Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18...
-
Upload
eileen-underwood -
Category
Documents
-
view
214 -
download
0
Transcript of Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18...
![Page 1: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/1.jpg)
Today’s Concepts:a) Static Equilibriumb) Potential Energy & Stability
Physics 211Lecture 18
Mechanics Lecture 18, Slide 1
![Page 2: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/2.jpg)
ACT
A (static) mobile hangs as shown below. The rods are massless and have lengths as indicated. The mass of the ball at the bottom right is 1 kg.
What is the total mass of the mobile?
A) 4 kg B) 5 kg
C) 6 kg
D) 7 kg
E) 8 kg 1 kg
1 m 3 m
1 m 2 m
Mechanics Lecture 18, Slide 2
Demo: planets mobileDemo: planets mobile
![Page 3: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/3.jpg)
ACT
In which of the static cases shown below is the tension in the supporting wire bigger? In both cases M is the same, and the blue strut is massless.
A) Case 1 B) Case 2 C) Same
Mechanics Lecture 18, Slide 3
M
300
L/2
T2
Case 2
M
300
L
T1
Case 1
![Page 4: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/4.jpg)
ML
T1
d1
Case 1
Balancing Torques
0sin1 LTMgL 0sin22 2 L
TL
Mg
sin1
MgT
sin2
MgT
It’s the same. Why?
Mechanics Lecture 18, Slide 4
M
T2 d2
Case 2
L/2
![Page 5: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/5.jpg)
CheckPoint
In which of the static cases shown below is the tension in the supporting wire bigger? In both cases the red strut has the same mass and length.
A) Case 1 B) Case 2 C) Same
Mechanics Lecture 18, Slide 5
300
L/2
T2
Case 2
300
T1
LM
Case 1
![Page 6: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/6.jpg)
CheckPointIn which of the static cases shown below is the tension in the supporting wire bigger? In both cases the red strut has the same mass and length.
A) Case 1
B) Case 2
C) Same
A) Case 1 because the lever arm distance is the largest there B) The torque caused by gravity is equal to the torque caused by the tension. Since in Case 1, the lever arm is longer, the force does not need to be as long to equalize the torque caused by gravity. C) The angles are the same so the tensions are the same, the lengths do not matter.
Mechanics Lecture 18, Slide 6
![Page 7: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/7.jpg)
Mechanics Lecture 18, Slide 7
Nice and clear; I like this explanation.Nice and clear; I like this explanation.Definition? Hmmm, I think I don’t know what this means.Definition? Hmmm, I think I don’t know what this means.
Good; I like this.Good; I like this.
Seems to me that a longer length means smaller tensionSeems to me that a longer length means smaller tension
It should seem VERY similar to a ladder problem.It should seem VERY similar to a ladder problem.But that is the But that is the questionquestion..
Location is important for torques.Location is important for torques.
But the question is about the force or tension.But the question is about the force or tension.
![Page 8: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/8.jpg)
CM
Mg
T1T2 T1 T2 TSame distance from CM:
T Mg/2So:
Mechanics Lecture 18, Slide 8
T1 T2MgBalance forces:
Homework Problem
![Page 9: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/9.jpg)
Mechanics Lecture 18, Slide 9
Homework Problem
![Page 10: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/10.jpg)
These are the quantities we want to find:
Mechanics Lecture 18, Slide 10
d
Mg y
x
T1
M
ACM
![Page 11: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/11.jpg)
What is the moment of inertia of the beam about the rotation axis shown by the blue dot?
A)
B)
C)
L
ACT
21
12I ML
2I Md
2 21
12I ML Md
Mechanics Lecture 18, Slide 11
d
M
![Page 12: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/12.jpg)
The center of mass of the beam accelerates downward. Use this fact to figure out how T1 compares to weight of the beam?
A) T1 Mg
B) T1 Mg
C) T1Mg
ACT
Mechanics Lecture 18, Slide 12
T1
Mgy
x
ACM
d
M
![Page 13: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/13.jpg)
The center of mass of the beam accelerates downward. How is this acceleration related to the angular acceleration of the beam?
A) ACM d
B) ACMd /
C) ACM / d
ACT
Mechanics Lecture 18, Slide 13
T1
Mgy
x
d
M
ACM
![Page 14: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/14.jpg)
Apply
Apply
Plug this into the expression for T1
ACM d
Use ACM dto find
ext CMF MA1 CMMg T MA
Iext CMA
Mgd I Id
2
CM
MdA g
I
1 CMT Mg MA
Mechanics Lecture 18, Slide 14
T1
Mgy
x
d
M
ACM
![Page 15: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/15.jpg)
Conserve energy:
After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction.
