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Homework Help: The point of force application

  1. Nov 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello everyone :smile:
    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
    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 :shy:

    Attached Files:

  2. jcsd
  3. Nov 21, 2012 #2
    You can get the lever arms from the force and torque values.
  4. Nov 21, 2012 #3
    yes. if you see I try this is in "The attempt at a solution"
    but its not possible fin Ix, Iy and Iz
  5. Nov 21, 2012 #4
    Isn't torque r crossproduct F?
  6. Nov 21, 2012 #5
    yes I call I the Position vector.. in your case r .. the 2 arethe same..
  7. Nov 21, 2012 #6


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    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
  8. Nov 22, 2012 #7
    if I understand you told me
    Ixyz . Fxyz = 0
    in order to find a colinear vector. ??
  9. Nov 22, 2012 #8


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    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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