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# Weight - Tank - Solution/Explanation

Sun 15 Mar, 2015 11:34 pm
An open tank has the following dimensions :

Length - 2 ft.
Width - 2 ft.
Depth - 1.5 ft.
Thickness - 1/8 in.

Lead weighs approx. .42 lbs. per cu. in.

Determine weight of tank.

Volume = length x width x height or depth

Solution provided with problem :

External dimensions of tank - 24.25 in. in length, 24.25 in. in width, 18.125 in. deep.

External volume - 24.25 x 24.25 x 18.125 cu. in.

Volume - 600 x 18 = 12000 cu. in. *

Enclosed volume - 24 x 24 x 18 = 12000 cu. in.

External volume - 10650 cu. in.

Enclosed volume - 10370 cu. in.

Volume occupied by lead - 10650 - 10370 = 280 cu. in. **

280 x .42 (cu. in. of lead) = 117.6 = 120 lbs.

My solution:

24.25 x 24.25 x 18 = 10585.125 = 10585 cu. in.

10585 x .42 (cu. in. of lead) = 4445.7 lbs.

Which is correct ?

* What is the 600 ?

** Why is the enclosed volume subtracted from the external volume to obtain volume occupied by lead ? The tank is completely made of lead.

Non-Homework Question.

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bahtah

1
Mon 16 Mar, 2015 05:43 pm
@Randy Dandy,
I took the question at face value in that your looking for the weight of the tank and not the tank with any type of liquid inside. I also used the dimensions in the question as the overall size of the tank. All I care about is the area of the tank:
The bottom 2'x2'=4sqft. The equal sides 2'x1.5'=3sqft x 4 = 12 sqft. So my measurements are +- 1/8". The area of lead is 16sqft with a thickness of 1/8". If I half that area to 8sqft and double the thickness to 1/4" I have the same amount of material. Then if I half that so I end up with 4sqft at 1/2" thickness I can then figure the amount of lead: 24"x24"x1/2" =288 Cubic In x .42= 120.96lbs.
Randy Dandy

1
Mon 16 Mar, 2015 06:50 pm
@bahtah,
Thanks.

What is the 600 ?

Is the internal dimensions subtracted from the external dimensions due to the empty space ?
bahtah

1
Mon 16 Mar, 2015 10:37 pm
@Randy Dandy,
The 600 is the 24.5 x 24.5 in your formula.

The difference between the external size of the tank and the internal size of the tank is the material that makes up the walls of the tank. If you had for example a one gallon paint can and set a smaller can of the same heigth inside it, then filled the space between the cans with concrete, the amount of concrete needed would be the difference between the size of the one gallon paint can and the smaller can now inside it. When I looked at the question I had enough information to figure the material needed because the thickness of the tank was given (1/8") so it was easy to just figure the total area of all the sides and bottom with a depth of 1/8". I knew that if I took half the area and twice the thickness I would have the same amount of material so I did that a couple times to have larger numbers to work with in my final calculation.
Randy Dandy

1
Mon 16 Mar, 2015 11:35 pm
@bahtah,
Thanks again.
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