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The drawing shows a wire tooth brace used by orthodontists. The topmost tooth is protruding slightly, and the tension in the wire exerts two forces $\overrightarrow{\mathbf{T}}$ and $\overrightarrow{\mathbf{T}}^{\prime}$ on this tooth in order to bring it back into alignment. If the forces have the same magnitude of $21.0 \mathrm{N},$ what is the magnitude of the net force exerted on the tooth by these forces?

11.6 $\mathrm{N}$

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

in this question, we have to calculate what is the magnitude off the resulting force? Well, for that, we begin by calculating the components off the resulting force. So each off these forces T and T Prime can be the composer in two components. One component that goes like this and another component that goes like this as the two forces. Our makes the same angle with the horizontal on have the same magnitudes. This The magnitude of this component is ego to the magnitude of this component, and the magnitude off both vertical components are equals. True to calculate the magnitude after its components, we can use direct and will triangle like that one. So here is 21 Newton's. Here is the X component off the prime on here is the Y component off the prime, and here is an angle off 16 degrees. Then, with that, we can calculate the X by using the co sign off 16 degrees because they go sign off. 16 degrees is equals to the address inside off the triangle, So T Prime X, divided by the high point in his 21. Therefore, T Prime X is close to 21 times they call Sign off 16 degrees. Now for the Y component. We have a following this sign off 16 degrees easy course to the opposite side, so t prime Why divided by the high Potter News which is 21 then t prime Why is he goes to 21 times this sign off 16 degrees? And this is also true for the components off the tee force. So we have to following Deasy's ah 21 times this sign off 16 degrees. These is also 21 times their sign off 16 degrees. These pointing to the left is equals to 21 times the co sign off 16 degrees and this is also doing 21 times they call sign off 16 degrees. Note that here on the horizontal axis, we have two forces off equal magnitude pointing in opposite directions. Therefore, they come so each order So this force cancel these force as a crossing points in the resulting force is a vertical force with a magnitude that is given by true times 21 times This sign off 16 degrees Therefore the magnitude off the resulting force Let me call it our is the close to two times 21 times this sign off 16 degrees, and this is approximately 11.6 neutrons, so these is the resulting force that acts on the tooth.

Brazilian Center for Research in Physics