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The point of force application

by The-alexandra
Tags: application, force, point
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The-alexandra
#1
Nov21-12, 08:14 AM
P: 15
1. The problem statement, all variables and given/known data
Hello everyone
I have got a cylindrical 6 axis force sensor (so I have the force Fx, Fy,Fz and torque Tx, Ty, Tz). Using these data I don’t know how I can find the point of force application.


2. Relevant equations



3. The attempt at a solution
I try this
The torque can be defined as the cross product I (Position vector) and F(force)
Tx,y,z=Ix,y,z ^Fx,y,z
So, I have 3 equations
Tx=IyFz-IzFy
Ty=IzFx-IxFz
Tz=IxFy-IyFx
But, when I solve the equation(in order to find Ix, Iy and Iz) , the variables in the equation vanish…

Do you have an idea how find this point.
Really thanks so much
Alexandra
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Basic_Physics
#2
Nov21-12, 08:21 AM
P: 358
You can get the lever arms from the force and torque values.
The-alexandra
#3
Nov21-12, 08:25 AM
P: 15
yes. if you see I try this is in "The attempt at a solution"
but its not possible fin Ix, Iy and Iz

Basic_Physics
#4
Nov21-12, 08:30 AM
P: 358
The point of force application

Isn't torque r crossproduct F?
The-alexandra
#5
Nov21-12, 08:54 AM
P: 15
yes I call I the Position vector.. in your case r .. the 2 arethe same..
haruspex
#6
Nov21-12, 03:46 PM
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P: 9,645
Quote Quote by The-alexandra View Post
But, when I solve the equation(in order to find Ix, Iy and Iz) , the variables in the equation vanish…
You can't expect it to give a specific vector for I. Suppose a solution is force G acting through the point r. Let s be any vector collinear with G. Then a force G acting through the point r+s is also a solution (indeed, the same solution really).
One way to fix that is to add the equation I.F = 0
The-alexandra
#7
Nov22-12, 06:47 AM
P: 15
hi.
if I understand you told me
Ixyz . Fxyz = 0
in order to find a colinear vector. ??
haruspex
#8
Nov22-12, 03:36 PM
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Quote Quote by The-alexandra View Post
hi.
if I understand you told me
Ixyz . Fxyz = 0
in order to find a colinear vector. ??
No, not in order to find a collinear vector; in order to find a perpendicular one.
Consider a force F acting through some point in an object. You could shift the point of application to anywhere in that same straight line and it would be exactly the same. I.e. a force acts through a line rather than through any specific point of the line. That's why your original equations were not enough to pin down a point.
Now, any I satisfying your equations would be a perfectly good answer. I merely proposed one way of selecting a specific point from that whole line of valid answers, namely, the point that made the I vector orthogonal to the F vector.


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