
Electrical Force
Magnetic Force
Newton’s First Law
Applications of Newton’s First Law
• Blood rushes from your head to your feet while
quickly stopping when riding on a descending
elevator.
• To dislodge ketchup from the bottom of a ketchup
bottle, it is often turned upside down and thrusted
downward at high speeds and then abruptly halted.
• Headrests are placed in cars to prevent whiplash
injuries during rear-end collisions.
• While riding a skateboard (or wagon or bicycle),
you fly forward off the board when hitting a curb or
rock or other object which abruptly halts the motion
of the skateboard.
Newton’s Second Law
This verbal statement can be
expressed in equation form as follows:
a = Fnet / m
The above equation is often
rearranged to a more familiar form as
shown below. The net force is equated
to the product of the mass times the
acceleration.
Fnet = m * a
• The acceleration is directly proportional to the
net force; Do not use the value of merely "any
'ole force" in the above equation. It is the net
force which is related to acceleration.
• The net force is the vector sum of all the forces.
If all the individual forces acting upon an object
are known, then the net force can be
determined.
The Fnet = m • a equation is often used in
algebraic problem-solving. The table below
can be filled by substituting into the equation
and solving for the unknown quantity. Fill in
the following table.
1.
2.
3.
4.
5.
Net Force
(N)
10
20
20
10
Mass
(kg)
2
2
4
2
Acceleration
(m/s/s)
5
10
• Compare rows 1 and 2
• A doubling of the net force results in a
doubling of the acceleration (if mass is
held constant).
• Compare rows 2 and 4
• A halving of the net force results in a
halving of the acceleration (if mass is held
constant).
• Acceleration is directly proportional to net
force.
• Compare rows 2 and 3
• Doubling the mass results in a halving of
the acceleration (if force is held constant).
• Compare rows 4 and 5
• Halving the mass results in a doubling of
the acceleration (if force is held constant).
• Acceleration is inversely proportional to
mass.
Applications of Newton’s Second Law
• The direction of the net force is in the same direction as
the acceleration. If the direction of the acceleration is
known, then the direction of the net force is also known.
• Consider the two diagrams below for an acceleration
of a car. From the diagram, determine the direction
of the net force which is acting upon the car
• Figure 1
• Figure 2
Instructions for Round Table
• Pass ONE paper and ONE pencil/pen
around the table clockwise.
• When the paper comes to you, solve the
sample problem while explaining your
reasoning OUT LOUD to your team
– The person to your right will be your coach
– The person to your left will be your accuracy
checker
– The person at the diagonal is your encourager
Warm-Up 10/27/09
Remember, next
week is the first ever
College In Colorado
College Application
Week. It will be held
from November 1 –
November 8, 2009,
and is a week during
which Colorado high
school seniors can
apply to college for
free! Log on to
www.CollegeInColorado.org
to apply.
Sample Problems – Round Table
1. Determine the accelerations which result when a 12-N
net force is applied to a 3-kg object and then to a 6-kg
object.
2. A net force of 15 N is exerted on an encyclopedia to
cause it to accelerate at a rate of 5 m/s2. Determine
the mass of the encyclopedia.
3. Suppose that a sled is accelerating at a rate of 2
m/s2. If the net force is tripled and the mass is
doubled, then what is the new acceleration of the
sled?
4. Suppose that a sled is accelerating at a rate of 2
m/s2. If the net force is tripled and the mass is halved,
then what is the new acceleration of the sled?
Solutions
1. A 3-kg object experiences an acceleration of 4
m/s/s. A 6-kg object experiences an
acceleration of 2 m/s/s.
2. Use Fnet= m * a with Fnet = 15 N and a = 5
m/s/s. So (15 N) = (m)*(5 m/s/s) And m = 3.0
kg
3. Answer: 3 m/s/s -- The original value of 2
m/s/s must be multiplied by 3 (since a and F
are directly proportional) and divided by 2
(since a and m are inversely proportional)
4. Answer: 12 m/s/s -- The original value of 2
m/s/s must be multiplied by 3 (since a and F
are directly proportional) and divided by 1/2
(since a and m are inversely proportional)
Finding Acceleration a = Fnet / m
Three major equations
• net force: Fnet = m*a
• gravitational force: Fgrav = m*g
• frictional force: Ffrict = μ*Fnorm
To solve for acceleration, the mass and the
net force must be known
Sample Problem 5: Individual Forces
• A rightward force is applied to a 6-kg object to move it
across a rough surface at constant velocity. The object
encounters 15 N of frictional force. Use the diagram to
determine the gravitational force, normal force, net
force, and applied force. (Neglect air resistance.)
