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Homework Help: Deformation of Solids

  1. Jun 23, 2008 #1
    A stainless-steel orthodontic wire is applied to a tooth. The wire has an unstretched length of 3.1 cm and a diameter of 0.22 mm. If the wire is stretched 0.10 mm, find the magnitude and direction of the force on the tooth. Disregard the width of the tooth, and assume that Young's modulus for stainless steel is 18 x 10^10 Pa.

    With the question there is a picture of a bent wire with a tooth sitting the the middle and two 30 degree angles on either side.

    I guess I don't understand this stuff as much as I thought I did because I cannot come up with the correct answer. Here is what I did:

    1. I converted everything to meters
    2. I found delta L by taking .031 m -.00010 m = 3.09 x 10 -2 m
    3. I found the A of the wire with 2(pie)r^2 + 2(pie)rh = 2(pie)(1.1 x 10 ^-4)^2 + 2(pie)(1.1 x 10 ^-4)(.031 m) = 2.15 x 10 ^-5 m^2
    4. Then I plugged all my info into this equation: F = yA/Lo * delta L
    and I can't come up with the right answer. I keep getting 3.86 x 10 ^-2 but I am supposed to get 22N. I am way off! Can somebody please help get me back on the right track!?
  2. jcsd
  3. Jun 24, 2008 #2


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    I see a few problems:

    1) delta L is the change in length.
    2) A is the cross-sectional area of the wire.
    3) you need to incorporate some trigonometry to account for those 30° angles.

    Good job converting everything to the same units; that's a good habit.
  4. Jun 24, 2008 #3
    I'm sorry, I can't figure out what trig to do. I found the cross sectional area of the wire which is 3.8 x 10 ^-2 but I'm stuck on the trig and the difference in length. Can anybody explain how to do this so I will know for future reference?

    P.s. thanks for your comment, it was really helpful
  5. Jun 25, 2008 #4


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    Hi okaymeka,

    I don't think that answer is correct for the area. Can you show what numbers you used to find the area? (Perhaps you did not convert mm to meters?) Also, what numbers are you using in the Young's modulus equation

    \frac{F}{A} = Y \frac{\Delta \ell}{\ell_0}

    to find the F in the wire?
  6. Jun 25, 2008 #5


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    The difference in length is the amount of stretch in the wire. The trig comes in because (if I understand the arrangement correctly) there are forces in different directions that need to be balanced.
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