Transcript 投影片 1

4. Newton's Laws
1.
2.
3.
4.
5.
6.
The Wrong Question
Newton’s 1st & 2nd laws
Forces
The Force of Gravity
Using Newton’s 2nd Law
Newton’s 3rd Law
Ans:
• Human body: contact force
• Wind & water: fluid pressure
• Gravity: action-at-a-distance
What forces govern the motion of the sailboard?
4.1. The Wrong Question
Q: Why do things move?
Aristotle (~350BC) : Because some forces are acting on them.
Galileo (~1600) :
Wrong question ( things move until stopped ).
If a ball is
released here …
… it always rises to its
starting height …
The right question:
Why do moving things change direction?
Newton: Because some forces are acting on them.
… so this ball
should roll forever.
4.2. Newton’s 1st & 2nd laws
Fnet  0  v 
Net force determines motion.
Newton’s 1st law of motion:
A body at rest or in uniform motion remains so
unless acted on by a nonzero net force.
Fnet = 0  v = const
GOT IT? 4.1.
On a horizontal tabletop is a curved barrier that exerts a force on a ball,
guiding its motion in a circular path.
After the ball leaves the barrier, which of the dashed paths shown does it follow?
Newton’s 2nd Law
Momentum:
pmv
(quantity of motion)
Newton’s 2nd law of motion:
The rate of change of momentum is equal to the net force.
dp
f
dt

For a constant mass:
d
dm
dv
m
v

v

m
 
dt
dt
dt
m
dv
f ma
dt
Mass, Inertia, & Force
Inertia: resistance to changes in motion.
1st law = law of inertia
f  mknown aknown

Loaded truck has
greater mass,
more inertia,
less acceleration.
 munknown aunknown
munknown
a
 known
mknown
aunknown
Operational
definition of mass
1 N (Newton)  force required to give a mass of 1 Kg an
acceleration of 1 m / s2.
1 N = 1 Kg m / s2.
1 Dyne = 1 g cm / s2.
Example 4.1. Accelerating Car
A 1200 kg car accelerates constantly in a straight line from rest to 20 m/s in 7.8 s.
a) What is the net force on it?
b) What is the net force on it if it rounds a bend 85 m in radius at v = 20 m/s?
f ma m
v
 20 m / s 
 1200 kg  
  3.1 k N
t
 7.8 s 
2
f ma m v
r
 1200 kg 
 20 m / s 
85 m
2
 5.6 k N
Inertial Reference Frames
Inertial Reference Frames: reference frame in which Newton’s 1st law works.
Examples of non-inertial reference frames:
• Accelerating planes.
• Car rounding a curve.
• Merry-go-round.
• Earth (rotating).
Examples of inertial reference frames:
Newton:
Distant “fixed” stars.
Einstein:
General relativity.
4.3. Forces
Fc = Fg
Examples of forces:
• Pushes & pulls.
• Car collided with truck & stopped.
• Moon circles Earth.
(action-at-a distance)
• Person sitting on chair. (contact force)
• Climber on rope. (tension force)
T = Fg
The Fundamental Forces
The fundamental forces:
• Gravity: large scale phenomena
• Electroweak force
• Electromagnetic force: everyday phenomena
1025
1036
• Weak (nuclear) force
• Strong (nuclear) force
1
1038
4.4. The Force of Gravity
m in f = m a is the inertia mass (same everywhere).
Weight = force of gravity on mass:
wmg
For a mass of 65 kg,
weight on Earth = (65 kg) (9.8 m/s2 ) = 640 N.
weight on Moon = (65 kg) (1.6 m/s2 ) = 100 N.
weight in outer-space = (65 kg) (0 m/s2 ) = 0 N.
Caution:
In daily use, weight is often measured in units of mass.
E.g., a person “weights” 65 kg.
Strictly speaking, m in w = m g is the gravitational mass.
Galileo experiment:
Einstein: mI = mG
mI = mG
(coincidence)
(exact: gravity is geometry)
Weightlessness
These astonauts feel “weightless” because
they are in freefall.
Their weight is about 93% of that on surface.
same a
4.5. Using Newton’s 2nd Law
Tactics 4.1. Free-Body Diagram
1. Identify object of interest & forces on it.
2. Object  dot.
3. Draw forces on object as vectors at dot.
Example 4.3. Elevator
A 740 kg elevator accelerates upward at 1.1 m/s2.
Find the tension force on the cable (of negligible mass).
T  Fg  m a
Ty  m g  m ay
Ty   740 kg   9.8 m / s 2  1.1 m / s 2   8.1 k N
GOT IT? 4.4.
How’s T compared to w = m g if the elevator
greater
a) moves upward starting at rest.
less
b) decelerates to stop while moving upward.
less
c) starts moving downward, accelerating from rest.
greater
equal
d) slows to stop while moving downward.
e) moves upward with constant speed.
Conceptual Example 4.1.
At the Equator
When you stand on a scale, the scale reading shows the force it pushes up to support you.
If you stand on a scale at Earth’s equator, is the reading greater or less than your weight?
Ans:
You’re in uniform circular motion so that the net force on you is centripetal.
v2
mam
R
ma
W=mg
 W  Fscale
Fscale  W  m a
Fscale
W
Parabolic Flight
4.6. Newton’s 3rd Law
Newton’s 3rd law: Action = Reaction
Forces of 3rd law pair act on different objects.
Hence, they don’t cancel each other out.
Horse-Cart dilemma
Example 4.4. Pushing Books
2 books lie on frictionless horizontal surface.
You push with force F on books of mass m1.
which in turn on book of mass m2.
What force does the 2nd book exert on the 1st ?
Acceleration of books:
Net force on 2nd book:
a
F
m1  m2
F12  m2 a 
Force exerted by 2nd book on the 1st
F21  F12  
m2
F
m1  m2
m2
F
m1  m2
GOT IT? 4.5
Is the net force on the larger block
(a) greater than 2 N , (b) equal to 2N or (c) less than 2 N ?
Normal force n : contact force acting normal to contact surface.
Not a 3rd law pair
Normal force
3rd law also applies to
non-contact forces
such as gravity.
Fnet  0
Measuring Force
Force can be measured using Newton’s 3rd law.
Ideal spring (Hooke’s law):
f sp  k x
k = spring constant
fsp = 0
fsp = fwall < 0
fsp = fwall > 0
Works only in inertial frame.
Example 4.5. Helicopter Ride
Helicopter rises vertically.
A 35 kg bag sits in it on a spring scale of k = 3.4 kN/m.
By how much the spring compresses
(a) when helicopter is at rest
(b) when it’s accelerating upward at 1.9 m/s2.
Fsp  Fg  m a
k xm g  m a
x
(a)
(b)
x
x
m a  g 
k
 35 kg   0  9.8 m / s 2 
 10 cm
3400 N / m
 35 kg  1.9 m / s 2  9.8 m / s 2 
3400 N / m
 12 cm
GOT IT? 4.6
Example 4.5:
Helicopter rises vertically.
A 35 kg bag sits in it on a spring scale of k = 3.4 kN/m.
By how much the spring compresses
(a) when helicopter is at rest
(b) when it’s accelerating upward at 1.9 m/s2.
Continued from Example 4.5:
No
(a) Would the answer to (a) change if the helicopter were
moving upward with constant speed?
No
(b) Would the answer to (b) change if the helicopter were
moving downward but still accelerating upward?