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Have I got this right? (Young Modulus) watch

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    The wire in a new guitar 'string' is made of a 90 cm steel wire of diameter 1 mm. When the string is fitted to the guitar, the string is put under a tension of 75 N and stretched by 0.5 mm.

    a) What is the stress in the wire?
    I did:
    Stress = force / area = 75 / π0.52 = 95.54 N/mm2

    b) What's the wire's strain?
    I did:
    Stain = change in length / original length = 0.5 / 900 = 5.56 x 10-4

    c) What is the Young Modulus for this kind of steel?
    I did:
    Stress / strain = 95.54 / 5.56 x 10-4 = 1.72 x 105

    I have a feeling I'm completely wrong..
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    (Original post by Draconis)
    The wire in a new guitar 'string' is made of a 90 cm steel wire of diameter 1 mm. When the string is fitted to the guitar, the string is put under a tension of 75 N and stretched by 0.5 mm.

    a) What is the stress in the wire?
    I did:
    Stress = force / area = 75 / π0.52 = 95.54 N/mm2

    b) What's the wire's strain?
    I did:
    Stain = change in length / original length = 0.5 / 900 = 5.56 x 10-4

    c) What is the Young Modulus for this kind of steel?
    I did:
    Stress / strain = 95.54 / 5.56 x 10-4 = 1.72 x 105

    I have a feeling I'm completely wrong..
    Looks good
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    (Original post by D-Day)
    Looks good
    Really? Yay :yep:

    I'm not sure how to answer these:

    a) 'In an experiment to find the Young modulus, the strain should not be more than 1 in 1000.' Explain what this means.

    b) Describe an experiment to determine the Young modulus for the metal in a wire. Taking into account the possible errors in measurements, explain why the limitation in a) is necessary.

    I really have no idea.
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    From what I can see, your radius is in mm.
    I don't know if you've converted afterwards, but I'd convert them all into cm before you do any equations,
    But you might have done this,
    Best wishes,
    Jess
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    Strain's a ratio, so I presume it just requires you to know this. Maybe the calculations aren't accurate if they're over this ratio? Check Wikipedia...
    For the second one, just describe experiments to measure stress and strain. Have you not done them in class?
    One of the errors is that you need to tie a knot in the wire to attach the weights, so this is where the stress is concentrated...
    Does that make any sense? :/
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    can someone teach me this young's modulus or list the different formulas etc


    iim really confused in class
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    (Original post by QuantumTheory)
    Strain's a ratio, so I presume it just requires you to know this. Maybe the calculations aren't accurate if they're over this ratio? Check Wikipedia...
    For the second one, just describe experiments to measure stress and strain. Have you not done them in class?
    One of the errors is that you need to tie a knot in the wire to attach the weights, so this is where the stress is concentrated...
    Does that make any sense? :/
    Nope, we haven't done them.
    I guess I'll just ask the teacher tomorrow. We haven't done any experiments on materials.
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    (Original post by dpwp)
    can someone teach me this young's modulus or list the different formulas etc


    iim really confused in class
    The Young's modulus is given by stress divided by strain. It's a ratio and has no units. It gives an indication of the stiffness of a material in its elastic region (i.e. before permanent deformation, some call it plastic flow). That's pretty much it.
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    For the calculation the area MUST be in  m^2

    and the length must be in metres

    All calculations are done in SI units.

    So I think your answer is wrong.

    Covert before doing your calculation

    I think the young modulus of steel is about  1.8 x 10^1^1 Pa.

    (Correct me if I'm wrong on that someone, I did it on memory!) (The value)
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    Callum Honisett's advice ^^^ is the most useful. For Physics, always always always use standard SI units. Because you haven't converted, both your stress and your strain are incorrect.
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    (Original post by callum Honisett)
    For the calculation the area MUST be in  m^2

    and the length must be in metres

    All calculations are done in SI units.

    So I think your answer is wrong.

    Covert before doing your calculation

    I think the young modulus of steel is about  1.8 x 10^1^1 Pa.

    (Correct me if I'm wrong on that someone, I did it on memory!) (The value)
    And given that her units are N/mm^2 then her answer is actually correct.
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    (Original post by TableChair)
    And given that her units are N/mm^2 then her answer is actually correct.
    In that sense it is, but young modulus is measured in pascals which ARE NOT

     N mm^-^2

    If she gave that answer in an exam it would be incorrect.

    All calculations and answers must be done/given in SI units.
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    (Original post by TableChair)
    And given that her units are N/mm^2 then her answer is actually correct.
    Firstly, we're looking for N/m^2 (which is equivalent to Pa), not N/mm^2.

    Also, she got the strain wrong, as she used different units for extension and original length.

    I'm going to sleep now, night.
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    Young's modulus is usually measured in GPa, which is what you get if you calculate the stress in N/mm^2. Also it doesn't matter what units you use as long as you're consistent.
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    (Original post by callum Honisett)
    In that sense it is, but young modulus is measured in pascals which ARE NOT

     N mm^-^2

    If she gave that answer in an exam it would be incorrect.

    All calculations and answers must be done/given in SI units.
    You can measure it in pascals if you want to. You don't have to. In this case she's measured it in MPa.

    If it doesn't specify which units you are to give the answer in, you can give it in whatever units you want.
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    (Original post by tommm)
    Firstly, we're looking for N/m^2 (which is equivalent to Pa), not N/mm^2.

    Also, she got the strain wrong, as she used different units for extension and original length.

    I'm going to sleep now, night.
    It doesn't say to give the answer in Pa, you can give it in MPa if you want.

    No she didn't. 90cm = 900mm.
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    I fail at units. Silly me.
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    (Original post by Draconis)
    The wire in a new guitar 'string' is made of a 90 cm steel wire of diameter 1 mm. When the string is fitted to the guitar, the string is put under a tension of 75 N and stretched by 0.5 mm.

    a) What is the stress in the wire?
    I did:
    Stress = force / area = 75 / ?0.52 = 95.54 N/mm2

    b) What's the wire's strain?
    I did:
    Stain = change in length / original length = 0.5 / 900 = 5.56 x 10-4

    c) What is the Young Modulus for this kind of steel?
    I did:
    Stress / strain = 95.54 / 5.56 x 10-4 = 1.72 x 105

    I have a feeling I'm completely wrong..
    please could you tell me the name of this textbook? my teacher gave me a photocopy from it and it's explainations are pretty good
 
 
 
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