Transcript Physics 106P: Lecture 1 Notes

Physics 103: Lecture 8
Application of Newton's Laws
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Survey results
Leftovers from Chapter 4
How to solve problems
Physics 103: Lecture 8, Pg 1
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Physics 103: Lecture 8, Pg 2
But, all is not well!
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Most requests
Please work out problems in class fully
» Will address this to some extent in class
» Better forum for that is the discussion
» Please ask questions
Some requests
Slow down and explain more
Most complaints
There is too much work for the course
Homework is too hard and takes too much time
Some complaints
We need more problems assigned for practice!
Some comments
Don’t go over pre-flights in lecture
Clarify the concepts (from pre-flights) in lecture
Thanks for taking time to answer
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Physics 103: Lecture 8, Pg 3
From Lecture 1: Course Philosophy
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Basic Course Philosophy
read about it (text)
untangle it (lectures)
play with it (labs)
challenge yourself (homework)
close the loop (discussion)
Do make use of discussion, office hours, etc.
Try the homework during the weekend so that you
can ask questions during the week
Homework is posted Friday and is due the following Friday.
Exam worries:
Will post practice exams by Monday
You can bring a single 8.5”x11” sheet with equations to exam
Physics 103: Lecture 8, Pg 4
Preflight 8: Question 1
Consider a person standing in an elevator that is accelerating
upward. The upward normal force N exerted by the elevator floor
on the person is
a) larger than
b) identical to
Example 4.11 in the book!
c) less than
the downward weight W of the person.
N
41%
55%
mg
3%
0%
20%
40%
60%
Person is accelerating upwards - net upwards force is non zero
Physics 103: Lecture 8, Pg 5
Preflight 8: Question 3
You are pushing a wooden crate across the floor at constant
speed. You decide to turn the crate on end, reducing by half
the surface area in contact with the floor. In the new
orientation, to push the same crate across the same floor
with the same speed, the force that you apply must be about
a) four times as great
b) twice as great
c) equally as great
d) half as great
e) one-fourth as great
f s  s N
f k  k N
as the force required before you changed the crate orientation.
Frictional force does not depend on the area of contact. It depends only on
the normal force and the coefficient of friction for the contact.
Physics 103: Lecture 8, Pg 6
Tension
Tension is a force along the
length of a medium
Tension can be transmitted around corners
If there is no friction in the pulleys,
T remains the same
Physics 103: Lecture 8, Pg 7
Example 8: Pulley Problem I
What is the tension in the string?
A) T<W
B) T=W
C) W<T<2W
D) T=2W
Same answer
W
W
W
Pull with
force = W
Physics 103: Lecture 8, Pg 8
Example 9: Pulley Problem II
What is the tension in the string?
A) T<W
B) T=W
C) W<T<2W
D) T=2W
a
2W
W
a
Physics 103: Lecture 8, Pg 9
Preflight 8: Question 5
In the 17th century, Otto von Guricke, a physicist in Magdeburg, fitted two
hollow bronze hemispheres together and removed the air from the
resulting sphere with a pump. Two eight-horse teams could not pull the
halves apart even though the hemispheres fell apart when air was
readmitted! Suppose von Guricke had tied both teams of horses to one
side and bolted the other side to a heavy tree trunk. In this case, the
tension on the hemisphere would be
a) twice what it was
b) exactly what it was
c) half what it was
Fh
2Fh
T
T
Fh
2T
Physics 103: Lecture 8, Pg 10
Reminder:
Procedure for Solving Problems
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Identify force using Free Body Diagram
Set up axes
Write Fnet=ma for each axis
Calculate acceleration components
Use kinematic equations
Solve!
the most crucial step!
Physics 103: Lecture 8, Pg 11
Problem 4.27
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What force does a trampoline have to apply to a 45 kg
gymnast to accelerate her up at 7.5 m/s2?
Fne t  m a
Ft
Fne t  Ft  m g
Ft  Fne t  m g m(a  g)
mg
Physics 103: Lecture 8, Pg 12
Problem 4.21
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A flea jumps by exerting a force of 1.2 x 10-5 N straight
down on the ground. A breeze blowing on the flea parallel
to ground exerts a force of 0.5 x 10 -6 N on the flea. Find
the direction and magnitude of the acceleration of the flea if
its mass is 6 x 10 -7 kg.
Fyne t  F  m g
Fxne t  f
ne t

F
F  m g
1
y 
1



direction,   t an
ne t  t an
 f 
Fx 
F
f
mg
magnitude, F ne t  (Fxne t )2  (Fyne t ) 2
F ne t
acceleration,a 
m
Physics 103: Lecture 8, Pg 13
Problem 4.33
Suppose you have a 120 kg wooden crate resting on a wood floor. (a) What
maximum force can you exert horizontally on the crate without moving it?
What kind of friction is impeding movement?
static
kinetic
N
N  mg
f s  sN  f s  smg
mg
F
fk
(b) If you continue to exert this force once the crate starts to slip
what will its acceleration be?
: F  s m g
What is the net force acting on the crate? Crat e st art s to slip
Force exerted
Force exerted - Static friction
Force exerted - Kinetic friction
F  fk  m a
f k  k m g
a
F  f k sm g k m g

 g(s  k )
m
m
Physics 103: Lecture 8, Pg 14
Example 3
T=50 N
=500
k = 0.2
M = 5 kg
Similar to example 4.8 in book
Find acceleration of block
Hints:
Draw FBD
Resolve T in x and y components
Normal force is smaller than when pulling along horizontal
Tension component along horizontal is also smaller
Fxne t Tx  k N Tx  k Mg  Ty  T cos  k Mg  T sin 
ax 
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

M
M
M
M
Answer: 6.0 m/s2
Physics 103: Lecture 8, Pg 15
Example 6
mg cos ()
mg sin ()
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Find acceleration and normal force
Choose axis and calculate components
Answers: a= g sin() FN=mg cos ()
Physics 103: Lecture 8, Pg 16
Problems 4.59 and 4. 60
A contestant in a winter games event pushes a 45 kg block of ice across a
frozen lake. Calculate the minimum force, F, he must exert to get the bock
moving. What is the force, F’, he needs to exert if he were to pull the
bock with a rope over his shoulder at the same angle above the horizontal?
Is F = F’?
F cos25o
F
25o
F’ cos25o
1) Yes
-F sin25o
2) No
F’ sin25o
25o
F’
s (ice-on-ice)=0.1
Physics 103: Lecture 8, Pg 17
Problem 4.59 and 4.60 Continued
Ice block start s to m ove when frict ion is just overcome,
i.e., F cos25o  f s
s m g
Case A : f s  sN  s (m g F sin25 )  F 
cos25o  s sin25o
sm g
Case B: f s sN  s(m g F sin25o )  F 
cos25o  s sin25o
o
F cos25o
F
25o
Case A
F’ cos25o
-F sin25o
F’ sin25o
N
N’
mg
mg
25o
F’
Case B
s (ice-on-ice)=0.1
Physics 103: Lecture 8, Pg 18