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Chapter 4: The Laws of Motion
Forces
 There
seem to be two kinds of forces in Nature:
Contact forces and field forces.
 A contact
force is transferred from one object to another
by physical contact.
 A field
force is transferred in space without physical
contact.
• However, according to modern physics contact forces are
field forces in disguise!
• All the fundamental forces in Nature are all field forces.
Forces
 Examples
of forces:
Forces
 Fundamental
forces
There are four know fundamental forces:
An example:
Free neutron decay
Forces
 Examples
of forces:
An example of weak interaction
-
Free neutron decay: n -> p + e- + ne
Newton’s First Law
 Galileo’s
thought experiment
• Galileo thought about an object moving on a frictionless surface.
• He posed a question: what will happen if no force is applied to the
object?
• He concluded that it’s not the nature of an object to stop, once set
in motion, but rather to continue in its original state of motion.
 Newton’s
first law of motion
An object moves with a velocity that is constant in magnitude and
direction, unless acted on by a non-zero net force.
• External forces come from the object’s environment. If an object’s
velocity is not changing in either magnitude or direction, then it’s
acceleration and the net force acting on it must both be zero.
• Internal forces originate within the object itself and cannot change
the object’s velocity (although they can change the object’s rate of
rotation).
Newton’s First Law
 Inertia
• It is much easier to throw a golf ball over a longer distance than
a bowling ball – a bowling ball resists to move more than a golf ball.
• Q: Why is that? A: Inertia.
• The tendency of an object to continue in its original state of motion
is called inertia.
 Mass
• Mass is a measure of the object’s resistance to changes in its due
to a force: more precisely inertial mass.
Newton’s First Law
 Examples
Ooops!
of inertia
Seat belt
motion of car
Newton’s Second Law
 Motion
of an object under influence of net non-zero force
• How does an object under influence of net force?
• Experiments show that, if you push an object with twice as strong
force, the object is accelerated twice as much….
 Newton’s
second law
The acceleration a of an object is directly proportional to the net
force acting on it and inversely proportional to its mass where the
constant of proportionality is one.
acceleration

a

F
m
net force


F

m
a
;  Fx  max ,  Fy  may ,  Fz  maz

inertial mass
Newton’s Second Law
 Units
of force and mass
• SI unit of force : newton (N)
• SI unit of mass : kilogram (kg)
• Newton’s second law: 1 N = 1 kg.m/s2 =0.225 lb.
Newton’s Second Law
 Examples
• Example 4.1: Airboat
m  3.50 102 kg
 

 F  Fprop  Fresist
 7.70 10 2 N
(a) Find the acceleration of
the airboat.
Fnet 7.70 102 N
ma  Fnet  a 

 2.20 m/s 2
2
m 3.50 10 kg const. force=const. accel.
(b) Starting from rest, find the time needed to reach a speed of 12.0 m/s.
v  at + v0  (2.20 m/s 2 )t  12.0 m/s  t  5.45 s
(c) After reaching the speed, the engine is turned off and drifts to a stop
over a distance of 50.0 m. Find the resistance force.
v 2  v02  2ax  0  (12 m/s) 2  2a(50.0 m)  a  1.44 m/s 2
Fresist  ma  (3.50 102 kg)(-1.44 m/s 2 )  504 N
Newton’s Second Law
 Examples
• Example 4.2: Horses and a barge
m  2.00 103 kg
F1  6.00 102 N
x-components:
F1x  F1 cos 1  5.20 10 N
30.0o
F2 x  F2 cos  2  4.24 102 N
-45.0o
2
Fx  F1x + F2 x  9.44 10 2 N
a x  Fx / m  0.472 m/s 2
y-components:
F1y  F1 sin 1  3.00 102 N
F2  6.00 102 N
acceleration:
F2 y  F2 sin 2  4.24 102 N
a  ax2 + a y2  0.476 m/s 2
Fy  F1 y + F2 y  1.24 102 N
 ay 
  tan    7.46
 ax 
a y  Fy / m  0.0620 m/s 2
1
Newton’s Second Law
 Gravitational
force
• Is the mutual attractive force between any two objects
• Is described by Newton’s law of universal gravitation:
.
Every particle in the Universe attracts every other particle with
a force that is directly proportional to the product of the masses
of the particles and inversely proportional to the square of the
distance between them.
m1m2
Fg  G 2
r
G= 6.67 x10-11 Nm2/kg2
universal gravitation constant
Newton’s Second Law
 Weight
• Is the magnitude of the gravitational force acting on an object of
mass m near Earth’s surface.
w  mg
g : accelerati on of the gravity
• SI unit : newton (N)
• Relation between g and G
M Em
w  mg  G 2
r
ME
ME
g G 2 G 2
r
RE
ME: mass of Earth
RE: radius of Earth
depends on environment
g  9.80 m/s 2 , g Sun  274 m/s 2 , g Moon  1.62 m/s 2
Newton’s Third Law
 Newton’s
3rd law
• Forces in nature always exists in pairs.
If object 1 and object 2 interact, the force F12 exerted by object 1
on object 2 is equal in magnitude but opposite in direction to the
force F21 exerted by object 2 on object 1.
action
reaction
Newton’s Third Law
 An
example of Newton’s 3rd law
• A TV on a table (TV at rest)
action
reaction


n  n '

'
Fg   Fg
normal force
gravitational force
TV at rest
Newton’s 2nd law
 

ma  0  Fg + n
Fg  mg, n  mg
Friction
 Friction and force
• Friction as a function of
applied force (see the graph)
• Static region (Object at rest)
f s  F static friction
 f s max   s n
coefficient normal force
of friction
• Kinetic region
(Object in motion)
f k  k n : kinetic friction
coefficient normal force
of friction
const.
Friction
 Coefficients of friction
Application of Newton’s Laws
 An
example of Newton’s 3rd law
• A piece of rope at rest under constant tension along it
T T'
Newton’s 2nd law
T  T '  ma  0  a  0
• A crate pulled by a man
No acceleration
Forces on
the crate
Newton’s 2nd law
max  T
may  n  mg  0
ax  T / m
Free-body diagram
Application of Newton’s Laws
 Objects
in equilibrium
• Objects that are either at rest or moving with constant velocity
are said to be in equilibrium.

a 0
Newton’s 2nd law
Net force is zero.

