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Newton's Laws of Motion
Dr. Robert MacKay
Clark College, Physics
Introduction
• Newtons 3 laws of motion
• 1. Law of inertia
• 2. Net Force = mass x acceleration
•
(F=MA)
• 3. Action Reaction
• Newton’s Universal Law of Gravity
Isaac Newton 1642-1727
Isaac Newton 1689
Knighted by Queen Anne 1705
Isaac Newton 1702
Isaac Newton 1726
Other topics
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Why do objects accelerate?
Why do objects not accelerate?
Forces in Balance (Equilibrium)
Forces out of Balance
Friction
Air resistance
Terminal Velocity
Law of inertia (1st Law)
• Every object continues in its state of rest, or
of uniform motion in a straight line, unless
it is compelled to change that state by forces
impressed upon it.
• acceleration = 0.0 unless the objected is
acted on by an unbalanced force
Law of inertia (1st Law)
• Inertia (The intrinsic tendency of an object
to resist changes in motion)
• Mass is a measure of an object’s inertia
• Mass is also a measure of the amount of an
object’s matter content. (i.e. protons,
neutrons, and electrons)
• Weight is the force upon an object due to
gravity
Newton’s 2nd Law
• Net Force = Mass x Acceleration
•
F=MA
Newton’s Law of Action
Reaction (3rd Law)
• You can not touch without being touched
For every action force there is
and equal and oppositely directed reaction force
Newton’s 2nd Law
• Net Force = Mass x Acceleration
• F=MA
M=2.0 kg
A= 6.0 m/s2
F=?
F=M A
= 2.0 kg x 6.0 m/s2
=12.0 Newtons
= 12.0 N
An object experiences a net force and exhibits an acceleration
in response. Which of the following statements is always true?
(a) The object moves in the direction of the force.
(b) The acceleration is in the same direction as the velocity.
(c) The acceleration is in the same direction as the force.
(d) The velocity of the object increases.
Newton’s 2nd Law
• Net Force = Mass x Acceleration
• F=MA
M=2.0 kg
A= ?
F=6.0 N
A=F / M
= ? m/s2
Newton’s 2nd Law
• Net Force = Mass x Acceleration
• F=MA
M=?
A= 20.0 m/s2
F= 10.0 N
M= F/A
= ? kg
Newton’s 2nd Law
• Net Force = Mass x Acceleration
• F=MA
M= 8.0 kg
A= 10.0 m/s2
F= ? N
F=M A
= ? N
Newton’s 2nd Law
• Net Force = Mass x Acceleration
• F=MA
A= 10.0 m/s2
M= 8.0 kg
F= ? N
F= weight
F=M A
= ? N
Weight
m= 6.0 kg
• W=mg
• g = 10 m/s2
weight is the force due to the
gravitational attraction between
a body and its planet
W= ?
Question 1: A force of 45 N pulls horizontally on a 15 kg
crate resting on a level frictionless surface.
(Actually the crate
has real good tiny wheels) What is the acceleration
of the crate?
1.
3.0 m/s/s
2.
30.0 m/s/s
3.
60.0 m/s/s
4.
0. 33 m/s/s
Question 2: A 15.0 kg crate is in contact with a 30.0 kg
crate on a level frictionless surface as shown. If the 15.0 kg
mass is pushed with a force of 45.0 N what is the acceleration
of the two masses?
A. 1.0 m/s/s
B. 1.5 m/s/s
C. 2.0 m/s/s
D. 3.0 m/s/s
Question 3: A 15.0 kg crate is in contact with a 30.0 kg
crate on a level frictionless surface as shown. If the 15.0 kg
mass is pushed with a force of 45.0 N what is the force that
the 15.0 kg mass exerts on the 30.0 kg mass?
15 N 6
20 N 2
25 N 1
30 N 12
28%
9%
4%
57%
Weight
• W=mg
8.0 kg
W= ?
A baseball of mass m is thrown upward with some
initial speed. A gravitational force is exerted on the
ball
(a) at all points in its motion
(b) at all points in its motion except at the highest
point
(c) at no points in its motion
Newton’s 2nd Law
• Net Force = Mass x Acceleration
• F=MA
M= 5.0 kg
F= 150.0 N
D =120.0 N
A= ? m/s2
Net Force = ?
Newton’s 2nd Law
• Net Force = Mass x Acceleration
• F=MA
F= 150.0 N
M= 5.0 kg
D=?
Net
Force
=
?
A= 20.0 m/s2
Newton’s 2nd Law
• Net Force = Mass x Acceleration
• F=MA
F= 150.0 N
M= 5.0 kg
D=?
