>> OK.
Our agenda today is to go over some problems
for homework that many people either email
me this weekend or I noticed that there was--
people had problems on a certain homework
problems.
So, we'll do that on Chapters 1 and Chapters
2.
So let's start off with a Chapter 1 problem.
This was assign maybe 5, let's see, maybe
5, I think it was.
It's a rotating cylinder inside of a fixed
cylinder, a picture on the board shows that.
I'll read it first.
The space between two 6-inch long concentric
cylinders is filled with glycerin, viscosity
is given.
The inner cylinder has a radius of 3 inches
and the gap between the cylinders is 1/10
inch.
One-tenth inch gap shown there.
Determine the torque and power required to
rotate the inner cylinder at 180 revolutions
per minute, RPM.
The other cylinder is fixed.
Assume the velocity distribution in the gap
to be linear.
OK.
We didn't put the 6-inch on there, so let
me put that on the picture there.
Six-inch from here to here.
All right, so there's the picture.
If you look at a view from the top, OK, there
is the outer cylinder, here is the inner cylinder.
Inner cylinder is rotating.
That's the radius r inside.
This is the radius r outside.
The outer cylinder is fixed.
The inner cylinder is rotating.
Yeah, omega.
The glycerin is between the two cylinders.
And this is gamma and mu.
OK.
So, we can start off.
Now we know-- let's take this assignment set.
This assumes the velocity distribution is
linear between the cylinder and the fixed
outer cylinder gap-- in the gap.
So it's go in this way so that means that
this would be the velocity here.
And then at the outside, the fixed side, the
velocity is zero.
So, it looks like this.
He said it's linear, so shown there as linear.
Now, first of all, a torque.
A torque is a force times a moment arm or
a force times distance.
So torque is equal to force times distance.
There's a force on the inner cylinder that's
rotating.
That force, let me show it here.
There's the force.
The distance, of course, is the radius r inside.
There's a force.
So, we can go ahead and say that the torque,
force times distance, and that's equal to
F from right there times a distance which
is r inside.
Now, force.
Glycerin is experienced in a shear stress.
And a shear stress, I'll put up here.
Shear stress is tau equal mu, du, dr.
The velocity u is a function of the radius
r in the picture.
U is the velocity anywhere in that distribution.
Capital V is the velocity at the inner cylinder.
Du, dr on the surface of the inner cylinder.
But, he said it's linear distribution.
So if you recall, it requires this with mu,
delta u over delta r.
And if we do that, we get mu.
Delta u is the velocity.
Here, V minus zero down here, V minus zero.
And then we have r outside minus r inside.
All right.
So, shear stress tau equal force over area.
So, the force, F equal tau times A. Tau times
A. V-- Put in tau first.
The area is the area that the glycerin touches
the inner cylinder.
Area that the glycerin touches the inner cylinder.
Circumference 2(pi)r times the height, l.
2(pi)rl.
R inside, l times the ri, the distance for
the fourth calculation.
So, that's equal to muV.
Let's put 2pi up from here.
2pi, mu, V, l over r outside minus r inside
times r inside cubed.
That's the torque, T. I'm not going to put
the numbers and you can do that straightforward.
OK.
Now, that answers the first part on the torque.
Now, the power.
OK, find the power, is equal to torque times
speed.
Torque is T. Speed is omega.
So took that answer, put it up there.
Omega, be careful, this has to be in radians
per second, OK.
So we have to convert-- what's the given to
us there?
RPM, 180 revolutions per minute into radians
per second.
Why?
Because the angle has to be dimensionless.
You can't put revolutions in there.
You got to put the angle rotation in radians
dimensions.
So if you do that, the torque down here is
going to be force times distance.
That would be pound-feet, multiply it by omega
in radians per second.
And don't forget, radians is dimensionless.
That's power.
That's power, work per time.
What's work?
Pound-feet.
Work is pound-feet, divided by time, what
do you get?
Power.
Yeah?
>> For the torque, should it be ri squared
because you have ri--
>> Oh, let's see here.
Yeah, hold on.
Dot, dot, dot, no, it's ri.
Thanks for bunging up.
I was doing the power.
