How Do You Calculate Forces and Sums with Uncertainties?

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To calculate forces with uncertainties, first determine the force using the mass's upper and lower error bounds, which in this case are 24.3 kg and 24.7 kg, along with the acceleration of -9.81 m/s². The resulting force will be expressed as a range, calculated by applying the formula F = ma to both bounds. For addition and subtraction, if the second mass (6.2 kg) has negligible error, simply add it to the midpoint of the first mass and adjust the error accordingly. If uncertainties are present in both values, combine them using the square root of the sum of their squares if they are independent. Understanding how to handle uncertainties is crucial for accurate calculations in physics.
jaydnul
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My textbook isn't very clear on this. Let's say i find the mass of an object to be 24.5kg +-.2kg. Then i want to use that answer to calculate F (F=ma) when acceleration is equal to -9.81m/s2. What would the final answer look like. Also what about addition and subtraction. For sake of simplicity, (24.5kg +-.2kg) + (6.2kg)=?

Thanks
 
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lundyjb said:
My textbook isn't very clear on this. Let's say i find the mass of an object to be 24.5kg +-.2kg. Then i want to use that answer to calculate F (F=ma) when acceleration is equal to -9.81m/s2. What would the final answer look like. Also what about addition and subtraction. For sake of simplicity, (24.5kg +-.2kg) + (6.2kg)=?

Thanks

For the first question you could calculate the force for each of the lower and upper error bar values (24.3 and 24.7 kg) and the midpoint value assuming the midpoint is your "best" or expected value. For the second question, are you assuming the error is negligible for the 6.2 value? If so you can simply add it to the first midpoint value and bring the error bars down to your sum. If the errors were non zero standard deviations you would generally square the errors, add them and then take the square root of the sum if the errors are independent.

If you're calculating a mean, then the individual observations would not have error bars. The standard error or standard deviation of the mean (they are not the same) of the sample would be calculated in the usual way.

EDIT: I think the reason your textbook is vague on this is because it depends on whether the errors are measured directly from some target value, or if they are calculated from a sample using statistical methods. In direct measurement of a contingent process, the errors are additive. In statistical samples the "error" is simply a measure of the variation in a sample around a mean or average value and uses some version of the mean squared error.
 
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If you only have one value that has any uncertainty, it's easy. If you are combining uncertainties, it's a little more difficult, and it depends on whether the uncertainties are independent. In your example, you only have one uncertain value, so if you multiply the value, you should multiply the uncertainty as well. Use your intuition here; I'd be surprised if that gave you the wrong answer.
 
Khashishi said:
In your example, you only have one uncertain value, so if you multiply the value, you should multiply the uncertainty as well. Use your intuition here; I'd be surprised if that gave you the wrong answer.

I'm not sure what you mean by "multiply the uncertainty". In the OPs example, there are error bars around the kg value. I suggested that he use the lower and upper limit values of the interval to calculate the interval for force. This is not multiplication of the uncertainty ( plus or minus 0.2kg).
 
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what I mean is a*(m +- e) = am +- ae
-9.81m/s^2 * (24.5kg +- 0.2kg) = -240.345N +- 1.962N
 
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