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Re: [OT] Forces.

by RichardK (Parson)
on Feb 14, 2016 at 12:31 UTC ( #1155203=note: print w/replies, xml ) Need Help??


in reply to [OT] Forces.

You could resolve force F to be 2 vectors one parallel to AB & one at 90 degrees. The parallel force F' can be ignored as it's just trying to stretch/compress the link AB, the other, F'', provides a turning moment around B.

sorry for the ascii graphics ;) F' <---- A-------B / | / | F V F''

Then as you iterate your solution F'' will reduce to zero, as you rotate AB in proportion to the magnitude of F'' for each step.

It's not entirely obvious to me that the body will rotate about A, it will depend on how the external forces are applied to it. So without more information I don't know if your intuition is correct.

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Re^2: [OT] Forces.
by bitingduck (Chaplain) on Feb 15, 2016 at 03:29 UTC
    It's not entirely obvious to me that the body will rotate about A, it will depend on how the external forces are applied to it. So without more information I don't know if your intuition is correct.

    Yes, it appears that there's missing (or hidden) information.

    If the forces from the wiggly red arrows don't have tangential components, then they won't spin up the disk/cylinder around A.

    If you aren't concerned about how fast it will get there, you don't need the masses and just need the component of the total force on the cylinder tangential to B (i.e. perpendicular to the A-B line, which will rotate about B). At equilibrium, that force will be zero and all the force will be parallel to the A-B axis (unless the force is strong enough to stretch or break things...).

    At risk of messing it up because I'm doing geometry in my head before eating, it looks like F_perp=F_x*sin(theta)+F_y*cos(theta), where theta is the angle measured in the little interior angle that you show in the second pic. When F_perp=0, then all the force is along F_parallel. For completeness that should be F_parallel=F_x*cos(theta)+F_y*sin(theta).

    You can check this by looking at the first picture, where theta=0. sin(0)=0 and cos(0)=1, so F_perp=F_x*0+F_y*1. If the linkage makes it all the way to 90 degrees, then F_perp=F_x*1+F_y*0, so I think that's right.

      Yes, it appears that there's missing (or hidden) information.

      I'm not hiding anything. At least not deliberately.

      If the forces from the wiggly red arrows don't have tangential components, then they won't spin up the disk/cylinder around A.

      I'm not party to any tangential forces involved. That is to say, the only knowledge I have of the forces acting is the bland, X-component/Y-component values returned from the integration.

      My thinking on why the body will rotate clockwise around A as the force causes A to rotate anticlockwise around B, is simply the conservation of momentum.

      That is: if we assume zero friction in the bearing at A (which we must), and an absence of viscosity in the medium (air, water, vacuum) surrounding the body (again, we must since we have no information), then the orientation of the mass of the body will tend to remain the same as the assembly rotates around B, simply because there is no force acting on it to cause it to change that orientation.

      The net affect of the body maintaining its orientation with respect to the universe(B, and the rigidly attached "wall"), as the point A rotates anticlockwise around B, is that the mass of the body appears to (and actually does) rotate clockwise with respect to A.

      If you aren't concerned about how fast it will get there, you don't need the masses and just need the component of the total force on the cylinder tangential to B (i.e. perpendicular to the A-B line, which will rotate about B). At equilibrium, that force will be zero and all the force will be parallel to the A-B axis (unless the force is strong enough to stretch or break things...).

      Agreed. The purpose of the freebody diagram, (and the static link from B to the "wall"), is to isolate this part of the mechanism from the rest of the Universe and so allow you to simplify the overall problem to one that calculates everything with respect B.

      At risk of messing it up because I'm doing geometry in my head

      I concur with your trig.


      With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
      Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
      "Science is about questioning the status quo. Questioning authority". I knew I was on the right track :)
      In the absence of evidence, opinion is indistinguishable from prejudice.
      ), is to isolate this part of the mechanism from the rest of the Universe and so allow you to simplify the overall problem to one that calculates everything with respect B.

        That is: if we assume zero friction in the bearing at A (which we must), and an absence of viscosity in the medium (air, water, vacuum) surrounding the body (again, we must since we have no information), then the orientation of the mass of the body will tend to remain the same as the assembly rotates around B, simply because there is no force acting on it to cause it to change that orientation.

        Yes. That's correct. A better way to think about it is that the body *won't* rotate with respect to the original coordinate system unless you apply a torque, which you apparently aren't doing. So if you put a little mark at the top if it, the mark will stay at the top all the way around the orbit. For convenience in calculating the force on the arm it's nice to have a second coordinate system that rotates with the arm (as described above).

        The thing that I implied above might be missing is details of the forces that might be tangential to the surface of the disk/cylinder that would produce rotation, and which it sounds like we can assume are zero.

Re^2: [OT] Forces.
by BrowserUk (Pope) on Feb 15, 2016 at 08:21 UTC
    You could resolve force F to be 2 vectors one parallel to AB & one at 90 degrees.

    The library routine actually provides me with the X & Y components; I have to combine them to derive the resultant vector.

    The parallel force F' can be ignored as it's just trying to stretch/compress the link AB, the other, F'', provides a turning moment around B.

    The problem with that is it implies that the two components of the force can be applied separately; but that doesn't work. For example, if you applied them in the other order, the turning moment followed by the parallel force, the result is different because after you've applied the turning moment, the parallel force is no longer parallel :)


    With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
    Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
    "Science is about questioning the status quo. Questioning authority". I knew I was on the right track :)
    In the absence of evidence, opinion is indistinguishable from prejudice.

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