Yup, thank you, I'll turn to power.
When you multiply it by that guy over there,
yeah.
Let's just make sure.
Tau up there is equal to mu, du, dr.
We've got V minus zero.
V, r minus r, zero mu.
Area is squared in feet, got it.
Circumference, pi times the 2(pi)r times what?
L, all right.
Mm-hmm.
>> Yeah, OK.
>> Thanks.
OK-- oh, to convert this guy?
This guy, you know, what's per revolution?
2pi radians, 2pi radians per revolution.
Sixty seconds in a minute.
Two conversion factors here to here.
OK.
So that was one that was in the Chapter 1
on the use of the shear stress and viscosity
mu.
You had three assignments in Chapter 1 with
shear stress.
So, this was the third one in that set.
OK.
[Inaudible] all of your notes as I talk about
the exam for a few minutes as I get working
to the last minute today.
All right.
It's going to cover Chapters 1 and 2 as I
mentioned.
It's going to be closed book.
I'm going to give you about four-page handout
of additional information.
In that handout-- it's five pages.
In that handout is going to be in the appendix
of properties of various fluids, Tables B1,
B2, B3, B4, and in the inside front cover
which we use Table 1.5, 1.6, 1.7, 1.8.
So, inside front cover, appendix under-- appendix
B, properties of water and air in both BG
and SI as a function of temperature.
OK.
You'll be given that.
You'll be given under fluid statics, the table,
it's Figure 2.18.
I'm going to give you Figure 2.18, locations
of centroids, movements of inertia, areas
of typical geometric shapes, Figure 2.18.
I'm going to give you at the end of each chapter--
Yeah.
It's called Chapter Summary.
Chapter Summary, a study guide.
In that section, there's a big red box that
has quite a few of the important equations
in that chapter.
We go to Chapter 2, do the same thing.
At the end of every chapter, there is a red
box and what the author thinks are the important
equations in that chapter.
Now, they're not the only ones but that's
what the author wants to emphasize.
So, I give you those in Chapter 1, Chapter
2, all the properties.
OK, I told you about fluids.
I'll give you the properties for moments of
inertia and centroids of different geometric
shapes from Figure 2.18.
You'll have that-- for this exam, that package
has everything to the whole course and the
finals.
So I'll pick them up when you're done, pick
them up and I'll give them back to you then
for the second exam and give them back to
you again for the final exam, the same package.
OK.
Now, besides these equations in the red boxes,
you're allowed to bring with you one sheet
of paper, 8.5 by 11, both the front and back
sides with anything else you think you'd like
to have for the exam.
It might be equations that aren't in those
red box equations at the end of the chapter.
It might be a homework problem you especially
had difficulty with.
It can be whatever you want in there, whatever
you want.
One sheet, both sides, 8.5 by 11.
I don't care if you photocopy every kind of
problem you've got.
Postage stamp sites and put on [laughter]
everything there.
Make yourself crazy.
It's just a crutch.
Sometimes people don't need it, but it's a
nice crutch if you weren't comfortable with
the building place, you keep it by your side.
That's what it's for, OK?
Yeah.
>> So we're going to be given the equations
from that book or the equations at the end
of chapter as well as the things in the back
of the book and then we can then [inaudible]
for additional things.
>> That's right.
Things you had difficulty with.
And, you know, maybe some things about stuff,
if I get it from like this, make sure I do
this, you know, all those stuff that would
make you feel better [inaudible].
OK.
Let's see.
What else?
Oh, probably somewhere 25, 30% of Chapter
1, 70-75% of Chapter 2.
Chapter 2 is much longer and it's much more
beefy.
So, of course, the emphasis is on Chapter
2, but there will be problems on Chapter 1.
Probably, four or five problems depending
on how long they are.
I gave you online, Blackboard, on that last
year's exam.
That's 50 minutes, three problems.
So, we have an hour and 15, so four or five
problems weighted by a different-- some might
be 30% and some might be 25 and some might
be 27.
OK, let's see, anything else on that.
Everything else that I missed, huh?
OK.
An hour and 15.
Oh, office hours.
