What Is the Hydrostatic Force on One End of a Tank Filled with Gasoline?

[sushifan]

1. Homework Statement
A large tank is designed with ends in the shape of the region between the curves y =(1/2)x^2 and y = 12, measured in feet. Find the hydrostatic force on one end of the tank if it is filled to a depth of 8 ft with gasoline. (Assume the gasoline's density is 42.0 lb/ft^3).

This is a section on applications.

2. Homework Equations

I'm not sure if this is precisely all I need.

Force F = ρgdA (g gravity, d depth, A area)

3. The Attempt at a Solution

(42.0)(9.8) ∫ sqrt(2y) (8-y) dy on [0,8]

823.2 ∫ 8 sqrt(2y) - ysqrt(2y) dy on [0,8]

823.2 [ 16sqrt(2)/3 y^(3/2) - 2sqrt(2)/5 y^(5/2)] on [0,8]

which evaluates to 56, 197.12 lb

 

[gneill]

You're mixing metric with imperial if you use 9.8 for g.

 

[sushifan]

You're mixing metric with imperial if you use 9.8 for g.

Oh, right! It's 32ft/sec^2, right?

 

[gneill]

Oh, right! It's 32ft/sec^2, right?

32.174 ft/sec²

 

[sushifan]

32.174 ft/sec²

Is there any way that you can tell me if my solution is in the right direction?

 

[gneill]

Is there any way that you can tell me if my solution is in the right direction?

Sure. It looks like you've got the right idea for the integration, and the equation being integrated looks appropriate.

 

[emailanmol]

Hey gneill and sushifan.

Really good problem.


How did you guys get dA as xdy?

My diagram suggests something else.May be am making a mistake :-(

 

[sushifan]

Hey gneill and sushifan.

Really good problem.How did you guys get dA as xdy?

My diagram suggests something else.May be am making a mistake :-(

I thought of my radius as x units as we traveled vertically on the y-axis, so I rewrote the given function in terms of y.

Could you share your set up of the problem? Maybe it's me who's making the mistake! I'm actually unsure of my solution.

 

[gneill]

Hey gneill and sushifan.

Really good problem.How did you guys get dA as xdy?

My diagram suggests something else.May be am making a mistake :-(

The contour is that of a parabola, y = (1/2)x² . So for a given y (corresponding to some height up from the vertex), the half-width is $\sqrt{2y}$ . The area element dA is then the total width by the height, dy, thus forming a thin rectangular area element. The "2" in "2y" must have units "feet" associated with it in order to keep units consistent.

 

[emailanmol]

I thought of my radius as x units as we traveled vertically on the y-axis, so I rewrote the given function in terms of y.
Could you share your set up of the problem? Maybe it's me who's making the mistake! I'm actually unsure of my solution.

Firstly, I would take pressure as Po (Atmospheric pressure)
+ phg as its not written to neglect the atmosphere. (Do this only if the answers not matching or not coming twice of what you got with the correct units).

Secondly, my area is coming exactly twice of what you got.(But it maybe wrong.Although I don't know why!)

The contour is that of a parabola, y = (1/2)x² . So for a given y (corresponding to some height up from the vertex), the half-width is $\sqrt{2y}$ . The area element dA is then the total width by the height, dy, thus forming a thin rectangular area element. The "2" in "2y" must have units "feet" associated with it in order to keep units consistent.

So the area of the small rectangle should be
2*(2y)^1/2*dy .

Right?

And really good observation on the units gneill :-)

 

[sushifan]

Secondly, my area is coming exactly twice of what you got.(But it maybe wrong.Although I don't know why!)

Ah, I have to take the integral twice to cover both sides of the parabola. That's what I forgot.

 

[emailanmol]

Does the answer match?

 

[sushifan]

Does the answer match?

Well I'm not sure what numerical value you got for your answer. You said you got twice my answer; do you mean twice my answer (56, 197.12) from my very first post?

Because that answer was wrong, since I used 9.8 for gravity and not 32.

So, I'm redoing the entire calculation.

It would help if you actually provided your set up or at least your final answer so I can compare.

 

[emailanmol]

No i just got my expression twice as what you got in terms of variables.

I was asking if your answer matches with that of textbook ?

I will post my answer in few mins .Let me calculate :-)

 

[sushifan]

I was asking if your answer matches with that of textbook ?

My textbook unfortunately does not come with an answer key. :[

And I'm doing my calculation right now, too.

 

[emailanmol]

See, i would refrain from posting the answer as its against the Guideline :-)

However,My expression is exactly double of your original post (with proper units of g offcourse)

 

[sushifan]

See, i would refrain from posting the answer as its against the Guideline :-)

However,My expression is exactly double of your original post (with proper units of g offcourse)

Didn't know that. I'm new here.

And apologies for the late reply, I fell asleep.

 

[gneill]

Didn't know that. I'm new here.

And apologies for the late reply, I fell asleep.

It's against the guidelines for those offering help to do the work for, or to simply give answers to the questioner. It's quite okay if the questioner does the work and comes to a correct answer