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Help on tension/force problem

  • Thread starter shizfest
  • Start date
1. Homework Statement

The drawing below shows an elastic cord attached to two back teeth and stretched across a front tooth. The purpose of this arrangement is to apply a force to the front tooth. (The figure has been simplified by running the cord straight from the front tooth to the back teeth.) If the tension in the cord is 1.6 N, what are the magnitude and direction of the force applied to the front tooth?

fig-047.gif






3. The Attempt at a Solution

I tried doing 1.6cos33 = magnitude. Apparantly this is wrong. If anyone can give me a hint or push me in the right direction I would be very grateful. Thanks.
 
Last edited by a moderator:

djeitnstine

Gold Member
614
0
You should account for the tension of both sides of the string since the tension everywhere is the same.
 
So would I try 2(1.6cos33) = magnitude ?

Thanks again for any help! Not giving up til I found the correct solution!
 
T=1.6
Θ=33

If the tension in the cord is 1.6N I assume that means on both sides of the tooth.

What I would try is first to draw a force diagram. I did this and then added up all of the forces that made an impact in the Y-direction (up-down). I came up with..

Sum of Y-Forces == -T(cos(Θ))-T(cos(Θ))

Leading me to believe the method may be..

2(-T(cos(Θ))

which would mean you've already got it, I hope so, heh~
 
Last edited:
Thanks for the input! To me that seems to be the only explainable way to do the problem.
 

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