Let's see, OK, tomorrow, you know, the regular
office hour, it's 9 to 11 but on Wednesday,
it's normally 12 to 1 so I'll go from 11 to
1.
Eleven to 1 on Wednesday entertaining your
last minute questions.
By that time, you just sit back and relax,
don't cram as you know.
You've been through too many [inaudible] test.
Don't cram for the last minute.
OK, anything else on that?
OK.
So, this was our only Chapter 1 review problem,
OK?
Now we're going to go on to Chapter 2.
OK, let me-- let's-- OK, I might give this
[inaudible], 2.115.
Two hemispherical shells are bolted together
as shown.
You've seen this picture.
The resulting spherical container weighs 300
pounds, is filled with mercury and supported
by a cable as shown.
Container is vented at the top.
If eight bolts are symmetrically located around
circumference, what's the vertical force that
each bolt must carry?
We have the bolts along here.
It's hanged up to the ceiling support here.
It's filled with mercury, 3 foot in diameter.
OK, let's see.
The container itself weighs 300 pounds.
Top and bottom.
OK, find the vertical force that each bolt
must carry.
OK.
Two ways to do it so let's see which way we'll
do first.
Let's do it this way first.
Here's the bottom hand for this thing.
We're going to look at this picture bottom
hand.
We're going to look at this picture, bottom
hand.
So it has mercury in here.
The shell is here.
OK.
There's pressure on the top here.
And we call that just-- it was called P. OK.
And now, we're going to draw a free-body diagram
of this.
So, take it again.
And this is my free-body diagram.
And we'll call the force due to-- no, so the
force is on this.
Bottom shell plus the mercury.
OK.
First of all, there's a force here downward
due to the pressure on the top.
We call F of P. There is a weight of the shell
and the weight of the mercury.
And then, of course, there's the bolt force.
And it depends on how you want to do it.
I'm going to show up like this, with a fat
arrow.
Force bolts.
So, there's our free-body diagram.
Then somebody says, you know what, I don't
like that pretty bad diagram.
I'm going to do it my own way.
So, the second way, draw a free-body diagram
of the-- here's the shell, OK, just the shell,
OK?
So, I'm going to draw a free-body diagram
that is now called a type of shell.
OK.
Number one, kids, I have the bolt force acting
up or holding this thing together.
And see how I label this thing.
There's the weight of the shell, OK, F-- I
call F sub B, the vertical force.
The weight of the shells acting down.
T shell.
And then, I have the force acting down on
the curve surface.
F sub B is the vertical force on the curve
surface.
We did that in Chapter 2.
Two different free-bodies.
One includes the mercury, one doesn't.
Of course, the bolt force acts at the center
of the hemispherical shell.
So, I put it in the middle there.
I can show up here.
I can show up there.
You could, but one could be that way.
OK, we'll do each one.
The top one up here.
Let's take this guy.
Summation of forces.
You know, you guys could call it Z upward
so we'll call it Z upward then.
So, and I'm going to call this just FB for
bolt force.
FB, this guy right here is equal to these
three guys moving down.
F due to P plus F due to the-- plus the weight
of the shell plus the weight of the mercury.
OK, let's do each one of these and separate
FP.
There's a pressure force.
What's the force equal to?
Pressure times area.
What's the area of what?
The shell or the mercury-- no, no, the mercury.
Mercury has a pressure.
OK.
So, the pressure times the area of what?
A mercury.
What is that?
A circle.
Pi r squared.
So, P times pi r squared.
The weight of a shell is the weight of the
shell.
Now, the weight of the mercury.
You're given gamma of the mercury.
Let's look at the problem.
Yeah, we know what mercury is.
So, a gamma Hg.
OK.
So, we have our weight for the mercury, gamma
Hg.
Sphere, pi, D cubed divided by 6, half of
F. Six times 2 is 12.
Volume of sphere is 1/6 pi, D cubed, pi over
6.
OK.
So, there's that guy.
The pressure right here or the pressure where
my fingertip is, the pressure where my fingertip
is, here's a keyword, it says, the container
is vented on top.
Oh, thank you very much.
The container is vented on top.
There's a hole right there.
The mercury is not squirting out, uh-oh.
The mercury is sit in nothing.
