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Unit 2
Newton’s Laws
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• Newton’s Laws of Motion1) Law of inertia: An object will remain in its
current state of motion unless acted upon by an external force.
2) ΣF = ma3) For every action (force) there is an equal and
opposite reaction (force). (Forces always come in pairs!)
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ΣF = ma orFnet = ma
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• Types of forces1) Field
a) Gravityb) Electromagnetic
2) Contacta) Tensionb) Normalc) Frictiond) Appliede) Resistance
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• Gravity
• Fg = mg
• Also known as “weight”
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• Tension
• T = mg for a suspended mass
• T < mg for a falling mass
• T > mg for a mass accelerating upward
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• Normal
• Normal means perpendicular
• N = mg on a horizontal surface*• N = mgcosθ on a slope of angle θ*• *Applied forces that are not parallel to the
surface will affect the normal force
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• Normal
• ΣF = ma
• mg + Fsinθ – N = ma
• But a = 0
• N = mg + Fsinθ
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• Friction
• f = μN
• When two surfaces slide relative to each other it is kinetic friction μk
• When two surfaces are stationary relative to each other it is static friction μs
• μ is almost always less than 1• For a given pair of surfaces, μk < μs
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• Applied
• Any external push or pull not already covered
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• Resistance
• Usually air or water resistance
• A VERY common topic for differential equations
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• A ball falls and is subject to a resistance force FR = –bv. Develop an equation for the velocity of the ball as a function of time. At t = 0, v = 0
mabvmg
dtdvmbvmg
bvmgdv
mdt
bvmgdv
mdt
Cbvmgb
tm
ln11
Cbvmgmbt
ln
Cbvmgmbt lnln
maF
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• A ball falls and is subject to a resistance force FR = –bv. Develop an equation for the velocity of the ball as a function of time. At t = 0, v = 0
bvmgCe mbt /
Cbvmgmbt lnln
bvmgCmbt
ln
bvmgCe mbt /
0/0 bmgCe mb
mgC mbtmgemgbv /
mbteb
mgb
mgv /
mbteb
mgv /1
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• Terminal velocity
• Constant velocity of a falling object
• What would be the terminal velocity in the previous example?
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• A ball falls and is subject to a resistance force FR = –bv. Develop an equation for the velocity of the ball as a function of time. At t = 0, v = 0
mabvmg
maF
0a
0 Tbvmg
bmgvT
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• Free Body Diagrams
• They show all external forces acting on an object
• A vector represents each force.– The head points in the direction of the force– The tail begins at the point of origin of the force– The length of the vector should be representative of
the magnitude of the force– When asked to draw a free body diagram on a test,
include only forces, not components– Label appropriately, using symbols from the problem
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• Draw a free body diagram for mass m2 in the diagram
m2g
N
Tf
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• Components of forces
• Always chose a coordinate system to minimize components
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• Components of forces
F
θ
xy
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• Components of forces
Fmg
f
N
mgcosθ
mgsinθ
xx maF
xmafFmg sin
yy maF
ymaNmg cos
θ
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• Applications of Newton’s third law
• Determine the force of m2 on m1
amFF m 12
m1 m2
F a
amFm 21
ammF 21
21 mmFa
21
12 mmFmFF m
21
12 mmFmFFm
21
12 mm
FmFFm
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amFm 21
21 mmFa
21
12 mm
FmFFm
21
21 mmFmFm
21
21 mm
FmFm
21
1?
21
2
mmFmF
mmFm
121
?
2 mmmm
21
1?
21
2 1mm
mmm
m
2112 mmmm
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Tarzan’s Tension!• Determine the tension in the 12 m vine when 90.
kg Tarzan has swung to the point where he makes a 30.° angle with the vertical and is moving with a linear speed of 5.0 m/s.
mg
30°T
mgcosθ
mamgT cos
rvmmgT
2
cos
rvmmgT
2
cos
N 967120.59030cos1090
2
T
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Jane’s Curves!• On the way to Nairobi one day to pick
up a few supplies, Jane was driving around a banked curve of radius r = 150 m. Some naughty monkeys had thrown banana peels onto the road making it essentially frictionless. If the road was banked at an angle of θ = 30.°, what speed would she need to drive to maintain her position (not slip up or down)?
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Jane’s Curves!
maF
maN sinmg
N
mgθ
θ
rvmN
2
sin
rvmmg
2
sincos
2sincos vgr
sincosgrv
30sin30cos15010v
m/s 25v
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Cheetah Dances!• On day, after he found Tarzan’s stash
of home-brewed Serengeti Special, and drank a couple bottles, Cheetah put on quite the show. If he completes two revolutions (4π radians) per second, determine the velocity of the empty bottle in his hand. The bottle is located 0.80 m from the center of his body, which is the axis of rotation.
m/s .1080.04122
rtdv
m/s .1080.044 rrv
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Elephant Power!• Tarzan’s mother-in-law came for
an extended visit and when she was finally ready to leave, her Jeep was stuck. Tarzan called an elephant friend named Shep to help out. If the powerful pachyderm was pushing with a force of 2500 N when he had the vehicle moving at 20. m/s, what was his power output?
FvP
W000,05202500 P
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• Vector Math
• Vectors are added head to tail
+ =a
a
b br
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• Vector Math
• Vectors are subtracted by reversing the second and adding
– =a
ab b
r
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• Vector Math
• Vectors are multiplied two ways: 1) Dot product2) Cross product
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• Vector Math
• Fundamentally,• A dot product is
• A cross product is
cos baba
n sin baba
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• Dot product
• Work is determined with the dot product:
• Dot products are scalar
dFW
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• Dot product
• Evaluate if• and
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• Cross product
• Angular momentum is calculated with a cross product:
• The direction of the cross product is given by the right hand rule.
• Cross products are vectors
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• Evaluate if•