Solution to Sample 5
Fnet = 0 N;
Fnorm = 58.8 N;
Fgrav = 58.8 N;
Fapp = 15 N
• When the velocity is constant, a = 0 m/s/s and
Fnet = 0 N
• Since the mass is known, Fgrav can be found:
Fgrav = m • g = 6 kg • 9.8 m/s/s = 58.8 N
• Since there is no vertical acceleration, the
normal force equals the gravity force.
• Since there is no horizontal acceleration, Ffrict =
Fapp = 15 N
Sample Problem 6
• A 5-kg object is sliding to the right and encountering
a friction force which slows it down. The coefficient of
friction ("mu") between the object and the surface is
0.1. Determine the force of gravity, the normal force,
the force of friction, the net force, and the
acceleration. (Neglect air resistance.)
• START BY SKETCHING A FREE BODY DIAGRAM
TO REPRESENT THE PROBLEM.
Solution to Sample 6
Fgrav = 49 N
Fnorm = 49 N
Ffrict = 4.9 N
Fnet = 5 N, left
a = 0.98 m/s/s, left
Fgrav = m • g = (5 kg) • (9.8 m/s/s) = 49 N. Since there
is no vertical acceleration, the normal force equals the
gravity force.
Ffrict can be found using the equation Ffrict ="mu"• Fnorm.
The Fnet is the vector sum of all the forces: 49 N, up
plus 49 N, down equals 0 N. And 4.9 N, left remains
unbalanced; it is the net force.
Finally, a = Fnet / m = (4.9 N) / (5 kg) = 0.98 m/s/s.
Finding Individual Forces
Fnet = m * a
• If the numerical value for the net force and
the direction of the net force is known,
then the value of all individual forces can
be determined.
• Remember: The net force is the sum of all
the forces acting on an object
Sample Problem 7
Free-body diagrams for four situations are
shown below. The net force is known for each
situation. However, the magnitudes of a few of
the individual forces are not known. Analyze
each situation individually and determine the
magnitude of the unknown forces.
Solution to Sample Problem 7
A = 50 N (the horizontal forces must be balanced)
B = 200 N (the vertical forces must be balanced)
C = 1100 N (in order to have a net force of 200 N, up)
D = 20 N (in order to have a net force of 60 N, left)
E = 300 N (the vertical forces must be balanced)
F = H = any number you wish (as long as F equals H)
G = 50 N (in order to have a net force of 30 N, right)
Inclined Planes: a tilted surface
• An object placed on a tilted surface will
often slide down the surface.
• The rate at which the object slides down
the surface is dependent upon how tilted
the surface is; the greater the tilt of the
surface, the faster the rate at which the
object will slide down it.
• Objects are known to accelerate down
inclined planes because of an unbalanced
force
Diagram of the Inclined Plane
• at least two forces acting upon any
object on an inclined plane
– force of gravity acts downward
– normal force acts in a direction perpendicular
to the surface
• Usually, any force directed at an angle to the
horizontal is resolved into horizontal and
vertical components BUT NOT IN THIS CASE!
Resolve the weight vector (Fgrav) into two perpendicular
components, one directed parallel to the inclined
surface and the other directed perpendicular to the
inclined surface.
The equations for the parallel and perpendicular
components are:
The perpendicular component of the force of
gravity balances the normal force
The parallel component of the force of gravity is
not balanced by any other force.
The parallel component of the
force of gravity is the net force
which causes this acceleration.
