Physics 211 Lecture 5 Today’s Concepts: a) Free Body Diagrams
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Transcript of Physics 211 Lecture 5 Today’s Concepts: a) Free Body Diagrams
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Physics 211Lecture 5
Today’s Concepts:a) Free Body Diagramsb) Force due to stringsc) Force due to springsd) Force due to gravity
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Classical Mechanics
Free Body Diagrams1)Choose system (object or group)2)Draw forces acting on object
-Gravity-Things touching (contact)
m
mg
T
FREE BODY DIAGRAMS! What can I consider a force?
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Free Body Diagrams
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Free Body Diagrams
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Free Body Diagrams
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Tension ForceDirection: Parallel to rope, always pulling.Magnitude: As much or little as needed for Fnet=ma
m
mg
T
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Checkpoint
A box of mass m is hung with a string from the ceiling of an elevator that is accelerating upward. Which of the following best describes the tension T in the string:
A) T < mg B) T = mg C) T > mg
a
Fnet = mass*acceleration. Because the elevator is accelerating upward, there is a net force directed upward as well. This means the force pulling the elevator up (tension) is greater than it is down (gravity).
BOX IN THE ELEVATOR
Since the elevator is accelerating upward, the tension on the box has to be greater than the gravitational force acting on the box, T - mg = ma
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Tension Clicker Question
A block weighing 4 lbs is hung from a rope attached to a scale. The scale is then attached to a wall and reads 4 lbs. What will the scale read when it is instead attached to another block weighing 4 lbs?
A) 0 lbs. B) 4 lbs. C) 8 lbs.
m m m
?
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Normal Force
"The normal force is simply what it has to be to do what it does.“ Oh God, what doe this mean??
Normal force is the force perpendicular to the surface that prevents the objects from passing through each other.1) Direction: Perpendicular to surface and out2) Magnitude: As much or little required keep objects separate.
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Checkpoint
A block sits at rest on a horizontal frictionless surface. Which of the following sketches most closely resembles the correct free body diagram for all forces acting on the block? Each arrow represents a force.
If an object is sitting at rest on a frictionless surface, then the only forces acting upon it are the gravitational force and the normal force
A B C D
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Spring ForceDirection: Parallel to spring: Push or pullMagnitude:F=-kx
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Springs Example
How to actually solve for something like the spring problem would be nice
A 5 kg mass is suspended from a spring with spring constant 25 N/m.How far is the spring stretched from its equilibrium position?
A) 0.5 m B) 2 m C) 5 m
F ma0k x Mg
Mgxk
5*9.825
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Checkpoint
A box of mass m is hung by a spring from the ceiling of an elevator. When the elevator is at rest the length of the spring is L = 1 m. If the elevator accelerates upward the length of the spring will be:
A) L = 1 mB) L < 1 mC) L > 1 m
m
aL
The spring must exert a greater force on the box because T>mg so its length will increase proportionally.
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Student Comment/Question
I feel really comfortable with these concepts. There is an interesting example that we encountered last year in high school physics with a person standing on a scale in an elevator. The question was when does the personal weigh the most (i.e. when does the scale read the heaviest)? I was wondering about the explanation for that question.
V
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Clicker Question
You are traveling on an elevator up the Sears tower. As you near the top floor and are slowing down, your acceleration
A) is upwardB) is downwardC) is zero
V
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Clicker Question
You are traveling on an elevator up the Sears tower, and you are standing on a bathroom scale.
As you near the top floor and are slowing down, the scale reads
A) More than your usual weightB) Less than your usual weightC) Your usual weight
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Clicker Question
A cart with mass m2 is connected to a mass m1 using a string that passes over a frictionless pulley, as shown below. The cart is held motionless.
The tension in the string isA) m1gB) m2gC) 0
m2
m1
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Clicker Question
A cart with mass m2 is connected to a mass m1 using a string that passes over a frictionless pulley, as shown below. Initially, the cart is held motionless, but is then released and starts to accelerate.
After the cart is released, the tension in the string is
A) = m1gB) > m1gC) < m1g
m2
m1a
a
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Universal Gravitation
Newtons relation of the earths gravity to the gravitational forces between two objects was difficult to follow.
2
MmF Gr
2earth
earth
M mF G
R
Near surface of earth r=Rearth
3 224
1126
5.97 106.67 10
6.4 10/
kgm kg s
mm
29.8 /m s m
Direction: Toward other object (down near earth surface)Magnitude:F=GMm/r2 (mg near earth surface)
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Most of it was really easy to understand.... the only problem are the gravity examples.. we should do more of those....
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rr
MmGFGravity ˆ2
2))12.2((||
rMmGFGravity
G*mmars/(3.2*rmars)^2
USE ALGEBRA!
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amFnet
rvm
rMmG
2
2
rGMv /
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rGMrt/
2
vdt
GMrt
3
2 Careful, check units. Asks for time in hours, not seconds!
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rGMv /
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GMrt
3
2
GMr
GMr
tt
30
31
0
1
2
2
2/3
0
1
0
1
rr
tt
3/2
0
101
ttrr