PHY 1053

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PHY 1053

• Chapter 4

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Free-Body Diagram

• A diagram that shows an isolated body and shows
  all forces acting on the body.
• Example A book resting on a table

W mg
Note the Earth pulls down on the book with a
force W mg. The table pushes up on the book
with a force N, called the Normal force.
N
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Newtons First Law

• A body at rest or moving at constant velocity
  will keep its state of motion indefinitely,
  unless it is acted upon by a net force.
• Use a free-body diagram to show all forces acting
  on the body

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Newtons Second Law

• If a body of mass m is acted upon by a net force
  F, then the body will accelerate in the direction
  of F, such that the acceleration will be
• a (F/m), or
• F m a

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Newtons Third Law

• If body A exerts force F on body B, then body B
  exerts an equal and opposite force, (-F), on body
  A.
• Action-reaction forces never act on the same
  body.
• To identify an action-reaction pair, always ask
  What interaction gives rise to a force?

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Linear Momentum

• If a body of mass m moves with velocity v, then
  the linear momentum of the body is
• p m v (kg m/sec) metric (sl ft/sec) engl.
• It is a vector
• It is very useful for analyzing collisions

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Conservation of Linear Momentum

• If no external forces act on a system (system is
  isolated), then the linear momentum of the system
  remains constant.
• p(initial) p (final)

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Perfectly Inelastic Collision
Car A of 1000 kg mass traveling at 12 m/sec rear
ends car B of mass 1800 kg, which was initially
at rest. If the cars lock up at impact, find the
final speed of the wreckage.
A
B
12 m/s
V
(1000 kg)
(1800 kg)
(2800 kg)
P(initial) p (final)
(1000)(12) (2800)V
V 4.29 (m/sec)
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Linear Impulse-Momentum Principle

• Let a force F act on a body over a time interval
  ?t, so that the bodys linear momentum changes
  from p(initial) to p(final). Then
• F (?t) p (final) - p (initial)
• Useful in analyzing impulsive forces (very large
  forces that act over very short time intervals)

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Example of Linear Impulse-Momentum Principle
A 1200 kg car, initially at rest, is pushed with
a constant force of 80 N for a time of 45
seconds. Neglect friction and find the speed of
the car after 45 seconds.
The net horizontal force acting on the car is 80
N. The initial linear momentum is 0. The final
linear momentum is m v(f). Then F ?t m
v(f) (80 N)(45 sec) (1200 kg) v(f) v(f) 3
m/sec
W
F
N
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Newtons Law of Gravitation
R
F
F
m(1)
m(2)
F G (m(1) m(2))/ R²
where G 6.67 X 10(-11) (N.m²/kg²)
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Uniform Circular Motion
v
R
Period T time for 1 complete revolution
Frequency f of revolutions / time to make them
f (1 / T ) Also, T (2pR)/v
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Centripetal Acceleration and Centripetal Force
v
R
a(c) v²/R
If a body moves with speed v in a circular path
of radius R, then a centripetal acceleration,
a(c) , acts on it pointing to the center of the
circle, such that a(c) v²/R
If the mass of the body is m, then the
centripetal force acting on the body points to
the center of the circle and equals
F(c) m v²/R
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Centrifugal Force

• If a body is in a rotating reference frame, a
  centrifugal force appears to act on it, pointing
  away from the center of the circle
• It is equal in value to the centripetal force,
  but opposite in direction to it
• It is a pseudo force that is observable only in
  a rotating reference frame
• Example Motion in a Merry-Go-Round

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Coriolis Force

• Surface of Earth is not an inertial frame
• Objects moving in the Northern Hemisphere
  experience a Coriolis Force tending to push them
  toward their right.
• Objects moving in the Southern Hemisphere
  experience a Coriolis Force tending to push them
  toward their left.
• Ocean currents and hurricanes are due to the
  Coriolis Force.