# Moment about an axis

A chiropractor performs a lumbar spinous pull manipulative procedure. Using instrumentation , she determines 3D coordinates of the axis of rotation of the contacted verebra at t = 15 msec relative to the local axis system as shown. A uniaxial load cell at the spinous process measures a force of Fs = 54.8 N during the thrust also at t = 15 msec. An optotrak camera system determines the 3D coordinates of two points along the line of force application relative to the local axis system again at t = 15 msec. All coordinate data (measured in metres as X, Y, Z) is shown below. Calculate the moment about the axis AB generated by the thrust at t = 15 msec. Assume bone acts as a rigid body.

Given: Point A along the axis of rotation: (.025, .134, .012)
Point B along the axis of rotation: (-.044, -.280, .048)
Point C at the spinous process: (-.003, -.032, -.082)
Point D along the line of force application: (.030, -.031, -.094)

okay i found this questions and have tried to figure it out. i know in theory what to do
1: calculate moment about a
get direction of c-d then get dc.
get length by pythagoreans and divide.
multiply by each component.
get position vector.
then i have force vector need to get cross product.
get moment about a-b figure out vector and need to know direction.
head-tail divide by length then have unit vector do cross product.

so i started to get moment about a.
i took all co-ordinates from a and got three co-ordinates for b,c and d. i multiplied all these co-ordinates by force 54.8 i added all co-ordinates and got -5.1 nm.

im just not sure i did the right thing. i dont want to move on until i know im on the right track.
any help would be appreciated
thanks

#### Attachments

• moments question.key.zip
124.3 KB · Views: 127