Transcript Chapter 5 Forces

Chapter 5
Forces
Force & Vectors
Forces
• Kinematics: Describes how an object
moves
• Dynamics: Describes why an object
moves
• Newton’s three laws of motion, describe
the “dynamics” of why objects move
Forces and Interaction
• Force – a “push or pull”
• Contact Force – you physically push on a wall
• Long-range Force – like magnets or gravity
• Force – a vector quantity. Has what two parts.
• SI Unit – the Newton abbreviated as “N”
• Superposition of Forces : forces combine
according to vector addition
4 Types Forces
• Gravitational forces
• The Earth pulls and holds the moon in orbit
• The moon pulls and causes tide changes
• Electromagnetic forces
• due to electric charges, both static and moving.
• Strong Nuclear Forces
• holds particles in the nucleus together. (Strongest
of the 4)
• Weak Nuclear Forces
• Radioactive decay
Newton’s Three Laws of Physics
Why
did the
The apple
was apple
at rest,
what
started it in
accelerate?
motion?
Newton’s 1st Law
Newton’s 1st Law
• An object at rest will stay at rest unless a force acts on it.
• Here, the girl is at rest until acted on by the force imposed
by the cannonball.
Newton’s 1st Law
• An object at rest will stay at rest unless a force
acts on it.
• Here, the skateboard/log combination is at rest
until acted on by the force imposed by the
rocket.
Newton’s 1st Law
• An object in motion will continue in a straight line until
acted on by some “outside” force.
• Consider a baseball that is pitched to a batter. What
force causes the ball to change direction?
Hit
Newton’s 1st Law
• An object in motion will continue in a straight line until
acted on by some “outside” force.
• Consider the space shuttle turning while in space flight.
What produces the forces?
Newton’s 1st law
• An object in motion will continue in a straight line
until acted on by some “outside” force.
• Consider the space shuttle when landing. What
produces the forces?
Multimedia
• The car and the wall
• The motorcyclist
• The truck and the ladder
Weight and Mass
Mass:
A measure of the amount of matter in an object has.
Symbol (m)
Unit (kg)
Weight:
The gravitational force exerted on a body
Symbol (W)
Unit (N Newton)
W  mg
Weight
A Motor cycle weights 2450N. What is its mass?
W  mg
m W / g
2450 N
m
9.8 sm2
m  250kg
Homework
• WS 5a 1-10
• WS 5a (b) 1-11
Newton’s 2nd Law
Newton’s 2nd Law
• Acceleration is directly proportional to the
magnitude of the net force.
• Acceleration is inversely proportional to the
mass of the object.
Newton’s 2nd Law
• Acceleration depends on both mass and
the net force
Fnet
a
m
F  ma
Acceleration depends on net force
• A force of 10N accelerates the box
Acceleration depends on net force
• A force of 20N accelerates the box twice
as fast
Acceleration depends on mass
• A force of 10N accelerates the box
Acceleration depends on mass
• A force of 10N accelerates the smaller box
faster
Free-body diagrams WS 5b #1
Draw the free-body diagram
for a book is at rest on a
table top.
Fnormal
Fgravity
Free-body diagrams WS 5b #2
A girl is suspended motionless from a bar which
hangs from the ceiling by two ropes. A free-body
diagram for this situation looks like this:
Ftension
Ftension
Fgravity
Free-body diagrams WS 5b #3
An egg is free-falling from a nest in a tree.
Neglect air resistance. A free-body diagram
for this situation looks like this:
Fgravity
A rightward force is applied to a book in order to
move it across a desk at constant velocity.
Consider frictional forces. Neglect air resistance. A
free-body diagram for this situation looks like this:
Fnormal
F friction
Fgravity
Fapp
If the net force is zero, there is
no acceleration
If the net force is not zero, there is
acceleration
Net Force
• If there is no movement
–The net force is zero
• If there is no acceleration
–The net force is zero
• If there is acceleration
–The net force is not zero.
Homework
• WS 5b (FBD)
– 3-10
Newton’s 1st Law Equilibrium Situation
• An object at rest will stay at rest unless a
force acts on it.
• An object in motion will continue in a
straight line until acted on by some
“outside” force.
Newton’s Second Law – Non-Equilibrium
  Situation
 

F  F1  F2  F3...  ma
• The result of a Non-Equilibrium situation is that the
body, will begin to move.
F1
F  F1  ma
F1
F2
F  F1  F2  ma
Newton’s Second Law – Non-Equilibrium
  Situation
 

