Work and Fluid Force: Calculus II

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Homework Help Overview

The discussion revolves around two calculus problems related to work and fluid force. The first problem involves calculating the work required to pump water from a tank formed by revolving a quadratic function around the y-axis. The second problem focuses on determining the fluid force on a submerged plate in a fluid with a specified weight density.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the setup of integrals for both problems, with some questioning the correctness of limits and expressions used. There are attempts to clarify the relationship between depth and the variables involved in the integrals.

Discussion Status

Several participants are exploring different interpretations of the problems and the setup of integrals. There is an ongoing exchange of ideas regarding the correct expressions for depth and the mass of elements involved in the calculations. Some guidance has been offered, but no consensus has been reached on the correctness of the approaches taken.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance they can receive. There is a focus on understanding the setup of integrals and the physical interpretations of the problems.

  • #61
LCKurtz said:
Yes. And if you look at the picture is should be "obvious" that the depth is -y.

I see that it could be -y by looking at the picture and where the coordinate system is set...

But I posted this question earlier [again] on the forums...because I actually have to turn this in before 5pm today and I didn't know if you would be online before then or not.

And someone else told me something different.

Heres the thread:

https://www.physicsforums.com/showthread.php?t=411314

So now I have 2 different answers...and I am even more confused. :[
 
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  • #62
think4432 said:
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?

What's weird about it? Might as well mulitply it on out to get 4879.1. And don't forget to put proper units on it.
 
  • #63
LCKurtz said:
What's weird about it? Might as well mulitply it on out to get 4879.1. And don't forget to put proper units on it.

Hah.

I don't know..

Maybe the decimal? Or something.

I was hoping for like a nice perfect rounded answer [because we did a couple in class and they were pretty nice numbers...so I just figured that this one would be nice and perfect too]

Haha.


4879.1 lb/ft^3

Thank you! Once again!
 
  • #64
think4432 said:
I see that it could be -y by looking at the picture and where the coordinate system is set...

But I posted this question earlier [again] on the forums...because I actually have to turn this in before 5pm today and I didn't know if you would be online before then or not.

And someone else told me something different.

Heres the thread:

https://www.physicsforums.com/showthread.php?t=411314

So now I have 2 different answers...and I am even more confused. :[

Halls is setting it up using a different coordinate system. So everything looks different including the equation of the slanted side. It should eventually give the same answer.

[Edit] He also has a typo on the second line, 5 - x should be 5 - y
 
  • #65
think4432 said:
4879.1 lb/ft^3

Thank you! Once again!

What are the units for work? And do you get them in this problem?
 
  • #66
LCKurtz said:
Halls is setting it up using a different coordinate system. So everything looks different including the equation of the slanted side. It should eventually give the same answer.

[Edit] He also has a typo on the second line, 5 - x should be 5 - y

I see!

Ok. But with my coordinate system.

The integral is (-y)(-7/4y - 7/4) dy from the limits [0,4]

So integrating 7/4(y^2) + (7/4)y

And get 7/24(y^2) (2y + 3)

154/3 as the final answer?
 
  • #67
LCKurtz said:
What are the units for work? And do you get them in this problem?

They gave us the units for the water density as lb/ft^3

Ohh.

But the units for work would be (ft)(lb)

Correct?
 
  • #68
think4432 said:
I see!

Ok. But with my coordinate system.

The integral is (-y)(-7/4y - 7/4) dy from the limits [0,4]

So integrating 7/4(y^2) + (7/4)y

And get 7/24(y^2) (2y + 3)

154/3 as the final answer?

The only thing you had wrong in your original post was the depth of 5-y instead of -y. Why are you now suddenly saying y goes from 0 to 4? Don't you remember your choice of coordinates?
 
  • #69
think4432 said:
They gave us the units for the water density as lb/ft^3

Ohh.

But the units for work would be (ft)(lb)

Correct?

Yes, and you should be able to explain how you get those units as the result of the units on the variables in your calculations.
 
  • #70
LCKurtz said:
Yes, and you should be able to explain how you get those units as the result of the units on the variables in your calculations.

Yes. Thats where I got the units from.
 
  • #71
LCKurtz said:
The only thing you had wrong in your original post was the depth of 5-y instead of -y. Why are you now suddenly saying y goes from 0 to 4? Don't you remember your choice of coordinates?

Oh! Oh! Oh!

Sorry! Sorry! I was looking at the other thread and got mixed up!

The limits are -5 to -1!

-154/3?

I did something wrong?
 
  • #72
think4432 said:
Oh! Oh! Oh!

Sorry! Sorry! I was looking at the other thread and got mixed up!

The limits are -5 to -1!

-154/3?

I did something wrong?

Do you have -5 as the lower limit and -1 as the upper like you should? It should come out positive and don't forget the water density.
 
  • #73
LCKurtz said:
Do you have -5 as the lower limit and -1 as the upper like you should? It should come out positive and don't forget the water density.

there's no pi in this problem, correct?

So the problem is just the limits a = -1 and b = -5 and integrating (7/24)(y^2) + (2y+3)

With the water density of 62.4 lb/ft^3 outside of the integral as a constant? Correct?
 
  • #74
I still get 154/3(62.4)

And then when I multiply that out I get 3203.2 ft(lb)
 
  • #75
LCKurtz said:
Do you have -5 as the lower limit and -1 as the upper like you should? It should come out positive and don't forget the water density.

think4432 said:
there's no pi in this problem, correct?

So the problem is just the limits a = -1 and b = -5...

Did you read my question: Do you have -5 as the lower limit and -1 as the upper like you should?
 
  • #76
LCKurtz said:
Did you read my question: Do you have -5 as the lower limit and -1 as the upper like you should?

Yes. Sorry I just had a typo A = -5 and B = -1

When you evaluate -5 [lower limit] you get -1225/24
And when you evaluate the -1 [upper limit] you get 7/24

So 7/24 - (-1225/24)

You get 154/3 times the weight density.

Which is 154/3 * 62.4

= 3203.2

?
 
  • #77
think4432 said:
I still get 154/3(62.4)

And then when I multiply that out I get 3203.2 ft(lb)

The numbers look OK now. But you need to quit guessing at the units. What was this problem asking for? What would the correct units for the answer be? Do you actually get those units?

I'm going offline now for the rest of the day. People arriving for big Father's Day party. Hopefully you can finish it up.
 
  • #78
LCKurtz said:
The numbers look OK now. But you need to quit guessing at the units. What was this problem asking for? What would the correct units for the answer be? Do you actually get those units?

I'm going offline now for the rest of the day. People arriving for big Father's Day party. Hopefully you can finish it up.

Ok! Thank you very much.

Happy Fathers Day to you! Hope you enjoy the rest of your day!

Thanks so much for your help though.

You are great.
 

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