Crazy mathematical question

[derran.m]

I think this is actually a redundant question without all of the information...

my boss just asked me this question:


if 1 cubic meter = 2300 kg, how many pounds per square foot is that?


Any math whiz's in here that can help?

From what I know ... cubic = L x W x H, whereas squared = LxW, thus you cannot convert cubic to squared without knowing the H

[ercchry]

you would need to know the footprint of the object in question

[Jlude]

Originally posted by ercchry
you would need to know the footprint of the object in question

doesn't he mean that one meter cube as in 3ftx3ftx3ft=2300kg or is it one meter cubed of a particular object.

If it's the former, then you have the information needed to find out the weight per sq ft.

[derran.m]

1 cubic meter @ 6" high ... does that help? should be able to just work it backwards from there, correct?

[ercchry]

Originally posted by derran.m
1 cubic meter @ 6" high ... does that help? should be able to just work it backwards from there, correct?

yeah that helps... if it was 1mx1mx1m then the force/sqft would be much greater then that of your 6" high object

[colinxx235]

1 m^3 = 35.315 ft^3

35.315/6 = 5.886 ft^2 at the base (assuming its 6 feet high?)

2300kg = 5070.63lbs


therefore your pressure (what he's asking you) in PSF (weird)

5070.63lbs/5.886 ft^2 = 861.47lb/ft^2

if it was 6 inches high than you are looking at...

71.79lb/ft^2

[V6-BoI]

Originally posted by derran.m
1 cubic meter @ 6" high ... does that help? should be able to just work it backwards from there, correct?

Hmm.. you will only know the height, you still need to know length and width. Unless you can assume it's they are the same, which will turn out to be about 8.16m?

[colinxx235]

Originally posted by V6-BoI


Hmm.. you will only know the height, you still need to know length and width. Unless you can assume it's they are the same, which will turn out to be about 8.16m?


why does one need to know the length/width?? given the height + volume, you can deduce the contact area at the bottom...

[Mibz]

Can Mibz do junior high math? Let's find out...

6" = .1524 meters so the surface area of the bottom would be 1/0.1524 = 6.562 square meters.
2300 / 6.562 = 350.5 kg/sqm = 772.9 lb/sqm = 71.83 lb/sqft with a lot of rounding

Did I do it right?

EDIT: *applause please*

[V6-BoI]

Originally posted by colinxx235



why does one need to know the length/width?? given the height + volume, you can deduce the contact area at the bottom...

That's a good point, the area will always be the same given the height and volume. Wasn't thinking for a second there.

[Tarrantula]

Originally posted by Mibz
Can Mibz do junior high math? Let's find out...

6" = .1524 meters so the surface area of the bottom would be 1/0.1524 = 6.562 square meters.
2300 / 6.562 = 350.5 kg/sqm = 772.9 lb/sqm = 71.83 lb/sqft with a lot of rounding

Did I do it right?

EDIT: *applause please*

:clap:

NOTE: I do not know if this is correct.

[derran.m]

Originally posted by Mibz
Can Mibz do junior high math? Let's find out...

6" = .1524 meters so the surface area of the bottom would be 1/0.1524 = 6.562 square meters.
2300 / 6.562 = 350.5 kg/sqm = 772.9 lb/sqm = 71.83 lb/sqft with a lot of rounding

Did I do it right?

EDIT: *applause please*

I was a bit slow but i got 71.79 so close enough ... told the boss lady 71.8 and apparently that works out .....

thanks guys!

[colinxx235]

Originally posted by derran.m


I was a bit slow but i got 71.79 so close enough ... told the boss lady 71.8 and apparently that works out .....

thanks guys!


yah... peep my answer fool, I got 71.79

[Tarrantula]

Originally posted by colinxx235



yah... peep my answer fool, I got 71.79

BETTER RECOGNIZE!

:werd:

[LLLimit]

I was wondering why you you needed weight/area from density, then I remembered typical density of concrete 2300 kg/m3

[ercchry]

i have a greeeat math problem myself right now...

i have a hand drawing of measurements of interior rooms of an office that i have not seen and do not have access to... only info i have to go off of is this drawing...

two rooms with an exterior combined dimension of 586" the interior lengths of the rooms are 16'7.5" and 28.5' how thick are the walls? ... 3 walls.... 44".... conclusion? measuring tape malfunction/failure :banghead:

so in conclusion i give up for the day... love work sometimes

[derran.m]

Originally posted by colinxx235
1 m^3 = 35.315 ft^3

35.315/6 = 5.886 ft^2 at the base (assuming its 6 feet high?)

2300kg = 5070.63lbs


therefore your pressure (what he's asking you) in PSF (weird)

5070.63lbs/5.886 ft^2 = 861.47lb/ft^2

if it was 6 inches high than you are looking at...

71.79lb/ft^2


Originally posted by colinxx235



yah... peep my answer fool, I got 71.79

So sorry lol when i refreshed I just went straight to the bottom to get the most recent response ... my bad, you win the nerd of the day award, not mibz ... lol ... just kidding

I do appreciate the help!
And just to answer the inquiry of why we needed to know PSF ... we're in trucking ... and our customer could not tell us exact weight of the unit, but did know the density of the material. We just needed to run the numbers to calculate the weight per square foot so we would know how heavy it actually is in its entirety.

[derran.m]

Originally posted by Tarrantula


BETTER RECOGNIZE!

:werd:
done :P LOL

[derran.m]

Originally posted by LLLimit
I was wondering why you you needed weight/area from density, then I remembered typical density of concrete 2300 kg/m3
:werd:
tis a concrete platform we are trying to figure out the weight of