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Physics 2011
Lecture 4: Newton’s Laws
S.Norr
Sir Isaac Newton
• Born: 1642 Died: 1727
• Philosophiae Naturalis Principia Mathematica
(Mathematical Principles of Natural Philosophy) (1687)
In Principia Mathematica:
Law 1: An object subject to no external forces is at rest or
moves with a constant velocity if viewed from an inertial
reference frame.
Law 2: For any object, FNET = F = ma
Law 3: Forces occur in pairs: FA-B = - FB-A
(For every action there is an equal and opposite
reaction.)
Netwon’s First Law:
Law 1: An object subject to no external forces is at rest or
moves with a constant velocity if viewed from an
inertial reference frame.
An Object in motion tends to stay in motion unless
acted upon by an external force
The application of an external force results in an
Acceleration of the object
What makes a good Inertial
Reference Frame?
• Acceleration must be negligible. Is Duluth,
MN a good IRF?
• Consider the UCM equations:
Calculating the Centripetal
Acceleration at Duluth:
• Duluth is on the surface of Earth, rotating with
a period of 1 Day, with a radius of less than ½
of Earth’s Diameter:
• T = 1 Day * 24 Hrs * 3600 Sec/Hr = 86400 Sec
• R = approx. 6 x 106 meters
Duluth is a fairly good IRF
BUT….Earth orbits the sun…
• T = 1 year = 365 * 24* 3600 = 31.6 Msec
• R = 150 x 109 meters
• a = 0.006 m/s2 - So, again a very small
acceleration DULUTH is an IRF
Newton’s Second Law
• The net resultant Force (the graphical sum of all
forces) acting on a Body is equivalent to the
product of its Mass and the Acceleration it is
experiencing.
Fnet = m*a
In other words, any imbalance of forces on an
object causes the object to accelerate.
That acceleration is directly proportional to the
net force and inversely proportional to its mass.
Defining Forces:
• Forces can be a push or a pull
• The SI unit of Force is the Newton (1 kg*m/s2)
• Forces act on a Body with a magnitude at some
direction:
A VECTOR!
(hold for applause)
Combining Forces:
• Forces add and subtract just like vectors,
as one might expect:
FNET(x,y,z) = F1(x,y,z) + F2(x,y,z) = [F1x + F2x] Î + etc
Adding Force Vectors Graphically
Superposition of Forces
• Superposition states that total reaction from
multiple disturbances is the algebraic sum of
the individual reactions from each
disturbance.
• Thus:
– Any force vector can be replaced by its component
vectors, all acting at the same point
– Any number of forces acting at the same point can be
replaced by a single Resultant Force equal to the vector
sum of the individual forces.
Component Vectors
Components and 2nd Law
• Components of F = ma :
FX = maX
FY = maY
FZ = maZ
• Suppose we know m and FX , we can solve for aX
and apply the things we learned about kinematics
1 2
over the last few weeks:
x = x0 + v 0 x t + ax t
2
v x = v 0 x + ax t
Example: Pushing a Box on Ice.
• A skater is pushing a heavy box (mass m = 100 kg)
across a sheet of ice (horizontal & frictionless). He
applies a force of 50 N in the i direction. If the box
starts at rest, what is its speed v after being pushed
a distance d = 10m ?
v
F
m
a
i
d
Calculations:
• Start with F = ma.
Or a = F / m.
– Recall that v2 - v02 = 2a(x - x0 ) from Chap.2
– So v2 = 2Fd / m ; where:
F = 50 N, d = 10 m, m = 100 kg
Thus, V = +/- 3.2 m/s2 and we’ll discard the –tive solution
v
F
m
a
i
d
Review: Newton's Laws
Law 1:
rest or
an
An object subject to no external forces is at
moves with a constant velocity if viewed from
inertial reference frame.
Law 2:
For any object, FNET = ma
Where FNET = F
Law 3:
FB ,A.
Forces occur in action-reaction pairs, FA ,B = -
Where FA ,B is the force acting on object A due to its
interaction with object B and vice-versa.
The Free Body Diagram
• Newton’s 2nd Law says that for an object
F = ma.
• Key phrase here is for an object.
• So before we can apply F = ma to any
given object we isolate the forces acting
on this object:
The Free Body Diagram...
