Calculating uncertainty in the final result (combining uncertainties)

Question

I'm struggling to determine the uncertainty in $F$ so it would match the textbook answer.

The problem statement is: A force F is obtained using the equation: $F = \frac{mv^2}{2\pi(x_2 - x_1)}$. The readings taken were: $m = 54.0 \pm 0.5\ \mathrm{kg}$, $v = 6.3 \pm 0.2\ \mathrm{ms}^{-1}$, $x_2 = 4.7 \pm 0.1\ \mathrm{m}$, $x_1 = 3.9 \pm 0.1\ \mathrm{m}$. Calculate the value of F and determine the uncertainty in your value.

Calculating the force: $F = \frac{(54.0\ \mathrm{kg}) \times (6.3\ \mathrm{ms}^{-1})^2}{2 \pi \times (4.7\ \mathrm{m} - 3.9\ \mathrm{m})} \approx 426.388\ \mathrm{N} = 430\ \mathrm{N} \text{(to 2 s.f.)}$. This agrees with the textbook answer.

Now, let $X = x_2 - x_1$. Then $X = 4.7\ \mathrm{m} - 3.9\ \mathrm{m} = 0.8\ \mathrm{m}$. The uncertainty in $X$ is $\delta X = \sqrt{(\delta x_2)^2 + (\delta x_1)^2} = \sqrt{(0.1\ \mathrm{m})^2 + (0.1\ \mathrm{m})^2} \approx 0.1414\ \mathrm{m}$. Thus, $X=0.8 \pm 0.1\ \mathrm{m}$ and the formula becomes $F = \frac{mv^2}{2\pi \times X} = \frac{1}{2 \pi} \frac{m v v}{X}$. This means I can now use another standard formula to calculate the uncertainty in $F$:

$\delta F = \sqrt{(\frac{\delta m}{m})^2 + (\frac{\delta v}{v})^2 + (\frac{\delta v}{v})^2 + (\frac{\delta X}{X})^2}$. And with values $\delta F = \sqrt{(\frac{0.5\ \mathrm{kg}}{54.0\ \mathrm{kg}})^2 + 2 \times (\frac{0.2\ \mathrm{ms}^{-1}}{6.3 \ \mathrm{ms}^{-1}})^2 + (\frac{0.1414\ \mathrm{m}}{0.8\ \mathrm{m}})^2} \approx 0.133$ or about 13%.

But the bloody textbook says it is 40% and quotes the answer as $430 \pm 180\ \mathrm{N}$.

I've tried some calculations with various values within the uncertainty and my result was always within 13% (or about 60 N) of 430 N, just as I would expect.

Where have I got wrong?

Answer 1

Both you and the book made a mistake, but the book's mistake is large, and an error of principle, while your mistake is just simple arithmetic.

First, you should get a feel for the errors involved: the mass error and the v error is negligible, because they are of order a percent or two, while the error in the difference in x, value .8m, is .14m, as you calculated, it is about 15%. This is something you should be aware of--- when you subtract approximately equal quantities, the errors amplify, because the fractional error is what is important, and the quantity becomes smaller.

In your expression,

${\delta F\over F} = \sqrt{(\frac{0.5\ \mathrm{kg}}{54.0\ \mathrm{kg}})^2 + 2 \times (\frac{0.2\ \mathrm{ms}^{-1}}{6.3 \ \mathrm{ms}^{-1}})^2 + (\frac{0.1414\ \mathrm{m}}{0.8\ \mathrm{m}})^2} \approx 0.133$

You didn't get the right answer. The answer is almost exactly equal to the square root of the last term, or

$${\delta F\over F} = {.14 \over .8} = .18$$

The actual error is 18%, not 13%. The remaining terms make this a little bit bigger, but not much. You made an error of arithmetic, which could have been avoided by noting that the last term, the error in $\Delta X$, is the only important one.

But the book did the following brain-damaged error estimate: they took the two values of X, and treated the plus/minus error as something you add or subtract to the quantity to find the biggest and smallest value it can have. Then they took the "boundary" values by adding/subtracting .1 from each, to get a largest/smallest value $\Delta X$:

$$\Delta X_s = (4.7 - .1) - ( 3.9 + .1 ) = .6$$ $$\Delta X_l = (4.7 + .1) - ( 3.9 - .1) = 1.0$$

This gives a 40% error. This procedure is wrong on principle, because the errors in the two x values are independent, and it is extremely unlikely that they will align to be exactly opposite. The correct estimate is that the error is 18%, for both $\Delta X$ and the final answer.

Answer 2

see WP Resistance_measurement Example

solution using 'WP interval arithmetic'
with sagemath package (free and online server at http://sagenb.org)

dM=0.5  
dv=0.2  
dx=0.1  
M=RIF((54-dM,54+dM))  
v=RIF((6.3-dv,6.3+dv))  
x2=RIF((4.7-dx,4.7+dx))  
x1=RIF((3.9-dx,3.9-dx))
F=M*v^2/(x2-x1)/2/math.pi  

F0=F.center()  
df=F0-F.lower()  
df/F0  

gives the result 0.182284511784079

18% is the answer

A PSE similar problem was solved with the (free) Euler Toolbox that also has implemented the interval arithmetic.