F  F1  F2  F3...  ma
• Two 50g masses are placed 40cm on either side of a fulcrum.
What is the Net force?
• If left mass slides 20cm right, what happens?
• If the left mass slides 20cm right and grows to 100g, what type
of situation is this?
Newton’s 2nd law e.g.
• What net force is required to accelerate a
1500kg race car at 3.0m/s2?
F  ma
F  1500kg (3.0m / s )
2
F  4500N
Newton’s 2nd law e.g.
A truck with a mass of
710kg starts from rest
and travels 40m in 3.0s
What Fnet acts on the
car?
F  ma
F  6248N
1 2
d  v1t  at
2
2d
a 2
t
2(40m)
a
2
3s
a  8.8
m
s2
Newton’s 2nd law e.g.
A 873kg dragster, starting from rest, accelerates to a
speed of 26.3m/s in .59s
a) Find the acceleration
b) Find the average force on the dragster
v2  v1  at
F  ma
v2  v1
a
t
a  44.6
F  38935N
m
s2
Newton’s 2nd law WS5d #1
• An artillery shell has a mass of 55kg. The
shell leaves the 1.5m long barrel at a
velocity of 770m/s.
• What is the force the gun applies to the
shell?
– Step 1
– Step 2
Find the acceleration
Find the Net Force
Newton’s 2rd Law WS5d #1
• What is the force the gun applies to the shell?
v2  v  2ad
2
2
1
v2 2  v12
a
2d
(770 ms )2  (0 ms )2
a
2(1.5)m
a  2 x10
5 m
s2
F  ma
F  55kg (2 x10
5 m
s2
F  1.1x10 N
7
)
Newton’s 2rd Law WS 5c #5
• A car, mass 1225kg traveling at 105km/hr slows
to a stop in 53m.
• What is the size and direction of the force that
acted on the car?
– Step 1
– Step 2
– Step3
Unit conversion km/hr to m/s
Find the acceleration
Find the Force
Newton’s 2nd Law WS 5d #5
105
km
hr
1000m 1hr
(
)(
)  29.2 ms
1km 3600 s
v2  v
a
2d
2
v2  v  2ad
2
2
1
(0 )  (29.2 )
a
2(53m)
m 2
s
m 2
s
a  8.0 sm2
F  ma
F  9800N
2
1
Homework
• WS 5c
– 1-5
• WS 5d
– 1,2
Newton’s 3rd Law
Newton’s 3rd Law
• When one object exerts a force on a
second object, the second exerts a force
on the first that is equal in magnitude but
opposite in direction.
• For every action there is an equal and
opposite re-action.
• Action – Reaction force pair
Newton’s 3rd Law
Example
• A book rests on a table
• The force from the weight
of the book pushes down.
• The table provides a
supportive force up.
N  mg
• Normal force (N) is
perpendicular to the
surface
W  mg
Friction
• Friction is a force like any other force
• Friction acts on materials that are in
contact with each other
• Friction slows down motion
• Forces due to friction are always in the
opposite direction of the motion.
Friction
• There are two types of friction:
– Kinetic (Sliding) friction: The force that
opposes the motion of a moving object
– Static friction: The force that oppose the start
of motion.
Friction
•
The force of friction depends on
1) The force pushing down (weight)
2) The surface materials in contact with
each other
Friction
Ff
force of
FN
Normal
friction
force
 coeffiecient of friction
Ff   FN
Friction Example
• A horizontal force of 30N pushes a 12kg crate across
a floor at a constant velocity. Find the coefficient of
Fnormal
sliding friction.
F friction
Ff
Fapp


