Problem in evaluating the absolute error

[patric44]

Homework Statement

an empty large box of mass (20+-0.01)kg , when a man sits inside the box , the mass of the box and the man together become (0.1+-0.001) ton, find the mass of the man.

Relevant Equations

m=m+-Δm

hi guys
i came across this simple question about evaluating the absolute error, the solution is very simple as following :
$$
m_{man} = M_{b+m}\pm\Delta\;M_{b+m}-M_{b}\pm\Delta\;M_{b}
$$
so the mass of the man alone is
$$
m_{man} = (M_{b+m}-M_{b})\pm(\Delta\;M_{b+m}+\Delta\;M_{b})
$$
which would equal
$$
80\pm1.01
$$
but the argument is if i tried to calculate this the other way around assuming that the mass of the man is $80\pm0.99$ every thing will hold, as
$$
m_{b+m} = (M_{m}+M_{b})\pm(\Delta\;M_{m}+\Delta\;M_{b})=100\pm1\;kg
$$
why is that?

 

[Steve4Physics]

Homework Statement:: an empty large box of mass (20+-0.01)kg , when a man sits inside the box , the mass of the box and the man together become (0.1+-0.001) ton, find the mass of the man.
Relevant Equations:: m=m+-Δm
.
$80\pm0.99$ every thing will hold, as
$$
m_{b+m} = (M_{m}+M_{b})\pm(\Delta\;M_{m}+\Delta\;M_{b})=100\pm1\;kg
$$
why is that?

I guess ‘80\pm 0.99’ should be $80\pm 0.99$.

You appear to have calculated ‘0.99’ by subtracting 0.01 from 1. But, when calculating the uncertainty of a sum or difference, you should always add the uncertainty values.

For example: x=10±1 and y=5±1, so x-y = 5±2.
But if you are given, with no prior knowledge, that x-y = 5±2 and y=5±1 then
x = (x-y) +y = 10±3 (not x = 10±1)

Some other points which might help:

1. A value should be specified to the same precision as its uncertainty.
E.g. (20±0.01) should be written as 20.00±0.01

2. The uncertainty of 0.01kg is very small compared to the uncertainty 1kg. And each uncertainty is specified to only 1 significant figure. The value of an uncertainty is itself an uncertain. So it doesn’t make much sense to say the combined uncertainty as 1.01kg – the combined uncertainty should be rounded to 1kg.

(Uncertainty values are typically only known to 1 or 2 significant figures. If you were told that the two uncertainties were 0.01kg and 1.00kg, then you would be justified in taking their sum as 1.01kg.)

 

[patric44]

I guess ‘80\pm 0.99’ should be $80\pm 0.99$.

You appear to have calculated ‘0.99’ by subtracting 0.01 from 1. But, when calculating the uncertainty of a sum or difference, you should always add the uncertainty values.

For example: x=10±1 and y=5±1, so x-y = 5±2.
But if you are given, with no prior knowledge, that x-y = 5±2 and y=5±1 then
x = (x-y) +y = 10±3 (not x = 10±1)

Some other points which might help:

1. A value should be specified to the same precision as its uncertainty.
E.g. (20±0.01) should be written as 20.00±0.01

2. The uncertainty of 0.01kg is very small compared to the uncertainty 1kg. And each uncertainty is specified to only 1 significant figure. The value of an uncertainty is itself an uncertain. So it doesn’t make much sense to say the combined uncertainty as 1.01kg – the combined uncertainty should be rounded to 1kg.

(Uncertainty values are typically only known to 1 or 2 significant figures. If you were told that the two uncertainties were 0.01kg and 1.00kg, then you would be justified in taking their sum as 1.01kg.)

thanks its clear now , but i have another question:
assuming that i started by the answer given for the mass of the man, m= 80.00±1.01 kg (neglecting the fact that the book had written dm as 1.01 not 1), and i already have the mass of the box as 20.00±0.01, the sum of their masses = 100.00±1.02 kg , doesn't equal the value that was given .
as if this calculation only work once and not the other way around!

 

[Steve4Physics]

assuming that i started by the answer given for the mass of the man, m= 80.00±1.01 kg (neglecting the fact that the book had written dm as 1.01 not 1), and i already have the mass of the box as 20.00±0.01, the sum of their masses = 100.00±1.02 kg , doesn't equal the value that was given .
as if this calculation only work once and not the other way around!

I think you are asking “why can’t you ‘reverse’ the process of combining uncertainties in order to recover the starting values?”.

For ‘reversal’, addition operations (on the magnitudes of uncertainties) would need to become subtraction operations. But it is meaningless to perform subtraction operations on the magnitudes of uncertainties

For example if two uncertainties happened to be equal (say both are ± 3) then subtracting the magnitude of one from the magnitude of the other would lead to zero uncertainty – which would not make sense.

You may need to work through the example I gave in post #2.

 

[patric44]

thank you so much, all clear now