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					 Physics 114A - Mechanics
Lecture 10 (Walker: 5.1-4)
         Newton’s Laws
          January 28, 2013

             John G. Cramer
 Professor Emeritus, Department of Physics
            Office: B451 PAB
       cramer@phys.washington.edu
                    Announcements
 Homework Assignment #3 and #4 are due at 11:59 PM
  on Thursday, January 31 and Thursday, February 7,
  respectively.
 As of today 210/213 clickers are registered.
 Exam 1 papers may be returned on Thursday. As soon
  as scores for the two parts of Exam 1 are available,
  they will be posted on WebAssign. For the multiple
  choice part of Exam 1, the average for 209 papers
  graded was 59.43±11.99 out of 75 points. The highest
  score was 75 and the lowest was 20.
 The Exam 1 solutions are available on the web from the
  “Exam 1” link on the Physics 114A Lecture Schedule.
 January 28, 2013      Physics 114A - Lecture 10      2/29
     Lecture Schedule (Part 2)
          28-Jan-13   10   Newton's Laws              14     29   5-1 to 5-4
          29-Jan-13   11   Common Forces              11     26   5-5 to 5-7
     4    31-Jan-13   12   Free Body Diagrams          -     24         -      HW3
                                                                                       1-D Dynamics

          1-Feb-13    13   Friction                    9     27   6-1
                                                                                     We are here.
          4-Feb-13    14   Strings & Springs          12     29   6-2 to 6-4
          5-Feb-13    15   Circular Motion             5     30   6-5                 Newton's Laws
     5    7-Feb-13    16   Work & Energy              11     23   7-1 to 7-2   HW4      Tension

          8-Feb-13    17   Work & Power                7     25   7-3 to 7-4
          11-Feb-13   18   Potential Energy           10     26   8-1 to 8-2
         12-Feb-13    19   Energy Conservation I      16     18   8-3 to 8-5
     6   14-Feb-13    R2   Review & Extension          -     49         -      HW5
                                                                                       Work-energy

         15-Feb-13    E2                EXAM 2 - Chapters 5-8
         18-Feb-13    H3                President's Day Holiday
         19-Feb-13    20   Momentum & Impulse          8     23   9-1 to 9-3
     7   21-Feb-13    21   Momentum Conservation      11     24   9-4 to 9-5   HW6
                                                                                     Mom. & Collisions

         22-Feb-13    22   Collisions & CM            10     22   9-6 to 9-8



January 28, 2013                        Physics 114A - Lecture 10                                        3/29
                           Force
       Kinematics vs. dynamics: what causes
 acceleration? Answer: force.
 Force: push or pull
      Force is a vector – it has magnitude and direction




January 28, 2013       Physics 114A - Lecture 10           4/29
                        Mass
   Mass is the measure of
how hard it is to change an
object’s velocity.
   Mass can also be thought
of as a measure of the
quantity of matter in an
object or the quantity of
inertia possessed by the
object.
  One liter of water has a
mass of 1 kg.
January 28, 2013    Physics 114A - Lecture 10   5/29
                   The Law of Inertia
   You push on an object and it moves. If you
stop pushing an object, does it stop moving?
   Only if there is friction! In the absence of
any net external force, an object will keep
moving at a constant speed in a straight line,
or remain at rest.
   This is Newton’s 1st Law, and it is also
known as the Law of Inertia.


January 28, 2013        Physics 114A - Lecture 10   6/29
   Motion and Inertial Frames



    In order to change the velocity of an object
– in magnitude or in direction – a net force is
required.
    An inertial reference frame is one in which
the first law is true. Accelerating reference
frames, e.g., a rotating frame, are not inertial.
January 28, 2013   Physics 114A - Lecture 10   7/29
                              Inertia
   If no force acts on an object, an
inertial reference frame is any frame in
which there is no acceleration on an the
object.
    In (a) the plane is flying horizontally
at constant speed, and the tennis ball
does not move horizontally.
   In (b) the pilot suddenly opens the
throttle and the plane rapidly gains speed,
so that the tennis ball accelerates toward
the back of the plane.
   Inertia is the tendency of mass to
resist acceleration, so that a force must
be supplied to overcome inertia and
produce acceleration.

January 28, 2013              Physics 114A - Lecture 10   8/29
Newton’s First Law of Motion




 Newton’s 1st Law:
 In the absence of external forces, an object
 at rest remains at rest; an object in motion
 remains in motion.
January 28, 2013   Physics 114A - Lecture 10    9/29
        Calibrating Spring Force
   Two equal weights exert twice the force of
one; this can be used for calibration of a spring:




January 28, 2013   Physics 114A - Lecture 10   10/29
           Acceleration vs. Force
   Now that we have a calibrated spring, we can
do more experiments.
   Acceleration is proportional to force:
                                  F1  a1




                                   2 F1  2a1




January 28, 2013   Physics 114A - Lecture 10    11/29
            Acceleration vs. Mass
Acceleration is inversely proportional to mass:
           M  a1




