- #1

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Is F=Y(change in L/L0)A the equation I would use? I know that Y for brass is 9.0 x 10^10 and Y for copper is 1.1 x 10^11 so would I add those together or what?

- Thread starter texasgrl05
- Start date

- #1

- 7

- 0

Is F=Y(change in L/L0)A the equation I would use? I know that Y for brass is 9.0 x 10^10 and Y for copper is 1.1 x 10^11 so would I add those together or what?

- #2

- 538

- 2

Length by which stack decreases= Decrease in length of copper+Decrease in length of brass.

BJ

BJ

- #3

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for copper i got:

6450=1.1*10^11(change in L/3).0022m = 7.9*10^-5

and for brass:

6450=9.0*10^10(change in L/5).0022m = 1.63*10^-4

when i added these together i got 2.43*10^-4

what am i doing wrong?

- #4

Astronuc

Staff Emeritus

Science Advisor

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Strain = Stress/E, where E = Elastic (Young's) Modulus.

What is the meaning of strain in terms of change in length?

- #5

FredGarvin

Science Advisor

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2) What are the units of your constants?

3) You didn't include a picture, so I am assuming that you left out that the cylinders are 3 m and 5 m in length?

- #6

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for copper i got:

6450N=1.1*10^11N/m^2(change in L/3cm).22cm = 7.99*10^-7

and for brass:

6450N=9.0*10^10N/m^2(change in L/5cm).22cm = 1.62*10^-6

when i added these together i got 2.43*10^-6 cm

i still don't think this is right?

- #7

FredGarvin

Science Advisor

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Your first equation should look like:

[tex]6450 N = (1.1 x 10^11 \frac{N}{m^2})(\frac{\Delta L}{.03 m})(\pi * (.022 m)^2)[/tex]

You can solve for [tex]\Delta L[/tex] which will be in meters for each metal.

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