Ff   FN
FN
30 N

12kg (9.8 sm2 )
  .25
Fgravity
Fapp  Ff
Friction e.g.
• A 4kg block has a
coefficient of friction of
.22 .
• What is the force of
friction for the 4kg block?
• What is the Fnet?
• What is the acceleration?
Fnet  ma
a  2.8 sm2
Fapp=20N
4kg
Ff   FN
Ff  .22(4kg )(9.8 sm2 )
F f  8.6 N
Fnet  Fapp  Ff
Fnet  20 N  8.6 N  11.3N
Homework
• WS 5e
• 1-3
• p. 102
• 13-16
• Quiz Newton’s 3 laws
Newton’s laws review
• Newton’s First Law - An object remains at
rest, or in uniform motion in a straight line,
unless it is compelled to change by an externally
imposed force.
• Newton’s first law describes an Equilibrium
Situation.
• An Equilibrium Situation is one in which the
acceleration of a body is equal to zero.
Newton’s laws review
• Newton’s Second Law – If there is a non-zero
net force on a body, then it will accelerate.
• Newton’s Second Law describes a Nonequilibrium Situation.
• A Non-equilibrium Situation is one in which the
acceleration of a body is not equal to zero.
Newton’s laws review
• Newton’s Third Law - for every action force
there is an equal, but opposite, reaction force.
• Newton’s Third Law says forces must come
in pairs.
• Paired force internal to a system have a Net
Force of zero.
Quiz
Friction Example
• A horizontal force of 35N pushes on a two block system as
shown. If the coefficient of sliding friction is .25, what is the
acceleration of the system?
Ff   FN
F f  22.1N
Fnet  Fapp  Ff
Ff  .25(4kg  5kg )(9.8 sm2 )
Fnet  35 N  22.1NFfriction
12.95 N
a
9kg
a  1.44
4kg
Two block system
5kg
m
s2
Fnormal
Fapp
Fgravity
Friction e.g.
• Two blocks are in contact with each other. The 4kg
block has a coefficient of friction of .22 . The two blocks
are accelerated together at 1.2m/s2.
• What is the force of friction for the 4kg block?
• What is the Fnet?
• What is the coefficient of friction for the gold block?

N
Ff

W
Fapp=30N
4kg
5kg
Fapp
Homework
• WS 5f
– #’s1-3
Newton’s 2nd law
• Three identical blocks of 15kg are connected as shown
and have a coefficient of friction of 0.3. If the system is
pulled to the right, what is the tension at T1?
• At T2?
• At T3?
Newton’s 2nd Law e.g
• A block rests on a table and is attached to a
hanging mass suspended by a pulley.
• Does the block move?
• What might stop the block from moving?
FPull
Newton’s 2nd Law e.g
• A 2kg block rests on a table and is attached to a hanging
.7kg mass suspended by a pulley.
• The coefficient of sliding friction is .25, if the block was set
in motion, would it continue to move?
• Use the direction of motion as positive

FPull
T
N
Ff
T
.7kg

W  m1 g

W  m2 g
Newton’s 2nd Law e.g
• A 2kg block rests on a table and is attached to a
hanging .7kg mass suspended by a pulley. The
coefficient of friction between the block and table is   0.25
Fnet  T  W  Ff  T
W  .7kg (9.8 sm2 )
W  6.8N
Ff  .25(2kg )(9.8 sm2 )
Ff  4.9 N
Fnet  6.8N  4.9N
.7kg
Newton’s 2nd Law e.g
Fnet  T  Ff  W  T
a(m1  m2 )  m2 g   m1 g
Fnet  W  F f
m2 g   m1 g
a
m1  m2
mtotal a  W  Ff

FPull
T
N
Ff
T
.7kg

W  m1 g

W  m2 g
• Two unequal masses are suspended on either side of a
pulley. What will happen?
• Draw the FBD for each mass.
Ftension
Ftension
Fgravity
Fgravity
Direction of positive
motion
Fnet1  m1 g  T
Fnet 2  T  m2 g
m1a  m1 g  T
m2 a  T  m2 g
T  m1 g  m1a
T  m2 a  m2 g
• Two unequal masses are suspended on either side of a
pulley. What will happen?
Direction of positive
motion
T  m1 g  m1a
T  m2 a  m2 g
m1 g  m1 a  m2 a  m2 g
m1 g  m2 g  m2 a  m1 a
g (m1  m2 )  a(m2  m1 )
a
g (m1  m2 )
(m2  m1 )
• Two unequal masses are suspended on either side of a
pulley. What will happen?
• Draw the FBD for each mass.
Ftension
Ftension
Fgravity
Fgravity
Fnet  W1  T  T  W2
Fnet  W1  W2
mt a  m1 g  m2 g
g (m1  m2 )
a
m1  m2
Direction of positive
motion
Homework
• WS 5g 4-5
• WS 5f 1-3
A spring scale hangs in an elevator and
supports a 4.9N package.
1) What upward force does the
mass exert on the scale
when the elevator is not
moving?
Fscale  W  4.9N
2) What force is exerted when
the elevator accelerates
upward at 1.5m/s2
500g
500g
500g
A spring scale hangs in an elevator and
supports a 4.9N package.
2) What force is exerted when the
elevator accelerates upward at 1.5m/s2
F
m
a
4.9 N
m
 500 g
2
9.8m / s
a  1.5 sm2
Felevator  .5kg (1.5m / s 2 )  .75 N
Fnet  .75N  4.9 N  5.65 N
500g
500g
500g
Newton’s 2rd Law e.g.
• A bowling ball and a baseball are simultaneously
dropped from the top of a football stadium.
• Use Newton’s 2nd Law to predict which ball will
hit the ground first.
Newton’s 2rd Law e.g.
• Use Newton’s 2nd Law to predict which ball will
hit the ground first.
– Step 1
– Step 2
– Step 3
draw the FBD for both objects
Calculate the net force for both objects
Calculate the acceleration for both
Newton’s 2nd Law e.g.
• Use Newton’s 2nd Law to predict which ball will
hit the ground first.
m
W

