Chapter 11, Problem 95PP

BIO Correcting Torsiversion

Torsiversion is a medical condition in which a tooth is rotated away from its normal position about the long axis of the root Studies show that about 2 percent of the population suffer from this condition to some degree. For those who do, the improper alignment of the tooth can lead to tooth-to-tooth collisions during eating, as well as other problems. Typical patients display a rotation ranging from 20° to 60°, with an average around 30°.

An example is shown in Figure 11-69 (a), where the first premolar is not only displaced slightly from its proper location in the negative y direction, but also rotated clockwise from its normal orientation. To correct this condition an orthodontist might use an archwire and a bracket to apply both a force and a torque to the tooth. In the simplest case, two forces are applied to the tooth in different locations, as indicated by F₁ and F₂ in Figure 11-69 (a). These two forces, if chosen properly, can reposition the tooth by exerting a net force in the positive y direction, and also reorient it by applying a torque in the counterclockwise direction.

Figure 11-69 Problems 92, 93, 94, and 95

In a typical case it may be desired to have a net force in the positive y direction of 1.8 N. In addition, the distances in Figure 11-69 (a) can be taken to be d = 3.2 mm and D = 4.5 mm. Given these conditions, a range of torques is possible for various values of the y components of the forces, F_1y and F_2y. For example Figure 11-69 (b) shows the values of F_1y and F_2y necessary to produce a given torque, where the torque is measured about the center of the tooth (which is also the origin of the coordinate system). Notice that the two forces always add to 1.8 N in the positive y direction, though one of the forces changes sign as the torque is increased.

95. •• Find the values of F_1y and F_2y required to give zero net torque of 0.0099 N · m. This is a torque that would be effective at rotating the tooth.

1. A.    F_1y = −1.7 N, F_2y = 3.5 N
2. B.     F_1y = −3.8 N, F_2y = 5.6 N
3. C.     F_1y = −0.23 N, F_2y = 2.0 N
4. D.    F_1y = −4.0 N, F_2y = −2.2N