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A random example from the top of my head:

Given a mass of $m=1.12\:\mathrm{kg}$ accelerated by $a=99.87465\:\mathrm{m\:s}^{-2}$, find the force $F=ma$.

Now, how many significant would I take this answer to? Is there a general rule to determine this? I know how to round it when the values have the same amount of significant figures but not when the values contain a different amount of significant figures.

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As a simple rule you use the number of significant figures of the least precise value. So in the example you would quote the force two the significant figures, because that is the precision with which you know the mass. You can't justify any more significant figures than that because of the implied uncertainty in the mass.

As to why this is the case, consider that the fact that the mass is quoted to 3 significant figures implies that it has been rounded to that value (since it isn't exactly 1.120000 Kg). So the 'true' value could be anywhere in the range $1.115-1.125$ Kg - all those values would be rounded to 1.12 Kg to 3 s.f.

That gives a range of possible values of the force:

$F=99.87465\times 1.115=111.3602 Kgms^{-2}$

$F=99.87465\times 1.125 = 112.3590 Kgms^{-2}$

Actually this illustrates why it is only a rule of thumb, rather than something more precise. You can't quite justify saying it is 112 Kg ($99.87465\times 1.12$ to 3 s.f.), since there is an uncertainty in the 3rd significant figure.

But in the absence of doing proper error analysis, it is a good enough. At least, it illustrates why you can't justify more significant figures than the least precise value.

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    $\begingroup$ It would be good to give some justification for this rule (as in, why does knowing the mass and force to 3sf and >3sf allow you to know the acceleration to 3sf, rather than, say, 2sf or 4sf?) $\endgroup$ – David Richerby Apr 17 '16 at 16:32
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Since it is a product of two quantities, the answer should have significant figures equal to that number which had the least number of significant figures of the two.

Example:

Say I have an object of mass $2.86\:\mathrm{kg}$, accelerated to $12.893\:\mathrm{m\: s}^{-2}$. Here the force on that object is equal to $$F=ma.$$

Here

\begin{align} m & = 2.86\:\mathrm{kg} &\text{(3 significant figures)} \\a & = 12.893\:\mathrm{m\:s}^{-2} &\text{(5 significant figures)} \end{align}

Now the answer, $F=ma=2.86 \mathrm{kg}\times 12.893\:\mathrm{m\:s}^{-2}=36.87398\:\mathrm{N}$ (7 significant figures)

But since the quantity with least significant figures is $m$ with 3, we can round off the answer to 3 significant figures.

Answer = 36.9 (round off to 3 significant figures)

As pointed out by David Richerby, The reason why Significant figures are used is given below.

The concept of Significant figures is used to make calculations easier without loosing the precision of the answer. Its good to use more significant figures but hard to do calculations with it unless you have a computer.

P.S: In your case the mass has least significant figures so the answer should also have significant figures not less that that of mass (3 significant figures)

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    $\begingroup$ It would be good to give some justification for this rule (as in, why does knowing the mass and force to 3sf and >3sf allow you to know the acceleration to 3sf, rather than, say, 2sf or 4sf?) $\endgroup$ – David Richerby Apr 17 '16 at 16:33
  • $\begingroup$ @DavidRicherby I don't know how much you know about Significant figures but the concept of Significant figures is used to make calculations easier without loosing the precision of the answer. Its good to use more significant figures but hard to do calculations with it unless you have a computer. $\endgroup$ – hxri Apr 17 '16 at 16:40
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    $\begingroup$ I'm familiar with significant figures and know why they're relevant. But the question was asked by somebody who wasn't sure how to combine values with different numbers of significant figures. Anyone who is asking that question will learn more if you not only tell them the answer but also tell them why it is the answer. $\endgroup$ – David Richerby Apr 17 '16 at 16:47
  • $\begingroup$ @DavidRicherby Alright I am adding it to my answer. Thanks for the response :D $\endgroup$ – hxri Apr 17 '16 at 16:55
  • $\begingroup$ Doesnt 36.87398 round up to 36.9 and not 36.8? $\endgroup$ – user11589 Apr 17 '16 at 20:12
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The general rule to determine the number of significant figures comes from the error propagation formula applied to the rounding errors :

$\Delta F=\sqrt{(\frac{\partial F}{\partial a}\Delta a)^2+(\frac{\partial F}{\partial m}\Delta m)^2}$

Writing m = 2.86 means that 2.86 - 0.005 < m < 2.86 + 0.005
Writing a = 12.893 means that 12.893 - 0.0005 < a < 12.893 + 0.0005

So the rounding errors are : $\Delta a = 5.10^{-4} \text{ and } \Delta m = 5.10^{-3}$

In this case we have $ F = m a$, this implies $\frac{\partial F}{\partial a}=m$ and $\frac{\partial F}{\partial m}=a$

Then $\Delta F=\sqrt{(2.86*5.10^{-4})^2+(12.893*5.10^{-3})^2}=0.06448$

So we have F=36.87398 =/- 0.06448

This can be rounded to F=36.9 +/- 0.1 and it follows that the result has 3 significant figures.

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protected by Qmechanic Apr 17 '16 at 22:50

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