So, I guess depth pressure is going to be
right there.
That exactly is true.
How far down are you from that hole r, [inaudible].
What's the pressure down there in Chapter
1-- Chapter 2, gamma times a distance, gamma
times r.
So, depth pressure gamma times r.
This guy right here is 2r, 2 times 2 is 4
times 2 is 8, 8 plus so, we have 2/3rds.
Right plus cubed, 2r, 8 plus r cubed.
That should be-- OK, got it, and gamma again
so, cubed in there.
We won't worry about, subscript gamma Hg because
it's just one gamma, it's gamma mercury plus
weight of the shell.
OK.
There's the answer.
I'll do it this way now.
All right, by the way, the answer is 1/8 of
that as he said what is the force in one bolt.
This is the force in all of the bolts, there's
eight of them so 1/8 of that thing.
All right, same thing, some force is in the
Z direction.
OK, what's going up?
F bolts?
OK, FB.
What's going down?
The weight of the shell plus FB.
OK.
FB, I'm doing it now by the way that I did
for a curve surface.
So, here's the free surface of the mercury.
Here's the curve surface of the mercury.
OK, what's the vertical force, OK, on that
curve surface?
It's gamma times the volume of the fluid above
the curve surface extending to the free surface
real or imaginary, take cringe of all vents,
go straight up, stop it.
That's the volume.
What is that volume?
And that volume plus that gamma.
That's the volume.
OK, that's the-- OK.
Gamma times the volume.
What's the [inaudible]?
What's the top, OK, circle, pi r squared.
How high is this r?
How high is this r?
OK, pi r squared, circle is this downward.
Pi r squared times r, pi r cubed.
This guy here is 1/2 of a hemisphere, 1/2
of hem.
Pi of 2.
Plus weight of shell.
If you want to convert that guy, this guy
right here is, again, d squared ri cube, 2r
cube 68, 8 divided by 12.
OK, 2/3 r cube, pi r cube.
You could say, same answer, same answer which
means you can play the game any way you want.
This is the way the solutions manual does
it.
This is the way I like to do it.
Nobody's wrong, nobody's right, everybody's
right.
I like this way.
How many terms do I have?
Two.
How many terms here?
This, this, this, three.
I don't care.
I don't care.
Just do it right.
What I'm saying though is there is two ways
to do it.
OK.
And one of the key words there was the cylinder
against and that the sphere is vented.
What does that mean, vented?
It means the pressure this atmosphere at the
vent point.
OK.
Now, let's go on and take a look at another
one.
Again, the answer is 8 times larger than the
portion of one goal, so divide it by 8.
All right, let's see.
Maybe I'll take the easier one first.
I think I will.
I'm going to work problem two for the last
midterm, OK.
Just answer [inaudible].
This is the problem two from last year's midterm.
It was multiple part problem, A, B and C.
I won't go through A, B, C. They're too simple.
I'll do part B. Here's the picture.
Water circle-- OK, let's see now.
We've got a radius here R and this was with
surface BC, and this was water, and this was
given as 1 meter.
And R was given as a half a meter.
And you're supposed to find, find the resultant
force on surface BC.
The width into the page is 2 meters.
And this?
OK.
I think of-- or I'll work with this way first,
OK.
I'm going to take a free-body.
This is the surface here.
Well that-- so the slides from water.
That's water.
OK.
The weight of the water is right there.
OK.
The horizontal force on the left hand side
would look like this.
Pressure varies linearly.
On top, it looks like this.
So the pressure here is P. There's two forces
here, there is F resultant vertical and F
resultant horizontal.
OK.
Now, if you want, let's forces in the Z direction--
-- of the Z.
That's how water does it.
OK, there they are.
What kind of forces are there.
There's a force through the pressure on top.
There's a weight acting down.
And there's a resultant force pushing up from
surface BC on to the water.
So, we have FR vertical equal the force due
to pressure on top, plus the weight.
The pressure on the top, here's the pre-surface
on part down on our weight one-- oh I'm sorry,
half.
It should be this is one, a half.
So the pressure where my fingertip is the
gamma water times an F, plus gamma water 
times the volume.
There's the volume right here.
