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?

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

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?

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:

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

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.

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.

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)

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.

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.

Re: Pressure/Area word problem of a oil tank help?

I get 12,240 lb.

- Hollywood

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.

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:

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