# Significant Figures in Physics

I'm a little confused on how to answer Physics questions in correct decimal places. In Math, I'm used to just giving a couple decimal places and that's fine. In Physics, you do all this rounding over and over and by the end you can be off by many whole numbers.

For instance, I took a simple problem and my answer was off quite a bit..

Problem was "If a static coefficient of static friction between a dresser and the floor is 0.45, what is the magnitude of minimum horizontal force needed to move a dresser of 28kg."

F = .45(9.8)(28)

Simple enough.. my answer was 123.48N

I look at the answer and it's 120N.

How and Why did they round like that?? Thanks.

## Answers and Replies

Andy Resnick
Science Advisor
Education Advisor
Pengwuino
Gold Member
Your worst number has 2 significant digits (actually all 3 have 2 significant figures). The reason we have this convention is that let's say you have to multiply 2 numbers, 4 and 6. Your answer is seemingly 24. However, your numbers could be up to 4.5 and down to 3.5... and 6.5 to 5.5 etc etc so your actual value could be at the highest 29.25 and at the lowest 19.25.

That's not a really good technical description but i hope it gives you a feel as to why we keep significant figures the way we do.

diazona
Homework Helper
Apologies to Pengwuino (whose answer was great, btw) but I have a semi-prepared rant on this ;-)

I'm a little confused on how to answer Physics questions in correct decimal places. In Math, I'm used to just giving a couple decimal places and that's fine. In Physics, you do all this rounding over and over and by the end you can be off by many whole numbers.

For instance, I took a simple problem and my answer was off quite a bit..

Problem was "If a static coefficient of static friction between a dresser and the floor is 0.45, what is the magnitude of minimum horizontal force needed to move a dresser of 28kg."

F = .45(9.8)(28)

Simple enough.. my answer was 123.48N

I look at the answer and it's 120N.

How and Why did they round like that?? Thanks.
Well... significant figures are actually not the whole story. Numbers in "real physics," e.g. experiments and published papers, are supposed to be accompanied by an uncertainty, which reflects the fact that it's impossible to measure anything exactly. So for example, you might have the height of a building measured as $$(45.0\pm 0.1)\mathrm{m}$$. The uncertainty of 0.1m reflects how precise the measurement is - it's basically a statement by the measurer that the actual height is most likely between 44.9 and 45.1 meters. Now, if you were to combine that with some other measurement(s), there's a rule to figure out what the uncertainty in the result is: if u depends on x and y, the rule is
$$\delta u = \sqrt{\left(\frac{\partial u}{\partial x}\delta x\right)^2 + \left(\frac{\partial u}{\partial y}\delta y\right)^2}$$
This is called the error propagation formula. An example would be multiplying by the area of the building to get its volume; presumably the area has its own uncertainty.

The thing is, keeping track of those uncertainties is a major pain in the butt. So for people who are just learning how to work with measurements (usually middle school or high school students), or people who don't need to be so precise about the uncertainties in calculated values, we have significant figures, which is a simplified way to keep track of uncertainties. The idea is to write down only those digits which are fixed by the range of the uncertainty; for example, the $$(45.0\pm 0.1)\mathrm{m}$$ measurement from before would just be written 45.0m, with an understanding that there's an uncertainty of 1 in the last digit written. (You could alternatively use the convention that the uncertainty is 5 in the next digit after the last digit written, so that 45.0m represents anywhere from 44.95m to 45.05m - basically anything that rounds to 45.0m. A lot of people do it this way, but it doesn't really matter which convention you use as long as you are consistent.)

Significant figures are easy because the rules for error propagation are easy to remember:
• When adding or subtracting, only keep the decimal places that are written in both numbers. For example, 45.0 + 2.545 = 47.5. The 45 at the end of 2.545 gets discarded because we don't know what the corresponding numbers in 45.0 are; it could be 45.044 or 44.975, making the answer 47.589 or 47.520 respectively, but you can be pretty sure it's 47.5<something>. So you write down 47.5.
• When multiplying or dividing, keep the smaller number of significant figures in each number. For example, 45.0 * 2.545 = 115; because there are 3 significant figures (3 written digits) in 45.0, you keep 3 significant figures in the answer. Again, you don't know whether the real value is 45.044 or 44.975, so the answer could be 114.63698 or 114.461375 or something in between, but it's probably going to be close to 115, so you write down 115.