               2M  1 a1
                    2




January 28, 2013           Physics 114A - Lecture 10   12/29
 Newton’s Second Law of Motion
   Combining these two observations gives
   F                                                 F
a  , and choosing appropriate units we can make a 
   m                                                 m

 Or, more familiarly, F  ma             or     F  ma

   This is the mathematical expression of
 Newton’s 2nd Law of Motion.
  Units: Mass has SI units of kg, and acceleration has
  SI units of m/s2. We define the SI unit of force as:
  1 newton = 1 N  1 kg m/s2.
January 28, 2013    Physics 114A - Lecture 10            13/29
    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  =         2
                                       1.5 kg
                        a1 (2.0 m/s )



 January 28, 2013             Physics 114A - Lecture 10                14/29
 Newton’s Second Law of Motion
    An object may have several forces acting on
it; the acceleration is due to the net force:

                                               (5-1)




January 28, 2013   Physics 114A - Lecture 10           15/29
                   Combining Forces

        Forces add vectorially.




                             n
    Fnet  F1  F2       Fi
                            i 1




January 28, 2013       Physics 114A - Lecture 10   16/29
              Clicker Question 1
        Two forces are exerted on an
     object. Which third force would
     make the net force point to the
     left?




               (a)             (b)               (c)   (d)


January 28, 2013     Physics 114A - Lecture 10         17/29
 Newton’s Second Law of Motion
Newton’s 2nd Law:
An object of a given mass
m subjected to forces F1,
F2, F3, … will undergo an
acceleration a given by:
                a = Fnet /m
where
Fnet = F1 + F2 + F3 + …
    The mass m must be
positive so that force and
acceleration are in the
same direction.

January 28, 2013     Physics 114A - Lecture 10   18/29
                   Typical Forces




January 28, 2013      Physics 114A - Lecture 10   19/29
 Newton’s Second Law of Motion
Free-body diagrams:
   A free-body diagram shows every force acting
on an object. To draw a free-body diagram:
 Sketch the forces
 Isolate the object of interest
 Choose a convenient coordinate system
 Resolve the forces into components
 Apply Newton’s second law to each coordinate
direction
January 28, 2013   Physics 114A - Lecture 10   20/29
2nd Law and Free-Body Diagrams
        Example of a free-body diagram:




January 28, 2013    Physics 114A - Lecture 10   21/29
                 Example:
       A Sliding Ice-Cream Carton
  A force exerted by a stretched rubber
  band produces an acceleration of 5.0
  m/s2 on an ice cream carton of mass 1.0
  kg. When a force exerted by an
  identical rubber band stretched the
  same amount is applied to a carton of ice
  cream of mass m2, it produces an
  acceleration of 11.0 m/s2.
(a) What is the mass of the second carton?
                                  (5.0 m/s 2 )
 F  m1a1  m2 a2 so m2  (1 kg)            2
                                                0.45 kg
                                 (11.0 m/s )
(b) What is the magnitude of the force
   applied by the rubber band?
F  m1a1 =(1 kg)(5.0 m/s 2 )  5.0 N
 January 28, 2013           Physics 114A - Lecture 10      22/29
          Example: A Space Walk
  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  v0t  1 at 2  1 at 2
               2        2
        2x 2(2.25 m)
    a 2                      0.50 m/s 2
         t       (3.0 s) 2
           ˆ
    a  axi  0.50 m/s2i  ˆ

                                 ˆ       ˆ
    F  ma  (68 kg)(0.50 m/s 2 )i  34 Ni

  January 28, 2013            Physics 114A - Lecture 10   23/29
              Clicker Question 2




January 28, 2013   Physics 114A - Lecture 10   24/29
          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?
F1  FM  FL  FC  3(80.5 N)  241.5 N
a1  F1 / m  (241.5 N) / (752 kg)  0.321 N/kg  0.321 m/s 2
F2  FM  FL  FC  80.5 N          a2  F2 / m  (80.5 N) / (752 kg)  0.107 m/s 2
  January 28, 2013                 Physics 114A - Lecture 10                             25/29
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.



January 28, 2013   Physics 114A - Lecture 10   26/29
Newton’s Third Law of Motion
Some action-reaction pairs:




January 28, 2013   Physics 114A - Lecture 10   27/29
Newton’s Third Law of Motion
     Although the forces are the same, the
  accelerations will not be unless the objects
  have the same mass.

Contact forces:
The force exerted by
one box on the other is
different depending on
which one you push.


 January 28, 2013   Physics 114A - Lecture 10    28/29
      End of Lecture 10
 Before the next lecture on Tuesday, read
Walker, Chapter 5.5-5.7.
 Homework Assignments #3 and #4 should
be submitted using the WebAssign system by
11:59 PM on Thursday, January 31 and
February 7, respectively.
 Exam 1 Solutions are available on the web
from the “Exam 1” link on the Physics 114A
Course Schedule.

				
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