Work and Fluid Force: Calculus II

[LCKurtz]

y-1, if you plug in all the values...it would be on the triangle, except for -5.

For -5 to be on it...the formula just has to be 'y'

Which is probably not right...?

You are just guessing. Fill in the table with the correct numbers and show it to me.

 

[think4432]

You are just guessing. Fill in the table with the correct numbers and show it to me.

The thing is I don't know where to plug in the numbers to fill in the table, which is why I am trying to think of a formula to plug in those numbers that would be on the triangle.

 

[think4432]

No, not even close. For one thing, a dy integral requires y limits.

You have this disk slab of water whose weight is wπ/3. This disk is at location y vertically. How far does it have to be lifted to get out the top? Because, remember, work is force times distance.

It has to be lifted the the height of the function? Which is just 3x^2?

But in terms of y,

x = y^(1/2)/root 3

So integration of a=0 to b=2 and the integration of wpi [ y/3 - y^(1/2)/root 3 ]

?

 

[LCKurtz]

The thing is I don't know where to plug in the numbers to fill in the table, which is why I am trying to think of a formula to plug in those numbers that would be on the triangle.

Draw the line y = -2 across your triangle. How deep is that line under the waterline? Put that number in the table for the actual depth. Do the other values of y. Once you have filled in the table, then see if you can figure out the formula. (It's obvious, you're going to kick yourself).

 

[LCKurtz]

No, not even close. For one thing, a dy integral requires y limits.

You have this disk slab of water whose weight is wπ/3. This disk is at location y vertically. How far does it have to be lifted to get out the top? Because, remember, work is force times distance.

It has to be lifted the the height of the function? Which is just 3x^2?

You were given a fixed parabolic container. It doesn't go on forever. How tall is it (a specific number)? How far does a slab of water at height y have to be lifted to get to the top?

So integration of a=0 to b=2 ..

Did you even read what I said about y limits? Those are still x limits.

 

[think4432]

Draw the line y = -2 across your triangle. How deep is that line under the waterline? Put that number in the table for the actual depth. Do the other values of y. Once you have filled in the table, then see if you can figure out the formula. (It's obvious, you're going to kick yourself).

If I draw a y = -2 line, it would be 7 ft under the top of the liquid.

So 7 ft is the actual depth.

And I get numbers like 6, 5, 4, 3, and 2. Which are all about the y = -2 line..and under the liquid, correct?

So would my height just by 7 + y?

Thats probably not it..because its not obvious to me, that it would be 7 + y

 

[think4432]

You were given a fixed parabolic container. It doesn't go on forever. How tall is it (a specific number)? How far does a slab of water at height y have to be lifted to get to the top?



Did you even read what I said about y limits? Those are still x limits.

Y limits...so I would just plug in the x limits in the y = 3x^2 and get [0, 12] for the limits?

So if the y limits are 0, 12 ...the distance would be 12-y?

 

[LCKurtz]

If I draw a y = -2 line, it would be 7 ft under the top of the liquid.

What picture are you looking at? Isn't it true in the picture you posted that no part of the triangle is more than 5 ft under the waterline??

 

[LCKurtz]

Y limits...so I would just plug in the x limits in the y = 3x^2 and get [0, 12] for the limits?

So if the y limits are 0, 12 ...the distance would be 12-y?

Yes for the distance of 12 - y. But the container isn't full of water to begin with and you don't have to lift water from where there isn't any in the first place. So the upper limit is ? And your final integral is what?

 

[think4432]

What picture are you looking at? Isn't it true in the picture you posted that no part of the triangle is more than 5 ft under the waterline??

Im looking at the same picture.

Maybe I am just really really slow at this or something, but I thought a line a y= -2 would be below the triangle...but that has nothing to do with it.

:[

Im glad you're very patient.

OH. If I do y + 5...and plug all those values in..then they would all be in that region! [above the triangle...not under like y + 7]

Right?!?

 

[think4432]

Yes for the distance of 12 - y. But the container isn't full of water to begin with and you don't have to lift water from where there isn't any in the first place. So the upper limit is ? And your final integral is what?

Upper limit is 4! Because 4 ft is where the water is filled too!

(62.4 ft/lb^3)pi from a = 0, b = 4 integrating [y/3 - (12-y)] << final integral. I hope.

 

[LCKurtz]

Im looking at the same picture.

Maybe I am just really really slow at this or something, but I thought a line a y= -2 would be below the triangle...but that has nothing to do with it.

When you started this problem, one thing you did was figure out the equation of the slanted side. When you did that you established your coordinate system. Judging from the equation you got I think your origin is at the water level 1 unit above the point of the triangle, with y positive upward. You have to use that coordinate system for all of your measurements. And use it to fill in that table.

Im glad you're very patient.

No problem, I had a little time to kill this afternoon. So show me what you get when you fill in the table now.

 

[LCKurtz]

Upper limit is 4! Because 4 ft is where the water is filled too!

(62.4 ft/lb^3)pi from a = 0, b = 4 integrating [y/3 - (12-y)] << final integral. I hope.

Close. Work is force times distance, not force minus distance.

 

[think4432]

When you started this problem, one thing you did was figure out the equation of the slanted side. When you did that you established your coordinate system. Judging from the equation you got I think your origin is at the water level 1 unit above the point of the triangle, with y positive upward. You have to use that coordinate system for all of your measurements. And use it to fill in that table.


No problem, I had a little time to kill this afternoon. So show me what you get when you fill in the table now.

I thought I had it with y + 5, Now I am back at square 1.

