# Young's Modulus of aquarium

1. Apr 2, 2008

### houseguest

1. The problem statement, all variables and given/known data

A large 3.00×10^4 L aquarium is supported by four wood posts (Douglas fir) at the corners. Each post has a square 3.80 cm x 3.80 cm cross section and is 80.0 cm tall.
By how much is each post compressed by the weight of the aquarium?

2. Relevant equations

deltaL = F*L_0/(Y*A)

Y (for Douglas fir) = 1 * 10^10 N/m^2

3. The attempt at a solution

This seems very straight forward, I just need to determine F.
F= ma = m(aquarium)g
but, what is the mass of a 3.00×10^4 L aquarium?
once I determine that, I figure it'll just be:

deltaL = ( m * 9.8 * .8)/(.144 * 10^10) meters
=> 5.44*10^10*m meters

How do I find m ??

Thanks for the help!

2. Apr 2, 2008

### Shooting Star

Since nothing else is given, consider the significant portion of the weight of the aquarium to be due to just water.

3. Apr 2, 2008

### houseguest

So I took the mass of water to be 1Kg per liter.
So the m = 3 * 10^4

Then the equation comes out as:

deltaL = ( 3 * 10^4 * 9.8 * .8)/(.144 * 10^10) meters
= 1.63 * 10^-4 m

But I am told this is the incorrect answer.

Any help???

Thanks!

4. Apr 2, 2008

### Mapes

What's the total area on which the weight acts, in m2?

5. Apr 2, 2008

### houseguest

4*.144 = .576 m^2

but, isn't it asking the deltaL for an individual post?

I thought that I might have to divide the force by 4, but that answer was also incorrect.

Thanks

6. Apr 2, 2008

### Mapes

Isn't it 0.0144 m2? Also, it's the pressure (stress) that causes the strain; you need to find the pressure on a single leg before you can calculate the strain. The way you originally had it, the compression distance was independent of the number of legs.

EDIT: Right, 0.00144 m2

Last edited: Apr 2, 2008
7. Apr 2, 2008

### houseguest

Hang on, I see the mistake I made earlier: 3.8 cm x 3.8 cm = .038m x .038m = .00144 m^2 (thanks btw), but wouldn't then total area then be 4 * .00144 = .00576 m^2 ?

Then the pressure would be 3 * 10^4 * 9.8 / .00576.

Then plug in that pressure for F in
deltaL = F*L_0/(Y*A)
Is that right?

Last edited: Apr 2, 2008
8. Apr 2, 2008

### houseguest

Although, it retrospect this can't be right since

deltaL = ( (3 * 10^4 * 9.8) / .00576 * .8) / (.00144 * 10^10) meters

= 2.84 m

and that is way to large to be realistic

9. Apr 2, 2008

### Mapes

You divided by area twice.

10. Apr 2, 2008

### houseguest

oh yeah, duh.

Thank you so much for your help!!! Seriously.