That volume is-- OK, here's a square, 1/2
by 1/2 and into the votex 2.
We won't put the 2 right now but it's 1/2
times 1/2, minus a quarter circle 
multiplied by 2 into the board, that's the
volume.
OK.
And there's the F pressure.
No, F pressure times area, pardon me, so the
area in there.
That does the pressure right there times the
area, I skipped a step.
Gamma times H, and H times area.
That area is the area over which the pressure
acts my hand.
OK, that area is the radius 1/2 times 2.
OK.
I think I have it going on.
OK.
You could put all the numbers in there, gamma
over 9810, 5954 OK.
OK.
Now, I'll do it another way.
Here's the same picture.
Now I take the force.
Now I take curve surface BC, pre-surface.
And then I say, you know-- by the way, all
I'm doing is finding the vertical force because
the horizontal one is too simple.
The horizontal one is Fh gamma H of Z times
A. That's the simple one.
Very few people get that one on, unless there's
a problem getting h of z.
OK, so that one is the simplistic one.
The vertical one is a tough one.
OK.
I'll do it the second way.
How is this one done again?
Take a free-body diagram of that little piece
of water.
Take a free-body diagram of the mercury and
the shell.
Now, I take a free-body here, what?
The shell all by itself.
A free-body diagram of what?
The surface BC all by itself.
Here's the rule in words.
The vertical force on curve surface BC is
the weight of the volume of water extending
vertically to the free-surface, real or imaginary.
OK.
There it is.
There is the volume, that whole area.
OK.
Vertical force is equal to gamma water times
the volume.
You're getting there, that's it.
You can-- you can check it out and see.
What's the volume?
This guy right here which is 1/2 times 1/2
times 2, a 1/2 times 1/2 times 2.
Let's do it.
It's easier.
This is easier.
Let's put the volume in here.
Let's make it easier for ourselves.
Let's take the whole thing.
We'll take this whole thing, this whole big
piece right here.
That's 1, OK, 1 there times 2, edge of the
board, OK.
So that this thing here, this is 1/2.
This is 1/2.
This is 1/2.
So 1 times 1/2 times 2, 1 times 1/2 times
two, yeah.
And what do I subtract out?
Subtract out that guy right there.
OK.
And gamma water and that, and the answer for
that-- Yeah.
It's darn close, right?
Calculator, round off.
Same thing, same thing, same thing.
So conclusion, either way works.
I don't care which way you work but just work
it right.
I prefer the bottom way but that's just me.
That's just me.
You pick up three or four textbooks.
A couple will work it this way.
Two or three look at that way.
So there's no consensus that one way is the
best way.
It's your choice.
Same thing over here.
OK.
Any question I had.
Go on.
OK.
Maybe let's go to the hard one from 2.122.
OK, 2.122.
Again, this is emerged surface.
But now, that surface is a gate.
So, let's draw that picture.
Let's see, 2.122.
We have a semi-cylinder OK.
And then attached to a hinge point here, second
point.
OK.
We have 4 feet and then 4 meters and 1 meter.
This guy is 1 meter.
This guy is 4 meters.
Free surface, the fluid is water.
When the water depth is greater than 4 meters,
the gate opens slightly.
And thus, the water flow 100.
OK.
This is atmospheric here.
This is atmospheric here.
Waters to rising the gate up to the top.
Determine the weight of the gate per meter
of length.
Now, he calls length into the board.
So if you see with their length, it means
into the board.
Here's the board width, that means it's the
board.
Here, the length means out of the board.
So per meter of length.
OK.
So I'm going to start that again.
And I'm going to do like I did up here.
OK.
So now, I'm going to take a free-body diagram
of-- and I'll show you here what it is without
some-- This is a gamma water and this is the
gate.
And I'm going to take the free-body diagram
of that whole thing.
OK.
So if I do that, here's the free-body diagram--
OK.
First of all, I have the weight of the water
acting down, weight of water.
Then I have the weight of the gate acting
down.
We should call that g for gate.
And then I have F horizontal acting over here.
The way we get that, you know, project the
surface on there and the-- in this case--
in this case, here's the water pressure here.
Here's the water pressure.