It's worth stressing that significant figures (and uncertainties) only affect how many digits you write down, not how many digits are in the actual results of the calculation. Say you're doing a calculation that has several steps, like your multiplication of .45(9.8)(28): you would first multiply .45 by 9.8 to get 4.41. If you're asked to write down this intermediate result, sure you can write down 4.4 because only two significant digits are justified, but when you continue with the calculation, you use the full-precision value 4.41. (If this is obvious to you, great, but a lot of people don't seem to get it)

Apologies to Pengwuino (whose answer was great, btw) but I have a semi-prepared rant on this ;-)

Well... significant figures are actually not the whole story. Numbers in "real physics," e.g. experiments and published papers, are supposed to be accompanied by an uncertainty, which reflects the fact that it's impossible to measure anything exactly. So for example, you might have the height of a building measured as $$(45.0\pm 0.1)\mathrm{m}$$. The uncertainty of 0.1m reflects how precise the measurement is - it's basically a statement by the measurer that the actual height is most likely between 44.9 and 45.1 meters. Now, if you were to combine that with some other measurement(s), there's a rule to figure out what the uncertainty in the result is: if u depends on x and y, the rule is
$$\delta u = \sqrt{\left(\frac{\partial u}{\partial x}\delta x\right)^2 + \left(\frac{\partial u}{\partial y}\delta y\right)^2}$$
This is called the error propagation formula. An example would be multiplying by the area of the building to get its volume; presumably the area has its own uncertainty.

The thing is, keeping track of those uncertainties is a major pain in the butt. So for people who are just learning how to work with measurements (usually middle school or high school students), or people who don't need to be so precise about the uncertainties in calculated values, we have significant figures, which is a simplified way to keep track of uncertainties. The idea is to write down only those digits which are fixed by the range of the uncertainty; for example, the $$(45.0\pm 0.1)\mathrm{m}$$ measurement from before would just be written 45.0m, with an understanding that there's an uncertainty of 1 in the last digit written. (You could alternatively use the convention that the uncertainty is 5 in the next digit after the last digit written, so that 45.0m represents anywhere from 44.95m to 45.05m - basically anything that rounds to 45.0m. A lot of people do it this way, but it doesn't really matter which convention you use as long as you are consistent.)

Significant figures are easy because the rules for error propagation are easy to remember:
• When adding or subtracting, only keep the decimal places that are written in both numbers. For example, 45.0 + 2.545 = 47.5. The 45 at the end of 2.545 gets discarded because we don't know what the corresponding numbers in 45.0 are; it could be 45.044 or 44.975, making the answer 47.589 or 47.520 respectively, but you can be pretty sure it's 47.5<something>. So you write down 47.5.
• When multiplying or dividing, keep the smaller number of significant figures in each number. For example, 45.0 * 2.545 = 115; because there are 3 significant figures (3 written digits) in 45.0, you keep 3 significant figures in the answer. Again, you don't know whether the real value is 45.044 or 44.975, so the answer could be 114.63698 or 114.461375 or something in between, but it's probably going to be close to 115, so you write down 115.

It's worth stressing that significant figures (and uncertainties) only affect how many digits you write down, not how many digits are in the actual results of the calculation. Say you're doing a calculation that has several steps, like your multiplication of .45(9.8)(28): you would first multiply .45 by 9.8 to get 4.41. If you're asked to write down this intermediate result, sure you can write down 4.4 because only two significant digits are justified, but when you continue with the calculation, you use the full-precision value 4.41. (If this is obvious to you, great, but a lot of people don't seem to get it)

***Warning*** That formula holds because its derived on a taylor series expansion. It doesnt hold if the deviation from the mean is large. If delta x or delta y are big, this propagation is wrong.

diazona
Homework Helper
***Warning*** That formula holds because its derived on a taylor series expansion. It doesnt hold if the deviation from the mean is large. If delta x or delta y are big, this propagation is wrong.
Yeah, there's an implicit assumption in all this uncertainty stuff that the uncertainties are small (plus I think the fact that values follow a roughly Gaussian distribution comes in somewhere...). The counterexamples don't come up very often in practice.

rcgldr
Homework Helper
significant figures ... accompanied by an uncertainty ...
(.45)(9.8)(28)
The same logic can be applied to significant figures:

(.45 +/- .005)(9.8 +/- .05) (28 +/- .5) = 123.52275 +/- 4.207125

with minimum and and maximum values of 119.315625 and 127.729875

I think this answer is a lot more informative than 120, which implies the answer is (120 +/- 5) or (120 +/- .5) the trailing zero could be significant in this case). To get around this the answer would need to be given as 12E1 (12 x 101). How does one distinguish between (120 +/- .5) versus (120 +/- 5) using significant digits format?

Still the goal of the class is to teach significant digits as a method, regardless if it leads to misleading results.

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