# Young's Modulus of aquarium

## Homework Statement

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?

## Homework Equations

deltaL = F*L_0/(Y*A)

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

## 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!

## Answers and Replies

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Shooting Star
Homework Helper
but, what is the mass of a 3.00×10^4 L aquarium?
Since nothing else is given, consider the significant portion of the weight of the aquarium to be due to just water.

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!

Mapes
Homework Helper
Gold Member
What's the total area on which the weight acts, in m2?

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

Mapes
Homework Helper
Gold Member
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:
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:
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

Mapes