Net
Force
=
?
A= 0.0 m/s2
Which of the following statements is most correct?
(a) It is possible for an object to have motion in the
absence of forces on the object.
(b) It is possible to have forces on an object in the
absence of motion of the object.
(c) Neither (a) nor (b) is correct.
(d) Both (a) and (b) are correct.
An object experiences no acceleration.
Which of the following cannot be true for the object?
(a) A single force acts on the object.
(b) No forces act on the object.
(c) Forces act on the object, but the forces cancel.
QUICK QUIZ 5.3
(end of section 5.5)
On Earth, where gravity is present, an experiment is performed on a puck on an air hockey
table, with negligible friction. A constant horizontal force is applied to the puck and its
acceleration is measured. The experiment is performed on the same puck in the far reaches of
outer space where both friction and gravity are negligible. The same constant force is applied
to the puck and its acceleration is measured. The puck’s acceleration in outer space will be a)
greater than its acceleration on Earth, b) less than its acceleration on Earth, c) exactly the
same as its acceleration on Earth, d) infinite since neither friction nor gravity are holding it
back?
Given :
q1=60 degrees
q2=45 degrees
T3=200 N (45 lbs)
Find: T1 and T2
X
Y
T1
-T1cos(q1)
T1sin(q1)
T2
+T2cos(q2)
T2sin(q2)
T3
0
R
0
-200N
0
30 kg
Find:
A
&T
T
20 kg
F = 200 N
30 kg
T
20 kg
Find:
a
&
T
30 kg
Friction=80 N
(mk= 0. 27)
Find:
a
&
T
T
20 kg
Newton’s 2nd Law
• Friction depends
Friction
?
• on surfaces in contact (roughness)
• contact force pushing surfaces together
F= 130.0 N
M= 5.0
A= 0.0 m/s2
Net Force = ?
Fnet=P-fK
P
fs,max=msN
f
fK=mKN
fs,max
fK=mKN
kinetic friction
(sliding friction)
Static friction
Applied Force=Static frictional force F=fs
F
Air ResistanceForce
• Depends on:
• velocity
• Air density
• Shape and aerodynamics of object
Terminal Velocity
• When air resistance force balances an
objects weight
Air Drag
Acceleration= 0.0
===>
Terminal velocity
w
Terminal Velocity
Acceleration = 0.0 ===>Terminal velocity
Air Drag
80 kg
w
Air Drag
10 kg
w
which has the greatest force of air resistance?
Not Terminal Velocity
Acceleration = ?
Air Drag = 240N
80 kg
W=?
which has the greatest force of air resistance?
Not Terminal Velocity
Acceleration = ?
Air Drag = 240N
80 kg
+x
W = 800 N
Not Terminal Velocity
Acceleration = ?
Air Drag = 240N
80 kg
SF = m a
+800N - 240N= 80kg a
+560N=80kg a
a=7.0 m/s2 Down
+x
W = 800 N
Which encounters the
greater force of air
resistance—
1. A falling
elephant, or
2. A falling
feather?
Which encounters the
greater force of air
resistance—
1. A falling
elephant, or
2. A falling
feather?
Newton’s 2nd Law Free Body
diagrams
A rock in Free Fall
Newton’s 2nd Law Free Body
diagrams
A rock in Free Fall
w
Free Body diagrams
A rock At the top of its parabolic trajectory
(no air resistance)
Free Body diagrams
A rock At the top of its parabolic trajectory
(no air resistance)
w
Newton’s 2nd Law Free Body
diagrams
A rock in Free Fall
w
Newton’s 2nd Law Free Body
diagrams
A rock Falling with air resistance
Newton’s 2nd Law Free Body
diagrams
A rock Falling with air resistance
R
w
Newton’s 2nd Law Free Body
diagrams
A rock on a string
Newton’s 2nd Law Free Body
diagrams
A rock on a String
T
For the FBD is
acceleration, a, up or
down ?
w
Newton’s 2nd Law Free Body
diagrams
A rock sliding without friction
Vo
Newton’s 2nd Law Free Body
diagrams
A rock sliding without friction
N
w
Newton’s 2nd Law Free Body
diagrams
A rock sliding with friction
Newton’s 2nd Law Free Body
+y
diagrams
A rock sliding with friction
N
Fk
+x
W=mg
w
A rock or car sliding with friction
 Fy
+y
 ma y
N  mg  0
N  mg  Fk  m k N  m k mg
N
Fk
+x
W=mg
w
A rock or car sliding with friction
F k  m k N  m k mg
+y
Fk
 F x  ma x
-F k  ma
N m k mg  ma
a  m k g (opposite
to v)
+x
W=mg
w
1. Sketch the problem
2. Draw a FBD with all forces
3. Label x and y axes
4. Resolve vectors into components
5. Use:
 Fy
 ma y
 Fx
 ma x
to help solve the problem
6. Think
Elevator
#50. A boy pulls on a box of mass 30 kg
with a Force of 25 N in the direction
shown.
