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The point of force application 
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#1
Nov2112, 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=IyFzIzFy Ty=IzFxIxFz Tz=IxFyIyFx 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 


#2
Nov2112, 08:21 AM

P: 358

You can get the lever arms from the force and torque values.



#3
Nov2112, 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 


#4
Nov2112, 08:30 AM

P: 358

The point of force application
Isn't torque r crossproduct F?



#5
Nov2112, 08:54 AM

P: 15

yes I call I the Position vector.. in your case r .. the 2 arethe same..



#6
Nov2112, 03:46 PM

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P: 9,829

One way to fix that is to add the equation I.F = 0 


#7
Nov2212, 06:47 AM

P: 15

hi.
if I understand you told me Ixyz . Fxyz = 0 in order to find a colinear vector. ?? 


#8
Nov2212, 03:36 PM

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P: 9,829

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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