By how much is each post compressed by the weight of the aquarium?

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A large 3.00×10^4 L aquarium is supported by four wood posts (Douglas fir) at the corners. Each post has a square 5.10 cm x 5.10 cm cross section and is 90.0 cm tall.

By how much is each post compressed by the weight of the aquarium?

In millimeters.

I have no idea how to do this. I’m not asking for just an answer necessarily (though I’d take that too), but if you could lead me through to the answer, that would be fantastic!

Thanks!

Ahhh… I see now that hanggle also stated that error. Ha, no bigz! Thanks to you both for all the help!

• Mass of aquarium: density of water volume = 3.0010^4 kg

Area of one post: 0.051m * 0.051m = 0.0026 m^2

Total area of post: 0.0026*4 = 0.0104 m^2

(delta)L = (F L)/(Y A) <– Y = 10^10 for wood

=(310^4 9.8 0.9)/(10^10 0.0104)

=0.0025 m

0.0025m = 2.5mm

BTW kustom,

3 10^4 9.8 supposed to be 294000

God Luck everyone, c u all on Wednesday 8am!

• Ok so we can use the equation given for this…

F/A = Y (delta L)/L

Where F is the force…

A is the total area where the force is being applied….

Y is the Young’s modulus, for fir its 10^10 N/m^2…

delta L is the length each post is compressed by…

L is the height of each post…

Solving for delta L gives us this…

delta L = FL / A Y

Force “F”: each Liter is a kilogram for water, thus the force is just 310^4kg 9.8m/s/s = 333200N.

L = .9m.

Area “A”: you want the total area, so the area of a post times 4..

A = 4 .051m .051m = .010404m^2

Y = 10^10 (should be given in a book or online ect…)

Solving FL / AY = (333200.9)/(.01040410^10) = .00288m

Multiply by 1000 for mm…

delta L = 2.88mm

Hope I explained it so it made some sense : )

Doing hmwk for Physics 201, WSU by any chance?

Saw this looking for help myself…

• Douglas Fir Posts