Transcript NewtonsLaws_1151

Newton’s Laws
Newton-1: Law of Inertia
• Newton’s First Law
• An object subject to no external forces is
at rest or moves with a constant velocity
if viewed from an inertial reference
frame.
– If no net forces act, there is no acceleration.
F net  0  a  0
Minimizing Friction
• One way to minimize friction is to float
objects on a cushion of air. You will use an
air track in lab this week.
• Any horizontal force exerted on the cart
is the net force acting on the cart.
Acceleration vs Force
• Experiments show that
Acceleration is proportional to force
F1  a1
2F1  2a1
Acceleration vs Mass
• Acceleration is inversely proportional to
mass
M  a1
2M  12 a1
Newton-2
• Combining these two observations gives
a
F
m
and by an appropriate choice of units
Or, more familiarly,
Units: Mass has SI units of kg,
and acceleration has SI units of m/s2.
The SI unit of force is:
1 newton = 1 N = 1kg m/s2.
F
a
m
Example: Accelerated Mass
m1
A net force of 3.0 N produces an acceleration of 2.0
m/s2 on an object of unknown mass.
What is the mass of the object?
F
(3.0 N)
m1  =
 1.5 kg
2
a1 (2.0 m/s )
Newton-2 (Second Law of Motion)
Fnet
a
m
Combining Forces
Forces add as vectors.
Fnet  F1  F2 
n
  Fi
i 1
Force Addition ACT
Two forces are exerted on an
object. Which third force would
make the net force point to the
left?
(a)
(b)
(c)
(d)
Free Body Reprise
Free-body diagrams:
A free-body diagram shows every force acting
on an object.
 Isolate the object of interest
Choose a convenient coordinate system
Sketch the forces
 Resolve the forces into components
 Apply Newton’s second law in each coordinate
direction
Free-body Diagram Example
Kinematics & Dynamics Combo
You are stranded in space, away from your
spaceship. Fortunately, you have a propulsion
unit that provides a constant net force F for
3.0 s. You turn it on, and after 3.0 s you have
moved 2.25 m.
If your mass is 68 kg, find F.
x  x0  v0t  12 at 2  12 at 2
2 x 2(2.25 m)
2
a 2 

0.50
m/s
t
(3.0 s)2
a  0.50 m/s 2
Fnet  ma  (68 kg)(0.50 m/s 2 )  34 N
Since there is only one force, we call that direction
positive x and only worry about magnitudes.
Free-body ACT
Example: Three Forces
Moe, Larry, and Curley push on a 752 kg boat, each exerting a 80.5 N force
parallel to the dock.
(a) What is the acceleration of the boat if they all push in the same direction?
(b) What is the acceleration if Moe pushes in the opposite direction from Larry
and Curley as shown?
Fnet1  FM  FL  FC  3(80.5 N)  241.5 N
a1  Fnet1 / m  (241.5 N) / (752 kg)  0.321 N/kg  0.321 m/s 2
Fnet 2  FM  FL  FC  80.5 N
a2  Fnet 2 / m  (80.5 N) / (752 kg)  0.107 m/s 2
Newton’s Third Law of Motion
Forces always come in pairs, acting on
different objects:
If Object 1 exerts a force F on Object 2, then
object 2 exerts a force –F on Object 1.
These forces are called action-reaction pairs.
Newton’s Third Law of Motion
Some action-reaction pairs:
Contact forces:
The force exerted by
one box on the other is
different depending on
which one you push.
Boxes 1 and 2 rest on a
12 N
frictionless surface.
What is the acceleration
in each case?
What is the force
between the boxes in
each of the cases?
3 kg
3 kg
1 kg 12 N
1 kg