• In the absence of friction and other forces
(tension, applied, etc.)…
acceleration of an object on an incline is the
value of the parallel component
a= gsin
(in the absence of friction and other forces)
In the presence of friction or other forces
(applied force, tensional forces, etc…
• As in all net force problems, the net force
is the vector sum of all the forces
TNS – Tilted Neck Syndrome
• All inclined plane problems can be simplified
through a useful trick known as "tilting the
head."
• tilt your head in the same direction that the
incline was tilted. Or better yet, merely tilt
the page of paper
Example 1: The free-body diagram
shows the forces acting upon a
100-kg crate which is sliding down
an inclined plane. The plane is
inclined at an angle of 30 degrees.
The coefficient of friction between
the crate and the incline is 0.3.
Determine the net force and acceleration of the crate.
1. Find the force of gravity acting upon the crate and the
components of this force parallel and perpendicular to
the incline.
2. Now the normal force can be determined.
3. The force of friction can be determined from the value of
the normal force and the coefficient of friction
4. The net force is the vector sum of all the forces.
5. The acceleration is 2.35 m/s/s (F /m = 235 N/100 kg).
Solution to Example 1
• The force of gravity is 980 N and the
components of this force are Fparallel = 490
N (980 N • sin 30 degrees) and Fperpendicular
= 849 N (980 N • cos30 degrees).
• Now the normal force can be determined
to be 849 N (it must balance the
perpendicular component of the weight
vector).
• ; Ffrict is 255 N (Ffrict = "mu"*Fnorm= 0.3 •
849 N).
Example 2:
• The two diagrams below depict the free-body
diagram for a 1000-kg roller coaster on the first
drop of two different roller coaster rides. Use the
above principles of vector resolution to determine
the net force and acceleration of the roller coaster
cars. Assume a negligible affect of friction and air
resistance. When done, click the button to view
the answers.
Solution to Example 2
Fgrav = m • g = (1000 kg) • (9.8 m/s/s) = 9800 N
The parallel and perpendicular components of the gravity
force can be determined from their respective equations:
Fparallel = m • g • sin (45 degrees) = 6930 N Fperpendicular = m •
g • cos (45 degrees) = 6930 N
Fnorm = Fperpendicular
Fnorm = 6930 N
Fnet = 6930 N, down
the incline
a = Fnet / m
= (6930 N) / (1000 kg)
a = 6.93 m/s/s, down
the incline
Ticket out the Door
1. Edwardo applies a 4.25-N rightward force to a 0.765-kg
book to accelerate it across a table top. The coefficient
of friction between the book and the tabletop is 0.410.
Determine the acceleration of the book.
2. In a physics lab, Kate and Rob use a hanging mass
and pulley system to exert a 2.45 N rightward force on
a 0.500-kg cart to accelerate it across a low-friction
track. If the total resistance force to the motion of the
cart is 0.72 N, then what is the cart's acceleration?
3. A rightward force is applied to a 6-kg object to move it
across a rough surface at constant velocity. The object
encounters 15 N of frictional force. Use a free body
diagram to determine the gravitational force, normal
force, net force, and applied force. (Neglect air
resistance.)
Ticket out the Door, cont.
4. A rightward force is applied to a 10-kg object to move
it across a rough surface at constant velocity. The
coefficient of friction between the object and the
surface is 0.2. Use the diagram to determine the
gravitational force, normal force, applied force,
frictional force, and net force. (Neglect air resistance.)
Warm-Up 11/02/09
What is the maximum weight of a truck
that can park on a hillside whose slope is
18.5o without beginning to slide. (use the
coefficient of static friction for rubber on
dry concrete – p. 124)
FN
Fgy =Fg cos
Fgx
Fgy
Y axis

X axis
Fg (weight)
Fgx =Fg sin
Ticket out the Door, cont.
5. Lee Mealone is sledding with his friends when he
becomes disgruntled by one of his friends
comments. He exerts a rightward force of 9.13 N
on his 4.68-kg sled to accelerate it across the
snow. If the acceleration of the sled is 0.815
m/s/s, then what is the coefficient of friction
between the sled and the snow
6. In a Physics lab, Ernesto and Amanda apply a
34.5 N rightward force to a 4.52-kg cart to
accelerate it across a horizontal surface at a rate
of 1.28 m/s/s. Determine the friction force acting
upon the cart.