I plugged in the values and got 4,3,2,1

Confused once again.

 

[think4432]

Close. Work is force times distance, not force minus distance.

Oh! So the integral is:

(62.4 ft/lb^3)pi from a = 0, b = 4 integrating [y/3 X (12-y)] << final integral. I hope.

or integrating [(12y-y^2)/3] from the limits 0,4 with 62.4pi out in front of the integral?

 

[LCKurtz]

I thought I had it with y + 5, Now I am back at square 1.

I plugged in the values and got 4,3,2,1

Confused once again.

Show me the whole table using your coordinate system and looking at your picture:

y = -2 depth = ?
y = -3 depth = ?
...

 

[LCKurtz]

Oh! So the integral is:

(62.4 ft/lb^3)pi from a = 0, b = 4 integrating [y/3 X (12-y)] << final integral. I hope.

or integrating [(12y-y^2)/3] from the limits 0,4 with 62.4pi out in front of the integral?

Yes. Ta Daaah! I hope you understand it well enough now that you could explain it to others.
(Do you have the right units on the 62.4?)

 

[think4432]

Yes. Ta Daaah! I hope you understand it well enough now that you could explain it to others.
(Do you have the right units on the 62.4?)

Once I read over this forum a couple of times, I will be able to explain it to others.

It says in the problem that w = 62.4 lb/ft^3

So that's what I use for the w, correct?

THANK YOU SO VERY MUCH FOR YOUR HELP ON THIS PROBLEM!

I don't know how many times I would have been wrong with your help!

Thank you so much...one problem down, one more to go!

 

[LCKurtz]

You're welcome. I'm off to dinner now. Will check back in later to see if you have the depth formula figured out.

 

[think4432]

Show me the whole table using your coordinate system and looking at your picture:

y = -2 depth = ?
y = -3 depth = ?
...

Would the depth be to the top of the liquid?

If that's the case, then when y = -2 then the depth would be 7...and that's not in the picture.

y = -3 then 8

But that's obviously not right...right?

 

[think4432]

You're welcome. I'm off to dinner now. Will check back in later to see if you have the depth formula figured out.

Thank you!

You are definitely my hero!

 

[LCKurtz]

Would the depth be to the top of the liquid?

If that's the case, then when y = -2 then the depth would be 7...and that's not in the picture.

y = -3 then 8

But that's obviously not right...right?

Here's a picture with the y values shown. You should be able to fill in the depth table by looking at it.

 

[think4432]

Here's a picture with the y values shown. You should be able to fill in the depth table by looking at it.

The depth from the top of the fluid to the end of the triangle is 5 feet
At -1: 4 feet
At -2: 3 feet
At -3: 2 feet
At -4: 1 feet
At -5: 0 feet

-5-y = formula?

-5 - 4 = -1
-5 - 3 = -2
-5 - 2 = -3
-5 - 1 = -4
-5 - 0 = -5

Very nice picture, by the way! Thank you!

 

[think4432]

Thank you!

You are definitely my hero!

I integrated this out and got 2(y^2) - (y^3)/9

And plugged in the limits from 0 to 4 with the pi and 62.4 out in front and came up with an answer of

224/9 pi (62.4)

And when I multiplied the 62.4 out it came out to be 1553.066 pi

Would this be correct? Just seems bit of a weird number?

Please check it out for me?

 

[LCKurtz]

The depth from the top of the fluid to the end of the triangle is 5 feet
At -1: 4 feet
At -2: 3 feet
At -3: 2 feet
At -4: 1 feet
At -5: 0 feet

-5-y = formula?

-5 - 4 = -1
-5 - 3 = -2
-5 - 2 = -3
-5 - 1 = -4
-5 - 0 = -5

Very nice picture, by the way! Thank you!

Now I see what you are thinking incorrectly. The term depth means how far underwater is that y value. The values in the table should be how far under water is it -- the length of the column of water above it, not below it. Try again.

 

[think4432]

Now I see what you are thinking incorrectly. The term depth means how far underwater is that y value. The values in the table should be how far under water is it -- the length of the column of water above it, not below it. Try again.

So

At -1: 1 feet
At -2: 2 feet
At -3 : 3 feet
At -4: 4 feet
At -5: 5 feet

?

 

[LCKurtz]

So

At -1: 1 feet
At -2: 2 feet
At -3 : 3 feet
At -4: 4 feet
At -5: 5 feet

?

Finally! Yes. So what is the formula; it isn't 5 - y. If I recall correctly, that is the only thing that was wrong with your setup.

 

[think4432]

Finally! Yes. So what is the formula; it isn't 5 - y. If I recall correctly, that is the only thing that was wrong with your setup.

I understand that it is the depth in the y direction and all...but I got to admit that I still do not know how to figure out the formula.

I know this is probably really bad...but I just don't know!

I also replied to the other question after I integrated it...could you please check that for me? The answer...

Because it seems really odd to me!

 

[think4432]

Finally! Yes. So what is the formula; it isn't 5 - y. If I recall correctly, that is the only thing that was wrong with your setup.

Wait, After staring at this for a while...

Would -y work?

-(-1) = 1 feet

-(-2) = 2 feet

-(-3) = 3 feet

-(-4) = 4 feet

-(-5) = 5 feet

?!?

 

[LCKurtz]

Wait, After staring at this for a while...

Would -y work?

-(-1) = 1 feet

-(-2) = 2 feet

-(-3) = 3 feet

-(-4) = 4 feet

-(-5) = 5 feet

?!?

Yes. And if you look at the picture is should be "obvious" that the depth is -y.