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Sapphire_Frenzy's avatar

Are my units correct?

Asked by Sapphire_Frenzy (90points) November 28th, 2012

I’m trying to work out the stress on a stretched piece of wire.

Stress=Force applied/cross sectional area

My calculation is: 16/9.079×10EXP-8

My answer has come out as: 176230862.4

Would this number be written as 1.76×10EXP8 N/m^2? It seems a bit wrong to me as the EXP started off as -8. My eyes are disbelieving the answer as it seems extremely large…

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13 Answers

bhec10's avatar

In short: yes, they are.

So, you’ve got:

Stress = 16N / 0.00000009079 m^2

you’re basically dividing a big number by a tiny tiny number, which is going to result in a huge number, and that’s why it is now EXP 8

which is 176230862.4 N/m^2

and that is the same as writing 1.76 * 10^8 N/m^2

Hope that helps.

You can actually test this by reversing your equation and using random numbers

I’ve multiplied: 184868383 * 0.00000007 and the result is a normal number: 12.94

Mariah's avatar

When you divide by a small number, your result is a large number.

LuckyGuy's avatar

We usually write it as 176 MPa. The number you have derived is the yield strength of wrought iron.
If you are curious, here is a chart of yield strength and tensile strength of different materials.

gasman's avatar

Your calculation is S = (16 Newtons) / ( 9.079×10^ -8 meters^2)
a) S = (16 / 9.079) / 10^ -8 (Newtons) / (meters)^2
b) S = 1.762 / 10^ -8 Nm^-2
c) S = 1.76×10^ 8 Nm^-2
Line a separates numbers from units—but keep track of both! In line b the number is calculated. I round to 4 significant figures to match the given data. In line c I switch a negative exponent denominator to a positive exponent numerator. (Make sense?) Line c also shows the SI abbreviations N for newtons and m for meters & standard notation for “newtons per meter squared.”

Note that 1 Nm^-2 equals 1 Pa (Pascal). The chart posted by @LuckyGuy shows the tensile strength of many materials, reported in MPa (megaPascal = 10^6Pa = 10^6Nm-2) so I guess you’re in the right ballpark numerically as well as unit-wise.

I offer this tip:

Dimensional units multiply & divide & form powers just as numbers do, which provides an important and convenient way to check for calculation errors, multiplying when you should be dividing, etc. For example, driving 50 miles in 2.5 hours is an average speed of:

(50 miles) / (2.5 hours)
= (50/2.5) (miles/hours)
= 20 mph
Note how miles divides by hrs to yield mph. Multiplying instead of dividing would yield units of mi-hr (“mile-hours”), which is meaningless, rather than mi/hr (“miles per hour).

This method works well for complicated calculations involving many unit conversions. For example how many seconds in a year?
1 y = (52 weeks) * (7 days/week) * (24 hours/day) * (60 min/hour) * (60 sec/min)
Now you calculate the numerical product – around 3.1×10^7
Meanwhile you collect the units together, like this
(weeks) * (days/week) * (hours/days) * (minutes/hours) * (seconds/minute)
Most of the units appear once in a numerator & once in a denominator, so they “cancel out.” All that remains unit-wise on the right-hand side of the equation is seconds (not 1/seconds or some other unexpected thing). The final result is 1 year = 3.1×10^7 seconds…and you are confident you multiplied & divided everything correctly because the units worked out exactly as shown.

Jeruba's avatar

I’m giving you a GQ just because this is a great example of how to ask a homework question: you tackle the assignment yourself, you get an answer of some kind, and you ask a specific question on whatever it is that’s giving you trouble.

That’s how to ask for help—as opposed to asking others to do the work instead of you.

Well done.

bhec10's avatar

@Jeruba Agreed. That was also the only reason why I gave him that long answer, his question showed that he had done it himself in the first place.

yankeetooter's avatar

One thing I would suggest to you in the future is to always include your units in the equation from the very beginning. If you do this, then you can’t go wrong, as the units that result in your answer will be a way of double-checking your work. For example, if calculating the velocity of an object, if your answer doesn’t contain distance units over time units, you’ve done something wrong. Doing the above will also help you with the issue you raised in this question…

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bhec10's avatar

@Sapphire_Frenzy I’m sorry. I’m sorry. I’m sorry.

I didn’t know, I hope you can forgive me :-)

I assumed that someone asking about forces and stresses would be a guy and not a girl. Sorry.

Sapphire_Frenzy's avatar


Haha, it’s alright :)

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