# Pressure/Area word problem of a oil tank help?

• Feb 24th 2014, 06:38 PM
canyouhelp
Pressure/Area word problem of a oil tank help?
A cylindrical oil storage tank 12 feet in diameter and 17 feet long is lying on its side. Suppose the tank is half full of oil weighing 85 lb per cubic foot. What's the total force on one end of the tank? The relevant relationships are that pressure http://static.k12.com/eli/ecollege/5...t_571426_1.jpgwhere p is fluid density and h is height below the surface of the fluid, andhttp://static.k12.com/eli/ecollege/5...t_575167_1.jpg where Area is the area of the surface under pressure p.

I spent hours on this problem and got a force of around 6,000lbs but apparently that's not right! How would you do this problem? What should I have gotten?
• Feb 25th 2014, 09:48 AM
hollywood
Re: Pressure/Area word problem of a oil tank help?
Let h be the distance below the surface of the liquid. You need to find an equation for the width of the tank as a function of h, call this w(h). Then take the integral from h=0, the surface of the liquid, to h=6, the bottom of the tank of dF, which is $\displaystyle \int\, dF = \int \rho h\,dA = \int_0^6 \rho h w(h) \,dh$.

- Hollywood
• Feb 25th 2014, 11:28 AM
canyouhelp
Re: Pressure/Area word problem of a oil tank help?
Quote:

Originally Posted by hollywood
Let h be the distance below the surface of the liquid. You need to find an equation for the width of the tank as a function of h, call this w(h). Then take the integral from h=0, the surface of the liquid, to h=6, the bottom of the tank of dF, which is $\displaystyle \int\, dF = \int \rho h\,dA = \int_0^6 \rho h w(h) \,dh$.

- Hollywood

I'm not sure how to find the width with this integral. When I try to integrate it I end up with some complicated number *w, So I know I'm doing it wrong. I'm not used to so many variables in one definite integral. How do you integrate it?
• Feb 25th 2014, 11:48 AM
ebaines
Re: Pressure/Area word problem of a oil tank help?
It may be easier to do this using polar coordinates. You can set $\displaystyle h = R(1-sin \theta)$, $\displaystyle w(h) = 2R\cos \theta$ and $\displaystyle dh = R \cos \theta d \theta$. So the integral becomes:

$\displaystyle F = \int_{-\pi/2} ^{\pi/2} 2 \rho R^3 (1-\sin\theta)\cos^2\theta d \theta$
• Feb 25th 2014, 12:01 PM
canyouhelp
Re: Pressure/Area word problem of a oil tank help?
Quote:

Originally Posted by ebaines
It may be easier to do this using polar coordinates. You can set $\displaystyle h = R(1-sin \theta)$, $\displaystyle w(h) = 2R\cos \theta$ and $\displaystyle dh = R \cos \theta d \theta$. So the integral becomes:

$\displaystyle F = \int_{-\pi/2} ^{\pi/2} 2 \rho R^3 (1-\sin\theta)\cos^2\theta d \theta$

Okay it took me a while, but I think it helped to put it in those terms. When I integrated this I got that F=pi*p is this right? I could've gotten several things wrong so it might be off.
• Feb 25th 2014, 01:00 PM
ebaines
Re: Pressure/Area word problem of a oil tank help?
I get $\displaystyle \pi \rho R^3$. Always check your units - your answer of $\displaystyle \pi \rho$ is not in units of force.
• Feb 25th 2014, 01:37 PM
canyouhelp
Re: Pressure/Area word problem of a oil tank help?
Quote:

Originally Posted by ebaines
I get $\displaystyle \pi \rho R^3$. Always check your units - your answer of $\displaystyle \pi \rho$ is not in units of force.

Okay, so when it's in units of force as pi(p)R^3 how do I find the p and R variables? (That may be a stupid question, but I've never worked out this kind of problem before. At least not from a word problem, it's really confusing me)
• Feb 25th 2014, 01:43 PM
ebaines
Re: Pressure/Area word problem of a oil tank help?
You've been given the data that $\displaystyle \rho$ (density) is 85 pounds/ft^3 and R = 6 ft.
• Feb 25th 2014, 01:52 PM
canyouhelp
Re: Pressure/Area word problem of a oil tank help?
Quote:

Originally Posted by ebaines
You've been given the data that $\displaystyle \rho$ (density) is 85 pounds/ft^3 and R = 6 ft.

Oh, duh! I can be so blind sometimes. When I use that I get that F= 57,679. But is that for just one side of the tank? It seems high.
• Feb 25th 2014, 06:50 PM
hollywood
Re: Pressure/Area word problem of a oil tank help?
I get 12,240 lb.

- Hollywood
• Feb 26th 2014, 07:34 AM
canyouhelp
Re: Pressure/Area word problem of a oil tank help?
Quote:

Originally Posted by hollywood
I get 12,240 lb.

- Hollywood

I'm not sure how I'm integrating this one wrong, but 12,240lbs would make a lot more sense. I'll ask my professor to explain this one to me.
• Feb 26th 2014, 09:40 AM
ebaines
Re: Pressure/Area word problem of a oil tank help?
Quote:

Originally Posted by ebaines
I get $\displaystyle \pi \rho R^3$.....

This is not correct - my apologies but I realize I calculated the force as if the tank was full, not half full. The integral that should be evaluated is:

$\displaystyle \int _{-\pi/2} ^0 2 \rho R^3(1-\sin \theta) \cos^2 \theta d \theta$

which yields:

$\displaystyle F = 2 \rho R^3 \left(\frac {\theta} 2 + \frac { \sin \theta \cos \theta} 2 + \frac {\cos^3 \theta} 3 \right] _{- \pi/2} ^0 = 2 \rho R^3(\frac {\pi} 4 + \frac 1 3 )$