Actually [inaudible] here and here.
OK.
And there's a pressure on top.
This-- this guy here, you know, like that
there is just gamma, h sub c times A, so that's
not a problem.
It was this pressure and that pressure force
is here, F due to pressure.
The pressure-- I'm sorry.
Not there, there's-- that's atmospheric.
On top, atmospheric, 0.
Left hand side, atmospheric, 0.
Bottom or water under gate, water adds pressure.
OK There they are.
OK.
Some moments above [inaudible] equals 0.
Point O is right here.
Let's write them down here.
I think I call clockwise positive.
So weight of the water and the gate are both
negative, minus weight of the gate, times
d2, minus weight of water, times d1.
That Fp crosses a clockwise and Fh crosses
a clockwise.
Plus Fpd3 plus Fhd4 equals 0.
Unknown, solve for weight of the gate.
D1 goes with weight of the water.
I'm not going to put all these guys in but
here's d1, weight of the gate times d2.
I'll take them all in.
F sub p times d3.
This guy, times d4.
Do I know all those?
Yeah, I can find them.
I can find them.
I could find all of those d's.
It's a bit of work but I can find them.
It's a bit of work.
When you're done, you get the weight of the
gate.
It's equal to 64.4 kilo newtons.
Now, I'm going to do it again.
This time I'm going to do it with a different
free-body.
OK.
This time I'm going to take free-bodies of
gate plus a little bit of water beneath it.
Now, I'm going to take free-body diagram of
the gate.
OK.
Now, that one on the top, atmospheric.
OK.
So, I have just like over there, the weight
of the gate back into the centroid.
I have the horizontal force here, again, H
sub c times A. And I have the vertical force
here.
OK.
Here's point O. Sum moments above point O
gives you zero.
So I have Fh plus Fv.
Fv times d1, plus Fh times d2, minus the weight
of the gate times d3 must be 0.
Fv has this guy over here and that's d1.
The weight to here, that's d3.
Horizontal times d2, that's here.
OK.
The unknown is here.
Solve for weight of the gate, 64.4 kilo newtons.
Over there, 64.4 kilo newtons.
Here we go again.
We've done it three times now.
I've shown you two ways to do all these problems.
If you want to do it-- By the way, the solution
on that, this is the way they do it in the
solution's manual.
There it is.
It's a lot of work.
Would you naturally think to make that to
control line?
I seriously doubt it.
I seriously doubt it.
You got to see someone work at that way and
then say, "Yeah, I think I understand them."
Well, maybe, you do and maybe you don't.
That way over there, I like, because it works
almost all the time, and it's much simpler
to see and understand than that way.
But I don't care if you see this better, that's
fine.
You'll get the right answer.
If you work it this way, you'll get the same
answer.
It makes no difference to me as long as you
do the work correctly.
Oh, there's work over here.
Oh yeah.
D2, that's easy.
D3, that's easy.
And they're both-- D1, that's not easy.
D1, that's easy.
Over here, you know, what do you to that guy?
Yeah, it requires a couple of lines, a couple
of lines, you know.
Not too serious but a couple of lines, you
know.
So yeah, and, you know, what's d3 half the
radius?
Not a problem.
But to draw that free-body of the gate and
the water, I guarantee you, most people would
never in a, you know, whole time just think
that, of course, that's the way you do it,
you know, right-- I don't know, right.
OK.
Just-- now, just to show you how you get--
I'll just show you how you get one [inaudible].
I'm going to show you how you get these distances
easier.
The distance from here to here, I'm going
to call that distance x.
Once you've got x, you can take 1 minus d1
which you'll need 1-- Yeah, take 1 minus x,
you get d1.
OK.
Here's how you get x.
All right.
There is a quarter circle.
It's up here.
V1, V2, the distance from here to here, the
distance from here to here.
Centroid, right here.
Centroid, right here.
I know where they're at.
OK.
Here to here.
The-- Take the area.
System seven [phonetic], OK?
This is a composite area.
Take the area times the [inaudible] arm from
anywhere you want.
I'm going to take it from the left hand side,
major to left hand side.
So the top 4 minus 1, this is 3 up to here.
OK.
