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Goals Equilibrium Weight The End
Physics 2514
Lecture 13
P. Gutierrez
Department of Physics & Astronomy
University of Oklahoma
P. Gutierrez (University of Oklahoma)
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Goals Equilibrium Weight The End
Goal
Goals for today’s lecture:
Application of Newton’s second law continue
Equilibrium Static and Dynamic.
Gravity and weight.
Continue with the application of friction Static and Kinetic
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Goals Equilibrium Weight The End
Clicker
An elevator suspended by a cable is moving downwards and
slowing to a stop. Which free-body diagram is correct?
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Goals Equilibrium Weight The End
Equilibrium
Newton’s Second Law.
~ 1, F
~ 2 , . . . will undergo
An object of mass m subjected to forces F
an acceleration ~a given by
~a =
~ net
F
m
~ net = Pn F
~
where F
i=1 i is the vector sum of all forces acting on
the object. The acceleration vector ~a points in the same
~ net .
direction as the force vector F
Condition for equilibrium.
X
~i = 0
F
⇒
~a = 0
i
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Goals Equilibrium Weight The End
Equilibrium
Equilibrium
P ~
i Fi = 0.
Static equilibrium system is at rest.
Dynamic equilibrium system moves with a constant
velocity; ~a = 0.
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Goals Equilibrium Weight The End
Example
A 50 kg steel box is in the back of a dump truck. The truck’s
bed, also made of steel, is slowly tilted. At what angle will the
file cabinet begin to slide?
Brief description: A 50 kg steel box is on a steel incline plane.
What is the maximum angle the incline can have for the box to
remain in static equilibrium?
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Goals Equilibrium Weight The End
Example
A 50 kg steel box is in the back of a dump truck. The truck’s
bed, also made of steel, is slowly tilted. At what angle will the
file cabinet begin to slide?
Brief description: A 50 kg steel box is on a steel incline plane.
What is the maximum angle the incline can have for the box to
remain in static equilibrium?
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February 23, 2011
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Goals Equilibrium Weight The End
Solution
Known
µs = 0.80, m = 50 kg
Unknown
angle θ of incline
Normal force n
Newton’s second law:
)
mg sin θ − µs n = 0
n − mg cos θ = 0
P. Gutierrez (University of Oklahoma)
⇒ µs = tan θ
Physics 2514
⇒ θ = tan−1 µs = 38.7◦
February 23, 2011
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Goals Equilibrium Weight The End
Solution
Known
µs = 0.80, m = 50 kg
Unknown
angle θ of incline
Normal force n
Newton’s second law:
)
mg sin θ − µs n = 0
n − mg cos θ = 0
P. Gutierrez (University of Oklahoma)
⇒ µs = tan θ
Physics 2514
⇒ θ = tan−1 µs = 38.7◦
February 23, 2011
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Goals Equilibrium Weight The End
Solution
Known
µs = 0.80, m = 50 kg
Unknown
angle θ of incline
Normal force n
Newton’s second law:
)
mg sin θ − µs n = 0
n − mg cos θ = 0
P. Gutierrez (University of Oklahoma)
⇒ µs = tan θ
Physics 2514
⇒ θ = tan−1 µs = 38.7◦
February 23, 2011
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Goals Equilibrium Weight The End
Example
Gravity and Weight
Mass is an intrinsic property of an object (objects inertia).
Gravity is a long range force acting between any two
masses. (Newton’s law of gravity.)
~ G = Gm1 m2 r̂
F
r2
Gravity near Earth’s surface (Newton’s second law.)
~G =
m~a = F
GmMe
r̂
Re2
⇒
g≡a=
GMe
= 9.8 m/s2
Re2
G = 6.67 × 10−11 N · m2 /kg2
Me = 5.98 × 1024 kg
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Re = 6.37 × 106 m
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Goals Equilibrium Weight The End
Example
Gravity and Earth
m~a =
y
⇒
GmMe
r̂
(Re + y)2
⇒
g≡a=
GMe
(Re + y)2
distance above the earth’s surface
y (km)
g (m/s2 )
0
9.830
Earth’s surface
10
9.799
Cruising altitude of a commercial aircraft
100
9.528
60 miles up
400
8.703
Height of space shuttle
400,000
0.003
Height of the moon
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Goals Equilibrium Weight The End
Example
Weight
Weight is a measurement
An objects weight results from the force exerted on a
spring scale could be a pulling or pushing contact force.
If the system is in static equilibrium (rest frame) w = mg.
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Goals Equilibrium Weight The End
Example
Weight
Weight is a measurement
An objects weight results from the force exerted on a
spring scale could be a pulling or pushing contact force.
If the system is in static equilibrium (rest frame) w = mg.
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Goals Equilibrium Weight The End
Example
Example
You are standing on a scale inside an elevator. When the
elevator is stationary, you find your weight is 680 N. What is
your weight if the elevator is moving downward with a velocity
of 10 m/s, but the accleration is upward at 3 m/s2 ?
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Goals Equilibrium Weight The End
Example
Example
You are standing on a scale inside an elevator. When the
elevator is stationary, you find your weight is 680 N. What is
your weight if the elevator is moving downward with a velocity
of 10 m/s, but the accleration is upward at 3 m/s2 ?
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February 23, 2011
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Goals Equilibrium Weight The End
Example
Solution
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Goals Equilibrium Weight The End
Example
Solution
X
i

Fyi = Fsp − mg = ma
w = N = Fsp
P. Gutierrez (University of Oklahoma)
⇒ w = ma+mg = m(a+g) = 888 N

Physics 2514
February 23, 2011
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Goals Equilibrium Weight The End
Example
Solution
X
i

Fyi = Fsp − mg = ma
w = N = Fsp
P. Gutierrez (University of Oklahoma)
⇒ w = ma+mg = m(a+g) = 888 N

Physics 2514
February 23, 2011
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Goals Equilibrium Weight The End
Example
Clicker
In the example the elevator is moving downwards at 10 m/s,
accelerating upwards at 3 m/s2 and the weight is found to be
888 N. Now assume that the elevator is moving upwards at
10 m/s and accelerating upwards at 3 m/s2 , what is the weight?
(Recall that for the stationary elevator the weight was 680 N.)
A) 888 N
B) 680 N
C) 471 N
D) 0 N
C) -888 N
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Goals Equilibrium Weight The End
Announcements
Read chapter 6.
Reading quiz on chapter 5 and 6; due by 11:59 PM today.
New homework available.
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