Calculating Paint Volume for a 100x100 Block Structure

  • Thread starter Thread starter mrShaun
  • Start date Start date
  • Tags Tags
    Paint
AI Thread Summary
To calculate the paint volume for a 100x100 block structure with each block measuring 97cm wide, 97cm deep, and 93cm tall, the total surface area must be determined. Each block has a surface area of 55,678 cm², leading to a total of 556,780,000 cm² for 10,000 blocks. Converting this to square meters gives 55,678 m². With 1 liter of paint covering 3 m², approximately 18,559 liters of paint are needed, which rounds up to about 185.5 million liters for the entire structure. The discussion highlights the importance of clarifying measurements and assumptions when calculating paint requirements.
mrShaun
Messages
1
Reaction score
0
Please help (wrote the question wrong) please help

Me and my friend have this question we were asked and we don't know it but if you have 100 blocks by 100 blocks 97cm wide 97cm deep 93 cm tall.

1 liter of paint covers 3 square meters how many liters of paint would you need rounded up.

Please help and thanks
 
Last edited:
Physics news on Phys.org
Assuming the measurements referr to the individual blocks, they each have 4 sides that are 97*97 and 2 that are 97*93. Hence each block is 55678 cm^2 and need 18559 1/3 liter of paint. You have ten thousand, so multiply that in too. You'll end up needing about 185.5 million liters (~49 million US gallon, roughly the total US house paint production during the past two months, btw. Hey, perhaps I should write pop science..).

While it's got to be some pretty strange paint to only cover 3 cm^2/l (~1.76 square inches per gallon. A gallon wouldn't be enough to paint the front and back of a postage stamp), that be the numbers..
 
LarrrSDonald said:
While it's got to be some pretty strange paint to only cover 3 cm^2/l
Uh...
3 square meters.
 
You have an array of blocks: 100x100, correct? That's a total of 10,000 blocks.

You want all six sides of each block painted? You don't specify.

Presuming you do, each block will have a surface area of
2x 97x97 + 4x 97x93 cm^2.
Now multiply that by the number of blocks: x10,000 - which equals your total surface area in cm^2 - then convert it to m^2 (i.e. coincidentally, dividing by 10,000).

You now know the total area to bre covered, in m^2, and you know 1 litre covers 3 m^2. How many litres will you need?
 
It was originally 1l/cm^2, the post has since been edited (with good reason) to reflect the actual problem (personally I usually put littled [EDIT] notes at the bottom explaining potential changes to avoid just this sort of problem, that way it's more obvious what was changed and why later posts don't necessarily reflect the new content). I was told this in a PM but since it smelled a bit of homework in my book I just went over the methods.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top