How Do You Calculate the Force Exerted by an Orthodontic Wire on a Tooth?

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SUMMARY

The calculation of the force exerted by a stainless-steel orthodontic wire on a tooth involves understanding Young's modulus and trigonometric principles. The wire has an unstretched length of 3.1 cm and is stretched by 0.10 mm, with Young's modulus for stainless steel being 18 x 1010 Pa. The correct formula to use is F = Y * A / L0 * ΔL, where A is the cross-sectional area and ΔL is the change in length. The discussion highlights the importance of incorporating trigonometry to account for the angles formed by the wire.

PREREQUISITES
  • Understanding of Young's modulus and its application in material science
  • Knowledge of trigonometry, particularly in resolving forces at angles
  • Ability to calculate cross-sectional area of cylindrical objects
  • Familiarity with unit conversions, specifically from millimeters to meters
NEXT STEPS
  • Learn how to calculate the cross-sectional area of a cylinder using the formula A = πr2
  • Study the application of Young's modulus in various materials, focusing on stainless steel
  • Explore trigonometric functions to resolve forces in systems with angles
  • Practice problems involving force calculations in elastic materials
USEFUL FOR

Students in physics or engineering, orthodontic professionals, and anyone involved in materials science or mechanical engineering will benefit from this discussion.

okaymeka
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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!?
 
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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.
 
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
 
Hi okaymeka,

okaymeka said:
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.

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

<br /> \frac{F}{A} = Y \frac{\Delta \ell}{\ell_0}<br />

to find the F in the wire?
 
okaymeka said:
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

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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