Calculating Moment About Axis: A Chiropractor's Procedure

[pinksunbeam]

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 don't want to move on until i know I am on the right track.
any help would be appreciated
thanks

 

[Nate810]

Your approach is correct, but you need to calculate the moment vector correctly. You can calculate the moment vector by taking the cross product of the position vector from A to C and the force vector from C to D. The position vector from A to C is given by: AC = (0.028, -0.166, -0.094)The force vector from C to D is given by: CD = (0.033, 0.001, 0.012)The moment vector is then given by: M = AC × CD = (0.071, -0.006, 0.272)The magnitude of the moment vector is then given by: |M| = √(0.071² + (-0.006)² + 0.272²) = 0.275 Nm.