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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?

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The book is probably using rule-of-thumb uncertainty estimates, while you are using more accurate ones. Throw the book away. –  Ron Maimon Sep 18 '11 at 21:58
@RonMaimon the odd thing is that the rule of thumb I have in this book supports my calculations. They say that percentage uncertainties should be added together and if there is a square of a value (like velocity in this case), then such percentage uncertainty should be added in twice. If I add percentage uncertainties for $m$, $v$, $x_1$ and $x_2$ I get about 12% as a result which is quite close to my calculation. They clearly had something in mind when they wrote 40% and I'd love to know what "rules of thumb" they have used as it might be helpful in the exam. –  Gruffalo Sep 18 '11 at 22:04
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2 Answers

up vote 4 down vote accepted

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,


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.

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Unless, of course, x1 and x2 are somehow not independent. –  Colin K Sep 19 '11 at 3:31
Thank you! This explains it. –  Gruffalo Sep 27 '11 at 17:42
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see WP Resistance_measurement Example

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



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.

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