Your Comments
Can you please explain how to do the suspended beam problem? Yes
Why are prelecture, lecture, and quiz material extremely easy? More importantly,
why are homeworks, discussion problems, and tests much more difficult?
I feel terrible about yesterdays exam. I feel confident with my practice exams and do
fairly well on them. I also feel that I know the concepts that I was supposed to
understand for the exam pretty well. However, when it comes to the real deal in the
exam hall, I freeze and I get confused. This depresses me. :( Its always harder by yourself
The pre-lecture didn't seem too explain how the do the torque portion of solving
these. Can you go over this part carefully? Yes
Are we supposed to be taking notes on the videos? Like is that the intention?
Because I seem to understand things during the videos but I can't recall them later
on, but at the same time i'm not sure what I could take notes on to help gauge my
need to figure out
memory when I need the information while doing homework. You
what works for you
I have a question. So, the second exam obviously was a very difficult scaled exam
and you knew that it was hard. I really enjoyed this class because you get the grade
you work for, but now I see that it is not the way I thought it was. My question is,
why did you make a very difficult exam knowing that the majority of students will do
bad on it ? I hope you are brave enough to answer and put this on your power point
tomorrow morning because I really dont want to lose the respect I have towards you
and this class. Thank you.
We didn't – we always try for the same difficulty
Mechanics Lecture 18, Slide 1
I have absolutely no idea how to do the "Three Masses" problem for the homework
in section 16 that's due Friday. … Could you please go over this in lecture, or at
least give a hint on how to start it.
T1  f  ms a
fRs  I s s
a  Rs s
T1
f
If you have the same
number of independent
equations as unknown
variables, then life is good

7
T1  ms a
5
T2  T1  Rd
 I d d
a  Rs d
T1
T2  T1  
T2
T2
mhg
mh g  T2  mh a
1
md a
2
Physics 211
Lecture 18
Today’s Concepts:
a) Static Equilibrium
b) Potential Energy & Stability
Mechanics Lecture 18, Slide 3
Clicker Question
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
1m
2m
D) 7 kg
1 kg
E) 8 kg
1m
3m
Mechanics Lecture 18, Slide 4
Clicker Question
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
T1
T2
300
300
L
M
M
Case 1
L/2
Case 2
Mechanics Lecture 18, Slide 5
It’s the same. Why?
Case 1
Case 2
d1
T1
T2
q
d2
q
L
M
M
L/2
Balancing Torques
L
L
Mg  T2 sin  0
MgL  T1L sin q  0
2
2
Mg
T1 
sinq
Mg
T2 
sinq
Mechanics Lecture 18, Slide 6
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
T1
C) Same
T2
300
300
L
L/2
M
Case 1
Case 2
Mechanics Lecture 18, Slide 7
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
A) magnitude of tension is directly related to the
length..
B) The lever arm is smaller, so the force must be bigger
to counteract the torque from weight.
C) both are Mg/2sin(theta).
Mechanics Lecture 18, Slide 8
Homework Problem
Balance forces: T1 + T2  Mg
T2
T1
CM
Same distance from CM: T1  T2  T
So: T  Mg/2
Mg
Mechanics Lecture 18, Slide 9
Homework Problem
Mechanics Lecture 18, Slide 10
These are the quantities we want to find:
ACM
T1

M
d
Mg
y
x
Mechanics Lecture 18, Slide 11
Clicker Question
What is the moment of inertia of the beam about the rotation axis
shown by the blue dot?
A) I 
1
ML2
12
B) I  Md
2
M
d
L
1
C) I 
ML2 + Md 2
12
Mechanics Lecture 18, Slide 12
Clicker Question
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) T1 < Mg
ACM
T1

M
d
Mg
y
x
Mechanics Lecture 18, Slide 13
Clicker Question
The center of mass of the beam accelerates downward. How is this
acceleration related to the angular acceleration of the beam?
A) ACM  d
ACM
T1

M
B) ACM  d /
d
C) ACM   / d
Mg
y
x
Mechanics Lecture 18, Slide 14
Apply
F
ext
 MACM
ACM  d
Mg  T1  MACM
T1  Mg  MACM
ACM
T1

M
Apply

ext
 I
ACM
Mgd  I  I
d
ACM
Md 2
g
I
d
Mg
y
x
Use ACM  d to find 
Plug this into the expression for T1
Mechanics Lecture 18, Slide 15
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?
Conserve energy:
1 2
Mgd  I 
2
2Mgd

I
T
M

d
y
x
CM demos
Mechanics Lecture 18, Slide 16
Applying
 Fext  MACM
T
T  Mg  M  2 d
T  Mg + M  2 d
Centripetal acceleration

d
ACM  2d
y
Mg
x
Mechanics Lecture 18, Slide 17
Another HW problem:
We will now work out the general case
Mechanics Lecture 18, Slide 18
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 f.
How does f depend on the angle of the ladder
and Bill’s distance up the ladder?
L
f
M
Bill
m
y
x
f
q
Mechanics Lecture 18, Slide 19
Balance forces:
x:
y:
Fwall  f
N  Mg + mg
Fwall
Balance torques:
L/2
L
mgd cos q + Mg cos q  Fwall L sin q  0
2
 d Mg 
Fwall   mg +
 cot q
L 2 

axis
Fwall  f
 d Mg 
f   mg +
 cot q
 L 2 
Mg
d
mg
q
f
y
N
x
Mechanics Lecture 18, Slide 20
This is the General Expression:
d Mg 

f   mg +
 cot q
L
2 

Climbing further up the ladder makes it more
likely to slip:
Making the ladder more vertical makes it less
likely to slip:
M
d
Lets try it out
f
m
q
Mechanics Lecture 18, Slide 21
If its just a ladder…
d Mg 

f   mg +
 cot q
L 2 

f 
Mg
cot q
2
Moving the bottom of the
ladder further from the wall
makes it more likely to slip:
Mechanics Lecture 18, Slide 22
CheckPoint
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 the biggest?
A) Case 1
B) Case 2
Case 1
C) Same
Case 2
Mechanics Lecture 18, Slide 23
CheckPoint
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 the biggest?
A) Case 1
B) Case 2
C) Same
A) When the angle between the ladder and the
Case 1
Case 2
wall is small, as in case 2, the normal force the
wall exerts on the ladder is smaller. Because
the frictional force must be equal to that
force, it is also smaller.
B) Case 2 has a greater frictional force
because the ladder in Case 2 puts more of a
force on the wall than the ladder in Case 1.
Mechanics Lecture 18, Slide 24
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
CheckPoint
Which of the following configurations best represents the
equilibrium condition of this setup?
A)
B)
C)
B) The center of mass is under the place where
the string is hung from.
C) C has the lowest center of mass (minimized
potential energy). Not to mention the fact that
the rope is perfectly vertical so there's no net
horizontal tension force.
Mechanics Lecture 18, Slide 26
Stability & Potential Energy
what is a footprint ??
footprint
footprint
Mechanics Lecture 18, Slide 27