1.0
kg
(9.8
)
m
W  5kg (9.8 )
s2
Wbowling
a
W  49N
F
a
m
5kg (9.8 sm2 )
5kg
s2
W  9.8N
F
a
m
Wbase
1.0kg (9.8 sm2 )
a
1.0kg
A 4500kg helicopter accelerates upward at 2m/s2.
What lift force is exerted?
Fnet  Flift  W
Flift
Fnet  ma
Flift  ma  W
W
Flift  4500kg (2 sm2 )  4500kg (9.8 sm2 )
Flift  53100 N
Inclined Plane Problems
• Draw the FBD for the piano on the inclined plane.
• Resolve the Normal Force (N) and the force of friction
(Ff) into their x and y components.
Inclined Plane Problems
• We can do this problem by resolving only one force.
• Rotate the FBD so that the N is in the y plane and the Ff
is in the x plane.
Inclined Plane Problems
N
W  W cos 
W  W sin 
W
W
W
Inclined Plane Problems
A 120kg create rest on an incline plane at 35o.


N
Ff
N
Ff
F  W cos 

W

W
F  W sin 
Inclined Plane Problems
A 120kg create rest on an incline plane at 35o.

N  W cos 
F  W sin 
F  120kg (9.8 sm2 )sin 35
F  674.6 N

N
Ff
F  W cos 

W
F  W sin 
F  W cos 
F  120kg (9.8 sm2 )cos35
F  963.3N
WS 5H #1
A hockey puck (m = 0.5kg) is placed on an icy incline (no
friction) shown below. The puck is then released and
allowed to slide.
•Draw the FBD
•Draw the modified FBD


N
N
Ff
24
W
F  W cos 
F  W sin 
WS 5H #1
A hockey puck (m = 0.5kg) is placed on an icy incline (no
friction) shown below. The puck is then released and
F  W sin 
allowed to slide.

N  W cos 

N
Ff
F  W cos 
24

W
F  .5kg (9.8 sm2 )sin 24
F  1.99 N
F  W cos 
F  .5kg (9.8 sm2 ) cos 24
F  4.47 N
F  W sin 
FNet  W  F f
FBD
• A boy pushing a lawn mower with 125N of force
at an angle of 25o ?
• Draw the Free Body Diagram.
Fnorm
Fnorm
Ffric
Ffric
Fapp
W=mg
W=mg
Fapp
Homework
• WS 5i 1-5
• Chapter test
4kg
Two block system
5kg
8kg
4kg
2kg
Inclined Plane Problems
F
F
FW
A car is coasting to the right and slowing
down. A free-body diagram for this situation
looks like this:
Free Body diagrams
• If the net forces are zero, the FBD is
balanced
Forces cont.
•
•
•
•
FBD’s to Fnet
Fnet to F=ma
Horizontal pulley example
Atwood’s pulley example
Home Work
• P106 # 5,7,13
Home Work
• P106 # 22,24
Homework
• P. 106 #’s 27,28
A college student rests a backpack upon his
shoulder. The pack is suspended motionless
by one strap from one shoulder. A free-body
diagram for this situation looks like this:
A skydiver is descending with a constant
velocity. Consider air resistance. A freebody diagram for this situation looks like
this:
A force is applied to the right to drag a sled
across loosely-packed snow with a rightward
acceleration. A free-boy diagram for this
situation looks like this
A football is moving upwards towards its
peak after having been booted by the
punter. A free-body diagram for this situation
looks like this:
Examples of zero net force?
• A book on a desk
• A sky diver with a parachute open
• A bike moving at a constant speed.
Free Body diagrams
• If the net forces are NOT zero, the FBD is
NOT balanced
Examples of non zero net force
• A book sliding across a desk
• A sky diver falling without a parachute
A
B
C