Free Fall and Air Resistance
• Why do objects which encounter air
resistance ultimately reach a terminal
velocity?
• In situations in which there is air
resistance, why do more massive objects
fall faster than less massive objects?
Falling Without Air Resistance
• Objects which are said to be undergoing
free fall, are not encountering a significant
force of air resistance; they are falling
under the sole influence of gravity. Under
such conditions, all objects will fall with the
same rate of acceleration, regardless of
their mass.
Consider the free-falling motion of a
1000-kg baby elephant and a
1-kg overgrown mouse.
The gravitational field strength is a
property of the Earth's gravitational field
and not a property of the baby elephant
nor the mouse. All objects placed within
Earth's gravitational field will experience
this amount of force (9.8 N) upon every
1 kilogram of mass within the object. All
objects free fall at the same rate
regardless of their mass.
Falling With Air Resistance
• Air resistance is the result of collisions of
the object's leading surface with air
molecules. The actual amount of air
resistance encountered by the object is
dependent upon a variety of factors.
• speed of the object : Increased speeds
result in an increased amount of air
resistance.
• cross-sectional area of the object:
Increased cross-sectional areas result in
an increased amount of air resistance.
Suppose that an elephant and a feather are
dropped off a very tall building from the
same height at the same time. We will
assume the realistic situation that both
feather and elephant encounter air
resistance. Which object - the elephant or
the feather - will hit the ground first?
Sample Problems – Round Table
In the diagrams below, free-body diagrams
showing the forces acting upon an 85-kg skydiver
(equipment included) are shown. For each case,
use the diagrams to determine the net force and
acceleration of the skydiver at each instant in time.
Solutions
1. The Fnet = 833 N, down
a = (Fnet / m) = (833 N) / (85 kg)
= 9.8 m/s/s down
2. The Fnet = 483 N, down
a = (Fnet / m) = (483 N) / (85 kg)
= 5.68 m/s/s down
3. The Fnet = 133 N, down
a = (Fnet / m) = (133 N) / (85 kg)
= 1.56 m/s/s down
4. The Fnet = 0 N
Terminal Velocity
• As an object falls, it picks up speed.
• The increase in speed leads to an
increase in the amount of air resistance.
• Eventually, the force of air resistance
becomes large enough to balances the
force of gravity and the object will stop
accelerating and fall with constant velocity.
• The object is said to have reached a
terminal velocity.
Individual Practice: Bronco Billy
and Skydiving
• Study the diagrams on p. 17 in your
packet
• Solve for acceleration at each stage of the
skydive
• Answer the questions 1-8
Newton’s Third Law
• If object 1 exerts force on object 2, then
object 2 exerts an equal and opposite
force on object 1.
• The two forces are acting on two different
objects, therefore even though the two
forces are equal and opposite, they do not
necessarily cancel each other.
• Often called the law of action and reaction
4.4 Newton’s Third Law of Motion
For every force (action), there is an equal and
opposite force (reaction).
Note that the action and reaction forces act on
different objects.
This image shows how a
block exerts a downward
force on a table; the
table exerts an equal and
opposite force on the
block, called the normal
force N.
Example 4.4
• A large truck collides head-on with a small
car and causes a lot of damage to the
small car. Explain why there is more
damage to the small car than to the large
truck.
Identify at least six pairs of action-reaction
force pairs in the following diagram.
The elephant's feet push backward on the ground; the ground
pushes forward on its feet. The right end of the right rope pulls
leftward on the elephant's body; its body pulls rightward on the
right end of the right rope. The left end of the right rope pulls
rightward on the man; the man pulls leftward on the left end of
the right rope. The right end of the left rope pulls leftward on the
man; the man pulls rightward on the right end of the left rope.
The tractor pulls leftward on the right end of the left rope; the
left end of the left rope pulls rightward on the tractor.
Homework:
• P. 134-137: 44, 48, 52, 53, 55, 60, 61, 62,
63, 65, 71, 74, 78
Homework
• ConcepTest Ch. 4
• Newton’s Laws Practice Problems
• Applications of Newton’s Laws Quiz
Tuesday, 11/11/08