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, FFNET = FF = maa
Law 3: Forces occur in pairs: FFA-B = - FFB-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*aIn 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
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.
Components and 2nd Law
• Components of FF = maa : 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 over the last few weeks:
tavv
ta21
tvxx
xx0x
2xx00
+=
++=
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 ii direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10m ?
d
FF
v
m a
ii
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
d
FF
v
m a
ii
Review: Newton's Laws
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, FFNET = maa
Where FFNET = FF
Law 3: Forces occur in action-reactionaction-reaction pairs, FFA ,B = - FFB ,A.
Where FFA ,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 FF = maa.
• Key phrase here is for an objectfor an object..
• So before we can apply FF = maa 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
F = floor
W = wall
E = earth
FFW,P
FFP,W
FFP,F FFP,E
FFF,P
FFE,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.
FFW,P
FFP,W
FFP,F FFP,E
FFF,P
FFE,P
The Free Body Diagram...
• The forces acting on the plank should reveal themselves...
FFP,W
FFP,F FFP,E
Aside...
• In this example the plank is not moving...– It is certainly not accelerating!– So FFNET = maa becomes FFNET = 0
– This is the basic idea behind statics, which we will discuss in a few weeks.
FFP,W + FFP,F + FFP,E = 0
FFP,W
FFP,F FFP,E
Example
• Example dynamics problem:
A box of mass m = 2 kg slides on a horizontal frictionless floor. A force Fx = 10 N pushes on it in the xx direction. What is the acceleration of the box?
FF = Fx ii aa = ?
m
y y
x x
Example...• Draw a picture showing all of the forces.• Isolate the forces acting on the block.
FFFFB,F
FFF,BFFB,E = mgg
FFE,B
y y
x x
Example...• Draw a picture showing all of the forces.• Isolate the forces acting on the block.• Draw a free body diagram.
FFFFB,F
mgg
y y
x 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 component.
– FX = maX
– FB,F - mg = maY
FFFFB,F
mgg
y y
x x
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.
• The vertical component of the forceof the floor on the object (FB,F ) isoften called the Normal Force Normal Force (N).
• Since aY = 0 , N = mg in this case.
FX
N
mg
y y
x 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?
m
(a)(a) N > mgN > mg
(b)(b) N = mgN = mg
(c)(c) N < mgN < mga
Solution
m
N
mg
All forces are acting in the y direction, so use:
Ftotal = ma
N - mg = ma
N = ma + mg
therefore N > mg
a
Tools: Ropes & Strings• Can be used to pull from a distance.• TensionTension (T) at a certain position in a rope is the magnitude of the
force acting across a cross-section of the rope at that position.– The force you would feel if you cut the rope and grabbed the ends.– An action-reaction pair.
cut
TT
T
Tools: Ropes & Strings...
• Consider a horizontal segment of rope having mass m:–Draw a free-body diagram (ignore gravity).
• Using Newton’s 2nd law (in xx direction): FNET = T2 - T1 = ma
• So if m = 0 (i.e. the rope is light) then T1 =T2
T1 T2
m
a x x
Tools: Ropes & Strings...• An ideal (massless) rope has constant tension along the
rope.
• If a rope has mass, the tension can vary along the rope– For example, a heavy rope
hanging from the ceiling...
• We will deal mostly with ideal massless ropes.
T = Tg
T = 0
T T
Tools: Ropes & Strings...
• The direction of the force provided by a rope is along the direction of the rope:
mg
T
m
Since ay = 0 (box not moving),
T = 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?
m = ?a = 12.2 m/s2
snap ! (a) 14.8 kg
(b) 18.4 kg
(c) 8.2 kg