(a) What is the acceleration of the box?
(b) What is the normal force?
(ignore friction)
#50. A boy pulls on a box of mass 30 kg
with a Force of 25 N in the direction
shown.
(a) What is the acceleration of the box?
(b) What is the normal force?
(ignore friction)
N
Fy=25sin(30)=12.5N
Fx=25cos(30)=21.7N
w
#50. A boy pulls on a box of mass 30 kg
with a Force of 25 N in the direction
shown.
(a) What is the acceleration of the box?
(b) What is the normal force?
(sliding WITH friction)
N
Fy=25sin(30)=12.5N
f=mkN
Fx=25cos(30)=21.7N
w
A car coasting without friction
A car coasting without friction
y
N
W=mg
w=mg
x
A car coasting without friction
y
N
Wx=Wsinq
Wy=Wcosq
q
wy
wx
W=mg
x
w=mg
q
A car coasting without friction
y
Wx=Wsinq
N
wy
Wy=Wcosq
W=mg
wx
x
w=mg
q
A rock or car sliding without friction
Wx=Wsinq
Wy=Wcosq
W=mg
wy
y
N
 Fx
 ma x
wx  ma
mg sin q  ma
a  gsin q
wx
x
w=mg
 Fy
 ma y
N  mg cosq  0
q
 N  mg cosq
A rock sliding with friction
A rock sliding with friction
y
Fk
N
W=mg
w=mg
x
A rock or car sliding without friction
y
Fk
wy
wx
N
Wx=Wsinq
Wy=Wcosq
q
W=mg
x
w=mg
q
A rock or car sliding without friction
y
Fk
wy
Wx=Wsinq
N
Wy=Wcosq
W=mg
wx
x
w=mg
q
A rock or car sliding without friction
Wx=Wsinq
Wy=Wcosq
Fk
wy
y
N
 Fk  m kN  mkmg cosq
wx
w=mg
 Fy
x
q
 ma y
N  mg cosq  0
 N  mg cosq
A rock or car sliding without friction
Wx=Wsinq
Fk
wy
y
N
wx
 Fx
 ma x
wx  F k  ma
mg sin q  m k mg cosq  ma
a  g(sin q  m k cosq)
w=mg
F k  mkmg cosq
x
q
1. Sketch the problem
2. Draw a FBD with all forces
3. Label x and y axes
4. Resolve vectors into components
Use:
 Fy
 ma y
 Fx
 ma x
to help solve the problem
5. Think
Atwood’s Machine
m1
m2
FBD2
Atwood’s Machine
FBD1
 Fx  ma x
 Fx  ma x
T - w2  m2a
T
m1
T
+m1g
w1 - w2  (m1  m2 )a
m2
-x
+x
w1  T  m1a
T - w2  m2a
-m2g
w1 - w2  (m1  m2 )a
m1g - m 2g  (m1  m2 )a
(m1 - m2 )g  (m1  m2 )a
(m1 - m2 )g
a
(m1  m2 )
T - w2  m2a
T  m2 (a  g)
(m1 - m2 )g
a
(m1  m2 )
(m1 - m2 )g (m1 + m2 )g 
T  m2 (a  g)  m2


(m1  m 2 ) (m1  m2 ) 

a
m 2m1
T 2
g
(m1  m2 )
1xg
If m1=m2=m
a=0
T=2m2/(2m)
T=mg
+y
Find the acceleration a
And
the tension T in the cord
connecting the two blocks
F=80 N
m2 = 6.0 kg
m1=4.0 kg
µk=0.0
+x
+y
Find the acceleration a
And
the tension T in the cord
connecting the two blocks
N2
N1
T
T
m2 = 6.0 kg
w2
F=80 N
m1=4.0 kg
µk=0.0
w1
+x
+y
Find the acceleration a
And
the tension T in the cord
connecting the two blocks
F=80 N
m2 = 6.0 kg
m1=4.0 kg
µk=0.30
+x
Find the acceleration a
And
the tension T in the cord
connecting the two blocks
N2
Fk2
N1
T
T
m2 = 6.0 kg
w2
N2=m2g
Fk2 = 0.3m2g
F=80 N
m1=4.0 kg
Fk1
µk=0.30
w1
N1=m1g
Fk1 = 0.3m1g