Permissible error

How to calculate the permissible error?
A force F is applied on a square plate of size L . If the percentage error in determination of L is 2% and that in F is 4% . What will be the permissible error in pressure?

rude man
Homework Helper
Gold Member
How to calculate the permissible error?
A force F is applied on a square plate of size L . If the percentage error in determination of L is 2% and that in F is 4% . What will be the permissible error in pressure?
EDITed:
Depends on how you define "permissible".
First, you say a "square" sheet. Is it really square? You have to assume "I don't know" so the area could be that of a general quadrilateral. That requires the measurement of 4 sides plus two opposite angles: http://keisan.casio.com/exec/system/1322718508
Plus a 7th independent measurement is for the force F.
So, assume the error in measuring each side's length and that of each of the two angles is 2%,
if the errors are random in all 7 cases, error propagation theory says to take the square root of the sum of the squares:
error = √(6(.022) + .042) = 0.063 = 6.3%.

But there is also the "worst-case" error which you would get if all your measurements are at their respective extremes, in which case the error would be (1.02)6(1.04) - 1 = 17.1%.

Last edited:
Thanks for the reply. Maybe you went for a higher level of explanation. I think that the question interconnects between force , area and pressure. As pressure = force/ area , can you precisely explain about the permissible error in pressure ?

rude man
Homework Helper
Gold Member
Thanks for the reply. Maybe you went for a higher level of explanation. I think that the question interconnects between force , area and pressure. As pressure = force/ area , can you precisely explain about the permissible error in pressure ?
If the error in force is 4.0% and the error in area is 2.0%, the error in pressure is √(.0402 + .0202) = 4.4%.

But L is likely to mean length, not area, so then it'd be √(.0402 + .0202 + .0202) = 4.9%.

haruspex
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If the error in force is 4.0% and the error in area is 2.0%, the error in pressure is √(.0402 + .0202) = 4.4%.

But L is likely to mean length, not area, so then it'd be √(.0402 + .0202 + .0202) = 4.9%.
We are told it is square. The two length errors are correlated?

rude man
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Gold Member
We are told it is square. The two length errors are correlated?
Good point. So it's just √(.022 + .042).

Evo
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Just a friendly reminder, if the template isn't used or the information required in the template isn't used, such as what the member has tried themselves to solve the problem, please do not reply, please send a report.

rude man
Homework Helper
Gold Member
Just a friendly reminder, if the template isn't used or the information required in the template isn't used, such as what the member has tried themselves to solve the problem, please do not reply, please send a report.
OK. Sorry. I thought after I wrote that it was probably excessive. Will be more on guard henceforth.
r m

Evo
Mentor
OK. Sorry. I thought after I wrote that it was probably excessive. Will be more on guard henceforth.
r m
You're cool, we're just trying to get things more even in responses, some members get turned back while some get answered.

haruspex