# Approximating Answers When Exact Numbers Are Involved

TRB8985
Homework Statement:
At a construction site, a 65.0 kg bucket of concrete hangs from a light (but strong) cable that passes over a light, friction-free pulley and is connected to an 80.0 kg box on a horizontal roof. The cable pulls horizontally on the box, and a 50.0 kg bag of gravel rests on top of the box.

Find the static friction force on the bag of gravel and on the box.
Relevant Equations:
W = mg
Good afternoon everyone,

I had a quick question on the correct way to report the answer of this problem.

I've solved it and came to the conclusion that the bag of gravel experiences no static friction, but the box experiences static friction with a magnitude equal to the weight of the bucket - 637 N. This was obtained by setting up a free-body diagram on each object and setting the acceleration equal to zero, as the combined weight of the box and bag of gravel is larger than the influence of tension.

This matches the answer in my solution manual. However, I noticed something a little weird.

The author seemed to have arbitarily used three significant figures to represent the acceleration due to gravity (9.80 m/s²). There's nothing else in the problem description besides what I've provided above to indicate that we were supposed to do such a thing.

Does this mean that, when I multiply the weight of the bucket (65.0 kg) by the acceleration due to gravity (9.8 m/s²), that I have to reduce the result of exactly 637 N to an answer containing two significant figures?
i.e. 640 N?

Thank you for your help! Have a great evening and a great upcoming 2023.

Last edited:

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Does this mean that, when I multiply the weight of the bucket (65.0 kg) by the acceleration due to gravity (9.8 m/s²), that I have to reduce the result of exactly 637 N to an answer containing two significant figures?
i.e. 640 N?
The ##65.0 \ kg## is giving you the mass of the object to three significant figures. So, probably ##637 \ N## is fine.

One thing you can do is calculate the error given the data. If the final formula is simple enough, you can taken to the maximum or minimum of each value to get the maximum or minimum final answer. E.g. ##65.0 \ kg## implies a mass in the range ##64.95 \ kg## to ##65.05 \ kg##.

• TRB8985
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If the author said that the acceleration of gravity is 9.81 m/s2 would you drop the "1"? Probably not. So why drop the "0"? I think the author expects you to report the answer to 3 sig figs, e.g. 637 N or 6.37×102 N. If the author said 9.8 m/s2 then you would report 6.4×102 N. This is not nit picking. Nowadays homework that is graded by algorithms often pays attention to sig figs in the answer that are tied to the input parameter accuracy.

• • TRB8985 and PeroK
TRB8985
Thank you both for your replies! I understand what you mean.

I guess the confusion here arises since, in every problem throughout this book up to this point, our solution manual uses the generic two-figure 9.8 m/s² quantity for g. It just seemed really weird that, now all the sudden, 9.80 m/s² is used without any forewarning anywhere and no explanation.

Does this mean we are allowed to introduce an extra significant figure in g in these kinds of situations? It doesn't seem very consistent, and I don't want to invent additional accuracy when I shouldn't be. Hopefully that makes sense!

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You’ve already had a couple of good replies from @PeroK and @kuruman. But since I’ve already written this...

9.81m/s² is typically used when a precision of 3 sig. figs. Is required. But 9.80m/s² is perfectly acceptable - the value of g varies with geographical location.

It is common practice to ‘relax’ the rules and allow final answers to have one (but no more than one) extra significant figure. Your example (using g=9.8m/s²) is a case in point. If I marked this for an exam’ (when I used to do such things), I would accept either 637N or 640N.
____________

A few other thoughts…

It the calculated value is then used in subsequent steps, the unrounded value should be used in these steps.

Rounding the answer to 640 N is correct. However, in this particular case, the trailing zero creates ambiguity: ‘640 N’ could represent (640±10)N or (640±1) N! You would need to write 6.4x102N to avoid the ambiguity. It’s not a good system.

Using g=9.8m/s²:
-the implied minimum possible weight of the bucket is 64.95 * 9.75 = 633 N
-the implied maximum possible weight of the bucket is 65.05 * 9.85 = 641 N
So, the final value is implied to be between 633N and 641N.

A more sophisticated approach would be use the explicit uncertainty of each value (e.g. m = (65.00±0.05)kg). An ‘error propagation’ technique would then be used to calculate the uncertainty of the result of the calculation. This is probably not required at the level you are currently working.

Happy 2023!

Edit - typos

• TRB8985
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I guess the confusion here arises since, in every problem throughout this book up to this point, our solution manual uses the generic two-figure 9.8 m/s² quantity for g.

This seems to imply that up to this point, the solution manual reported results to 2 figures at most. (Well, those answers resulting from using ##g##.)

• TRB8985
TRB8985
This seems to imply that up to this point, the solution manual reported results to 2 figures at most. (Well, those answers resulting from using ##g##.)
Indeed it did! At least, as far as I saw.

Sounds like I may have been a bit too strict adhering to significant figure rules in this case. Thank you all for the replies! It's been very insightful. Best wishes to everyone.