The top piece, well, up in here, 3, 1, radius
of 1.
So that's 1 out here left to right.
OK, 3 times 1?
This is the area.
That's the area.
How far from the left hand side of the centroid?
Half.
Plus same thing down here, take the area,
1/4 pi r squared.
R is 1.
Times the distance from left hand side to
where?
The centroid, 4r over 3 pi.
That's one moment of the areas of each individual
piece.
Equal, the total area times the, we'll call
it, x-bar.
To the centroid of the total area, call it
xr.
Total area, pi 1 squared divided by 4 plus
3 times 1.
Put that total area in there.
Solve for x-bar, give x-bar equals 0.484.
So that means that d1 then 1 minus x-bar--
I didn't compute that up.
That's what it is.
That's the only difficult calculation, difficult
to perform is the review of the 2214.
So now, what if it didn't ask you for-- what
if we gave you the weight [inaudible] and
he didn't ask you for the weight of gate.
Let's say, that he gave it to you.
You only need to take a moment to spot something
maybe.
Up here, this method becomes-- how many forces
of free-body?
One, two, three, four.
Forces of the free-body?
One, two, three.
Yeah, I like him better.
Right there, I like him better.
Less forces to worry about.
So no matter what they-- this might be just
a solid-- let's say the surface is solid.
The surface is solid here, you know.
It's not pivoted here.
It's just a solid surface.
OK.
You got to do it that way or that way?
Oh, that's-- that's pretty simple.
You got Fh, Fd, take the result in forces,
the square-- the sum of the squares.
Over here, you first of all have to-- have
to worry about, you know, this guy and this
guy.
And once you do that, then you go and take
the square root of some of the squares.
It's OK.
It's not that much more work.
But it's just not as understandable to the
average person as that way.
So I prefer that way.
And I don't care if you do this on the exam.
I don't care.
OK.
Now, let's take one last thing about these
surfaces and let's see [inaudible] is here.
[ Inaudible Remark ]
It doesn't matter.
It would show up.
OK.
Here's the surface.
I'm going to change it just for simplicity
sake [inaudible] better for me.
Sometimes people say, "Well, I don't understand
why you put this water above this surface
here.
What are you doing putting up water out there
to the free surface?
That doesn't make a lot of sense to me."
Well, maybe this will make more sense to you.
On either side-- OK, let's assume there's
water here.
OK.
The pressure here at that point is gamma times
h.
Now, let's assume fictitiously that there
is water over here.
Fictitiously-- There's not water there.
Let's assume there was.
What would the pressure of that water be right
there?
Oh, that's a trick question of course.
They're the same, gamma times h.
One is fictitious and one is real.
So even though I know there's no water over
here, could I fictitiously assume there's
water going from the surface where?
To the free surface.
Would I get the same pressure on that surface
as I get by looking at the right hand side?
Of course I do.
That's enough proof for why you do what I
did.
OK.
It doesn't matter because this is a fictitious
layer of water.
The pressure here would be the same as the
pressure here but in a different direction
of course.
So it makes no difference.
That's where it comes from, real or fictitious
free surface.
It doesn't matter.
It gives you the same pressure at that point.
OK.
Now, let's look at one-- last one.
Anything on that before I [inaudible]?
OK.
Let's put them right here.
Yes.
OK.
Let's do this as I think-- yeah [inaudible].
OK.
Let me read the problem 2137, 2137.
If you are missing any homework, here is all
the homework not picked up from [inaudible]
submitted.
It's graded.
It's in my dead homework file.
Nobody picked it up.
If you want to study for the midterm, it'd
be a [inaudible].
This is only for today.
This won't do me any good of course.
I can [inaudible].
They won't do me any good tonight, tomorrow
night.
I don't want to see it.
You might want to see it for Wednesday.
So I'll get back to you.
I'll turn it back in Wednesday, exam day that's
just to make sure you get it done by Monday
so you can study Tuesday of exam.
OK.
I'll read it.
I turn it in with the exam on Wednesday.
OK.
Let's see where we're working.
Oh I see, problem three-- once the135, 136,
137 were all assigned.
I'm going to work 2137, 2137.
OK.
Let's draw a picture here.