 F  0   Fx  0,  Fy  0 (,  Fz  0 if 3D)
Application of Newton’s Laws
 Examples
• Example 4.5: A traffic light at rest
Application of Newton’s Laws
 Examples
• Example 4.5: A traffic light at rest
F
y
 0  T3 + Fg  0
T2y
T3  Fg  1.00 102 N
F
x
T1y
 T1 cos 37.0 + T2 cos 53.0
0
cos 37.0
 1.33T1
cos 53.0
 Fy  T1 sin 37.0 + T2 sin 53.0
T2  T1
 1.00 10 2 N  0
T1 sin 37.0 + (1.33T1 ) sin 53.0
 1.00 10 2 N  0
T1  60.1 N, T2  1.33T1  79.9 N
T2x
T1x
T3y
Application of Newton’s Laws
 Examples
• Example 4.6: Sled (at rest) on a frictionless hill
   
 F  T + n + Fg  0
F
x
 T + 0  mg sin 
 T  (77.0 N) sin 30.0  0
 T  38.5 N
F
y
 0 + n  mg cos 
 n  (77.0 N) cos 30.0  0
 n  66.7 N
Application of Newton’s Laws
 Examples
• Example 4.8: The runaway car
(a) Determine the acceleration of the car.
  

ma   F  Fg + n
max   Fx  mg sin 
F
y
 mg cos  + n  0
ax  g sin   (9.80 m/s 2 ) sin 20.0
 3.35 m/s 2
(b) Time taken for the car to reach the bottom?
x  (1 / 2)a x t 2  25.0 m
 t  3.86 s
(c) velocity at the bottom?
v  at  (3.35 m/s 2 )(3.86 s)  12.9 m/s
 =20.0o
Application of Newton’s Laws
 Examples
• Example 4.9: Weighing a fish in an elevator
ma   F  T  mg
 T  ma + mg  m(a + g )
(a) Find the weight when a>0.
T  (4.08 kg)(2.00 m/s 2 + 9.80 m/s 2 )
 48.1 N
(b) Find the weight when a<0.
T  (4.08 kg)(-2.00 m/s 2 + 9.80 m/s 2 )
 31.8 N
(c) Find the weight when the
cable breaks (a=-g).
T  (4.08 kg)(-9.80 m/s 2 + 9.80 m/s 2 )  0
a=2.00 m/s2
a=-2.00 m/s2
Application of Newton’s Laws
 Examples
• Example 4.10: Atwood’s machine
m1a1  T  m1 g
T  m1a1 + m1 g
m2 a2  T  m2 g


a2  a1
m2a1  T + m2 g
(m1 + m2 )a1  m2 g  m1 g
 m2  m1 
 g
a1  
 m1 + m2 
 2m1m2 
 g
T  
 m1 + m2 
m2>m1
Application of Newton’s Laws
 Examples
• Example 4.11: A block on a ramp (static friction)
f s max   s n
F
F
(1)
x
 mg sin    s n  0 (2)
y
 n  mg cos   0 (3)
(3)  n  mg cos 
(2)( 4)  tan    s
(4)
Application of Newton’s Laws
 Examples
• Example 4.12: A sliding hockey puck (kinetic friction)
v 2  v02 + 2ax
v 2  v02
a
 1.67 m/s 2
2x
F
y
 n  Fg  n  mg  0
n  mg
ma   Fx   f k  k n
 k mg
a
 k    0.170
g
v0=20.0 m/s -> v=0
x=1.20x102 m
Application of Newton’s Laws
 Examples
• Example 4.13: Connected objects
(a) Find acceleration and tension I.
For Object 1:
F
F
x
 T  f k  m1a1 (1)
y
 n  m1 g  0
(2)
(1)  T  k m1 g  m1a1 (3)
For Object 2:
F
y
(2)-(4)
 m2 g + T  m2 a2
 m2 a1
(4)
m1=4.00 kg
m2=7.00 kg
k =0.300
m2 g  k m1 g  (m1 + m2 )a1
m2 g   k m1 g
a1 
 5.17 m/s 2 , T  32.4 N (from (1))
m1 + m2
Application of Newton’s Laws
 Examples
• Example 4.13: Connected objects
(b) Find acceleration and tension II (system approach).
For a system made of
Object 1+2:
(m1 + m2 )a  m2 g   k n
 m2 g   k m1 g
m2   k m1 g
a
m1 + m2
m1=4.00 kg
m2=7.00 kg
k =0.300
Application of Newton’s Laws
 Examples
• Example 4.14: Two blocks and a cord
What is the max. force by the string
without causing the top object to slip?
For top object 1:
ma   s n1
0  n1  mg
n1  mg  ma   s mg
 a  s g
For bottom object 2:
Ma   s mg + T
M s g  T   s mg
T  (m + M ) s g  51.5 N
m=5.00 kg
M=10.0 kg
s=0.350