Calculating Compression of Aquarium Support Posts

AI Thread Summary
The discussion revolves around calculating the compression of wood posts supporting a large aquarium. The aquarium's volume is 3.00×10^4 L, leading to a mass of approximately 3.00×10^4 kg, given that water has a density of 1 kg/L. Participants clarify the correct cross-sectional area of each post, which is 0.00144 m², and the total area for four posts is 0.00576 m². The correct approach involves calculating the pressure on a single post before determining the compression using the formula deltaL = F*L_0/(Y*A). Miscalculations regarding area and force distribution are addressed, emphasizing the importance of correctly applying the principles of pressure and strain in the context of the problem.
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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!
 
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houseguest said:
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!
 
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
 
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
 
You divided by area twice.
 
  • #10
oh yeah, duh.

Thank you so much for your help! Seriously.
 

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