# A certain orthodontist uses a wire brace to align a patient's crooked tooth as in the figure below.Figure: http://www.webassign.net/sercp8/p4-18.gif The tension in the wire is adjusted to have a...

A certain orthodontist uses a wire brace to align a patient's crooked tooth as in the figure below.

Figure:

http://www.webassign.net/sercp8/p4-18.gif

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

18.5cos14 =17.95

Now I'm not sure what else I'm supposed to do?

Thanks

### 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.

=18*2^(1/2)sqrt[ 1-cos28]

=18*2^(1/2)[1-(1-2sin^2(14)]

=18*2.sin14.

=36sin14.

=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.)