What is the angular speed when the meter stick is vertical?
2Mgd
I
21
2Mgd I
Mechanics Lecture 18, Slide 15
M
y
x
d
T
![Page 16: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/16.jpg)
y
x
Applying
Mg
dACM 2d
Centripetal acceleration
ext CMF MA2T Mg M d
2T Mg M d
Mechanics Lecture 18, Slide 16
T
![Page 17: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/17.jpg)
Another HW problem:
We will now work out the general case
Mechanics Lecture 18, Slide 17
![Page 18: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/18.jpg)
General Case of a Person on a Ladder
Bill (mass m) is climbing a ladder (length L, mass M) that leans against a smooth wall (no friction between wall and ladder). A frictional force f between the ladder and the floor keeps it from slipping. The angle between the ladder and the wall is .
How does f depend on the angle of the ladder and Bill’s distance up the ladder?
L
m
f
y
x
MBill
Mechanics Lecture 18, Slide 18
![Page 19: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/19.jpg)
Balance forces:
x: Fwall f
y: N Mg mg
Balance torques (measured about the foot):
Mg
mg
L/2
f y
xN
Fwall
daxis
cosmgd cos2
LMg sinwallF L 0
cot2
Mg
L
dmgFwall
cot2
d Mgf mg
L
wallF f
Mechanics Lecture 18, Slide 19
![Page 20: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/20.jpg)
f
d
M
m
Climbing further up the ladder makes it more likely to slip:
Making the ladder more vertical makes it less likely to slip:
This is the General Expression:
cot2
d Mgf mg
L
Mechanics Lecture 18, Slide 20
![Page 21: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/21.jpg)
cot2
d Mgf mg
L
Moving the bottom of the ladder farther from the wall makes it more likely to slip:
cot2
Mgf
If it’s just a ladder…
Mechanics Lecture 18, Slide 21
![Page 22: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/22.jpg)
In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground bigger?
A) Case 1 B) Case 2 C) Same
CheckPoint
Case 1 Case 2Mechanics Lecture 18, Slide 22
![Page 23: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/23.jpg)
In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground bigger?
A) Case 1 B) Case 2 C) Same
Case 1 Case 2A) Because the bottom of the ladder is further away from the wall.
B) The angle is steeper, which means there is more normal force and thus more friction.
C) Both have same mass.
Mechanics Lecture 18, Slide 23
CheckPoint
![Page 24: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/24.jpg)
Mechanics Lecture 18, Slide 24
But what about the friction force?But what about the friction force?
I don’t know what this means.I don’t know what this means.
okayokay
A Force can not A Force can not be equal tobe equal to a torque. The frictional force will be a torque. The frictional force will be smallersmaller with a smaller with a smaller angle between the ladder and the wall.angle between the ladder and the wall.
![Page 25: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/25.jpg)
CheckPoint
Suppose you hang one end of a beam from the ceiling by a rope and the bottom of the beam rests on a frictionless sheet of ice. The center of mass of the beam is marked with an black spot. Which of the following configurations best represents the equilibrium condition of this setup?
A) B) C)
Mechanics Lecture 18, Slide 25
![Page 26: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/26.jpg)
CheckPointWhich of the following configurations best represents the equilibrium condition of this setup?
A) B) C)
If the tension has any horizontal component, the beam will accelerate in the horizontal direction.
Objects tend to attempt to minimize their potential energy as much as possible. In Case C, the center of mass is lowest.
Mechanics Lecture 18, Slide 26
DemoDemo
![Page 27: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/27.jpg)
Mechanics Lecture 18, Slide 27
But the wire is NOT perpendicular to the beam in any of the situations.But the wire is NOT perpendicular to the beam in any of the situations.
okayokayI don’t know what this means.I don’t know what this means.
GoodGoodOkayOkay
What’s a “footprint” when something is What’s a “footprint” when something is suspendedsuspended??
GoodGoodVery goodVery good
? ? ?? ? ?
What’s a “footprint” when something is What’s a “footprint” when something is suspendedsuspended??When things aren’t clear, please When things aren’t clear, please come see me — always!come see me — always!
That does That does notnot seem “intuitive” to me. But it seem “intuitive” to me. But it isistrue — from analyzing the forces and torques.true — from analyzing the forces and torques.
![Page 28: Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Physics 211 Lecture 18 Mechanics Lecture 18, Slide 1.](https://reader036.fdocuments.in/reader036/viewer/2022070413/5697bfe91a28abf838cb703b/html5/thumbnails/28.jpg)
footprint
footprint
Stability & Potential Energy
Mechanics Lecture 18, Slide 28
CM DemosCM Demos