• Consider the following case
– What are the forces acting on the plank ?
P = plank
FP,W
F = floor
FW,P
W = wall
E = earth
FP,F
F
P,E
FF,P
FE,P
The Free Body Diagram...
• Consider the following case
– What are the forces acting on the plank ?
Isolate the plank from
the rest of the world.
FP,W
FW,P
FP,F
FF,P
FP,E
FE,P
The Free Body Diagram...
• The forces acting on the plank should
reveal themselves...
FP,W
FP,F
FP,E
Aside...
• In this example the plank is not moving...
– It is certainly not accelerating!
– So FNET = ma becomes FNET = 0
FP,W
FP,W + FP,F + FP,E = 0
FP,F
FP,E
– This is the basic idea behind statics, which
we will discuss in a few weeks.
Example
• Example dynamics problem:
A box of mass m = 2 kg slides on a
horizontal frictionless floor. A force Fyx = 10
F=F
i the x a
N pushes on
itx in
direction.
What is
=?
x
m
the acceleration of the box?
Example...
• Draw a picture showing all of the
forces
y
FB,F
F
x
FF,B
FB,E
FE,B
Example...
• Draw a picture showing all of the
forces.
• Isolate the forces acting on the
block.
y
FB,F
F
x
FF,B
FB,E =
mg
FE,B
Example...
• Draw a picture showing all of the
forces.
• Isolate the forces acting on the y
block.
• Draw a free body diagram.
FB,F
F
mg
x
Example...
•
•
•
•
Draw a picture showing all of the forces.
Isolate the forces acting on the block.
Draw a free body diagram.
Solve Newton’s equations for each y
component.
– FX = maX
– FB,F - mg = maY
x
FB,F
F
mg
Example...
•
FX = maX
– So aX = FX / m = (10 N)/(2 kg) = 5 m/s2.
• FB,F - mg = maY
– But aY = 0
– So FB,F = mg.
N
y
FX
x
mg
• The vertical component of the force
of the floor on the object (FB,F ) is
often called the Normal Force (N).
• Since aY = 0 , N = mg in this case.
Example Recap
N = mg
y
FX
a X = FX / m
mg
x
Normal Force
• A block of mass m rests on the floor of an
elevator that is accelerating upward. What
is the relationship between the force due
to gravity and the normal force on the
block?
(a) N > mg
(b) N = mg
(c) N < mg
a
m
Solution
All forces are acting in the y direction,
so use:
N
Ftotal = ma
a
m
N - mg = ma
N = ma + mg
therefore N > mg
mg
Tools: Ropes & Strings
• Can be used to pull from a distance.
• Tension (T) at a certain position in a rope is the
magnitude of the force acting across a crosssection of the rope at that position.
– The force you would feel if you cut the rope and
grabbed the ends.
– An action-reaction pair.
T
cut
T
T
Tools: Ropes & Strings...
• Consider a horizontal segment of rope
having mass m:
–Draw a free-body diagram (ignore gravity).
m
T1
a
T2
x
• Using Newton’s 2nd law (in x direction):
FNET = T2 - T1 = ma
Tools: Ropes & Strings...
• An ideal (massless) rope has constant
tension along the rope.
T
T
T = Tg
T=0
• If a rope has mass, the tension can vary
along the rope
– For example, a heavy rope
hanging from the ceiling...
Tools: Ropes & Strings...
• The direction of the force provided by a
rope is along the direction of the rope:
T
Since ay = 0 (box not moving),
m
T = mg
mg
Force and acceleration
• A fish is being yanked upward out of the water
using a fishing line that breaks when the tension
reaches 180 N. The string snaps when the
acceleration of the fish is observed to be is 12.2
m/s2. What is the mass of the fish?
snap !
(a) 14.8 kg
(b) 18.4 kg
a = 12.2
m/s2
(c)
m=?
8.2 kg
Solution:
T
• Draw a Free Body Diagram!!

Use Newton’s 2nd law
in the upward direction:
a = 12.2 m/s2
m=?
FTOT = ma
T - mg = ma
mg
T = ma + mg = m(g+a)
m
T
g a
m
180 N
 8.2 kg
9.8  12.2  m s
2