A picture given in test book, so here is a
container of some liquid.
The height here is h.
The area of this cylindrical container is
A. And the fluid is gamma one, has specific
weight, gamma one.
It says that the cylinder is now put into
the liquid, partially submerged in the liquid,
so the second picture then looks like this.
And of course the water level-- water-- here's
the water-- rises.
Here's the cylinder.
OK.
That cylinder in the textbook, he shows that
the volume V, specific weight gamma two, and
the water level rises by an amount in delta
h.
And then he says, show that this equation
for delta h is true.
Show that the equation in textbook for delta
h is true.
So in other words, provide delta h in terms
of some of these variable things in this problem.
That cylinder is moving like that, so it's
in a buoyant state.
And if we do a free-body diagram on the cylinder--
-- of course, we have the weight acting down
and we have the buoyant force acting up.
So we have our weight of the cylinder equal
F buoyant, And let's see, the weight of the
cylinder.
Oh, [inaudible] here it is.
The weight of the cylinder is the volume times
gamma 2.
The whole volume of the cylinder, V slash,
the buoyant force equal gamma 1 H2A2 equal
gamma 1V2.
And I'm going to find H2.
The 
buoyant force is the weight of the liquid
displaced.
The weight is equal to gamma times the volume.
So-- And A2, let's put that up here, A2 is
there.
So it's gamma 1 V plus V2, V2-- I'll draw
a picture here.
There's V2.
It's a portion of cylinder that's under water.
No, no, no.
OK.
So-- So now we have to note that those volumes.
The 
volume in the tank final-- The volume initial
in the tank plus the volume of the cylinder
submerged.
Who's going to answer?
This is the left hand side.
The volume, the final volume in the tank,
that's right there between [inaudible].
That's equal to the-- look at the water around
here.
That's the amount of water that was over here
to start.
It didn't leaked out.
Nobody pour more water in.
The water around here is the same as the water--
I'll call it water-- there's a water here
[inaudible], plus the volume of the cylinder
submerged, this much, which is V2.
OK.
So it's just a statement about the volumes.
OK.
Final volume of the tank is the initial volume
of water in the tank plus the volume of the
submerged cylinder.
OK.
Now, let's go ahead and put those numbers
in there.
Final volume of the tank is equal to, OK,
H plus delta H times A, H times A plus V2.
Submerged volume V2, half of H is here.
The water in the tank is H times A. Final
volume of the tank is between my hands and
the top area is A. Capital H [inaudible].
Capital HA, capital HA, gone.
So from this, I get that P2 delta H times
A.
[ Inaudible Remark ]
Yeah, yeah, H times A, they'll turn into [inaudible].
Now, take this guy here, take this guy here,
put that V2.
And for that V2, here's the equation then
right here.
This much right here, OK, solve for that delta
H. I'll put them in here, gamma 1H2A2 equal
gamma 1 times V2, delta H times A. Gamma,
gamma cancels out.
Oh no, they don't.
Pardon me.
They don't cancel out [inaudible] here.
This gamma over here is gamma 2.
This gamma over here is gamma 1, so gamma
1.
This is gamma 2.
Pardon me.
OK.
And so, we're going to solve for delta H.
OK.
Let's see what my volume is here.
[ Multiple Speakers ]
OK.
Where did-- OK.
This wasn't in here when I started.
This was in here when I started.
This I suppose-- yeah, that's in there.
This wasn't in there and that wasn't in there.
So you have to define other things that we
defined for you to solve the problem, and
that's a tough part of the problem.
How do I know if I put an H here?
Because you want a volume?
How do I know to put an H2 here?
Because you want a volume.
How do I know to put an A2 here?
Because you're going to need that A2 to get
a volume.
So, all those things go into volume.
They're not shown in the sketch in the textbooks,
so that's where people have their difficulty
is that they don't really see where that comes
from.
OK.
So I think that's all on that one.
Let me see if I got the weight right.
Gamma 2 which is the cylinder times the volume,
right.
Gamma 1 is the water from the [inaudible],
H2A2, H2A2.
That's where I missed.
No wonder.
OK.
That is a minus sign right here, a minus sign.
