A certain orthodontist uses a wire brace to align a patient's crooked tooth as in the figure below.
The tension in the wire is adjusted to have a magnitude of 18.5 N. Find the magnitude of the net force exerted by the wire on the crooked tooth.
What I have got so far:
- I believe I have found the two components?
18.5sin14 = 4.47
Now I'm not sure what else I'm supposed to do?
1 Answer | Add Yours
The two equal tensions are the on a particular crooked teeth acting at (180-2*14degree) = 152 degree.
Therefore the resultant force = sqrt(18.5^2+18.5^2+2*18.5cos152 degree) =sqrt[2*18.5^2(1-cos28), by the law of parallelogram of forces.
=8.7092 N approximately.
(What you have done is component of forces along and perpependicular to the outer surface of the crooked tooth.
The forces alons the outer surfaces are (18cos14)N and (18cos14 ) N in opposite directions and so get cancelled.
The two equal components of the tension force perpendicular to the surface of the crooked teeth each of (18sin14) N gets added and is equal to 36sin14 Neton = 8.7092 N.)
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