In a traction system three forces are applied to the femur as shown. The femur itself is positioned at an angle of 15° with the horizontal.

- What is the resultant force of this traction system?
- What impact, if any, will it have on the position of the femur?

In order to determine the resultant force R, which is a
vector, we have to calculate both its magnitude and its
direction. First, let us compute the magnitude of R. We
accomplish this by resolving each force into its *x* and
*y* components, and then summing up all the *x*
components and summing up all the *y* components as
follows:

Force | x-component | y-component |
---|---|---|

F1 = 10 lb | 10 cos(-19°) = 10 × 0.9455 = 9.455 | 10 sin(-19°) = 10 ×(-0.3256) = -3.256 |

F2 = 10 lb | 10 cos(17°) = 10 × 0.9563 = 9.563 | 10 sin(17°) = 10 × 0.2924 = 2.924 |

F3 = 10 lb | 10 cos(61°) = 10 × 0.4848 = 4.848 | 10 sin(61°) = 10 × 0.8746 = 8.746 |

Resultant R | Rx =
23.866 |
Ry = 8.414 |

Since we have calculated the *x* and *y* components
of the resultant R, we can find its magnitude by using the
Pythagorean theorem:

R2 = Rx2 + Ry2 = 23.8662 + 8.4142 = 569.586 + 70.795 = 640.381

which gives us:

R = 25.306 or rounded to 2 significant digits, the same as the original measurements: R = 25 lb.

The angle of the resultant can be calculated as follows:

From this:

Since the femur had an original orientation of 15° we can conclude that the effect of this traction system on the femur will be a rise of 4.42°.

Written by Erika Crema, October 10, 1997