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Simple torque question

  1. Jul 29, 2010 #1
    in this diagram i made:

    [PLAIN]http://img201.imageshack.us/img201/1856/physicsqa.png [Broken]

    left: simple application of linear force at center of mass
    middle: application of linear force offset from center of mass gives rise to torque around center of mass
    right: how to explain the forces and torques here?

    most books just say that the contact point with the fixed object becomes the fulcrum
    but i want to understand how that happens

    i have drawn about 20 diagrams - and none of them make sense

    the closest to an explanation i have got:
    as the torque is a force applied to all points in the mass
    the contact force is 1/2 the torque
    an equal and opposite torque reaction somehow adds this 1/2 to the original ...

    can anybody explain this?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 30, 2010 #2
    shameless bump
    can anybody help with this?
     
  4. Jul 30, 2010 #3
    most of the books are wrong then.
     
  5. Jul 30, 2010 #4
    I doubt they are wrong :-)

    But they don't explain the physics behind the pivot

    I am trying to understand the hidden force diagram "behind" the typical force diagram in books
     
  6. Jul 30, 2010 #5
    with a force of 10N and a distance from CoM of 2m the torque would be 20Nm
    but with a pivot 2m from the center of mass (4m from the force) the torque is 14Nm
    that is what I don't understand
     
  7. Jul 30, 2010 #6

    kuruman

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    The right diagram, if I understand it correctly, shows basically what you do when you open a door. If you understand how a door opens, you should understand this. The fixed point marked by "X" is the hinge (call it fulcrum if you wish). As you apply a force away from the hinge, you generate a torque that rotates the center of mass about the hinge. Note that there is just the right force exerted at the hinge to keep the CM in a circular path centered at the hinge. Did I answer your question?
     
  8. Jul 30, 2010 #7

    kuruman

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    You need to understand that, because torques generally depend on the point about which you calculate them. A notable exception is a "couple", i.e. equal and opposite forces, in that the torque they generate is independent of the reference point. Here you have a single force so it matters about what point you calculate the torque. It is best to calculate the torque about the fixed point, then it would be 40 Nm (not 14 Nm). Why is is the best point? Because then you will be able to calculate the angular acceleration of the object using Newton's 2nd Law for rotations and from this the linear acceleration of the CM.
     
  9. Jul 30, 2010 #8
    [PLAIN]http://img441.imageshack.us/img441/3527/physicsqc.png [Broken]

    so in my physics simulator - when it says Torque = 14.054054 Nm
    it is measuring the torque around the center of mass

    whereas the torque around the pivot will be 40 Nm

    the problem here is that I know how to work out the answer according to the book
    but I don't want to use that approach

    I want to use forces only - so a pivot would be modeled as a reaction force etc.

    that is what I am struggling with - that is why I drew a diagram with 3 rectangles
    I want to understand the way forces and torques work when adding a pivot

    not how to calculate things given a pivot

    Am I making any sense?
     
    Last edited by a moderator: May 4, 2017
  10. Jul 30, 2010 #9

    kuruman

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    So you need to replace the pivot with a pair of forces that will do the same job as adding a pivot in the simulation. Is that it? Can you post exactly what is given in this problem, i.e. dimensions of object, forces applied etc. I can only guess what these are.
     
    Last edited by a moderator: May 4, 2017
  11. Jul 30, 2010 #10
    Please provide references where the statement you claimed had been made.
     
  12. Jul 30, 2010 #11
    thanks for taking a look

    I = 10.8333 (at center of mass)
    mass = 5 kg
    force applied 2 m away from CoM
    pivot 2 m away from CoM
    force, CoM and pivot are colinear
    force applied is 10 N (perpendicular to force-CoM)

    I am planning how to build a physics simulator
    but there is no concept of a pivot - only rigid bodies (pos, vel, mass, moi, friction)
    so a pivot could be modelled as (vel = 0, mass = moi = 9e99)
     
  13. Jul 30, 2010 #12

    kuruman

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    You mean "about" the center of mass. I assume the length of the rectangle is 4.0 m, correct? If the rectangle represents a uniform rod, then its moment of inertia about the CM should be
    I = ML2/12 = 5*42/12 = 6.67 kg m2, not 10.833. So what is this I = 10.833?
    Also, as the rectangle rotates, does the force stay perpendicular to the rectangle (i.e. rotates with it) or is its direction fixed?
     
  14. Jul 30, 2010 #13
    the rectangle is 5x1
    the force and pivot are inset 0.5
    10.833 is the default Icm calculated by my (2D) simulator
    maybe the simulator treats it as 5x1x1 I am not sure
     
  15. Jul 30, 2010 #14
    [PLAIN]http://img826.imageshack.us/img826/2307/physicsqd.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  16. Jul 30, 2010 #15

    kuruman

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    OK, so the length of the rectangle is 5.0 m not 4.0 m as I originally thought. Now I agree with the moment of inertia given by the simulation. What about the force? What is its direction as the object rotates? Also, is gravity in the picture or does this happen on a horizontal table? Finally, if I were you, I would turn friction off, understand the simpler problem, then turn friction back on.
     
  17. Jul 30, 2010 #16
    there is no gravity, no friction and the force *will* rotate with the object
    (but I only care about this single point in time so that shouldn't be a factor?)
     
  18. Jul 30, 2010 #17

    kuruman

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    Good. This is what you do.

    1. Use the parallel axis theorem to find the moment of inertia about the pivot.
    2. Use Newton's Second Law for rotations to find the angular acceleration α.
    3. Use α to find the linear acceleration of the center of mass, aCM.
    4. Use aCM in Newton's Second Law for linear motion to find the horizontal force at the pivot. Note that the pivot exerts no vertical force because all the forces are horizontal.
     
  19. Jul 30, 2010 #18
    yes - as per the book - and i can do that already - i did it to check the sim calculations

    but try and derive the solution without the concept of a pivot ...

    or to put it another way - imagine going back in time - you know basic newtonian physics
    and I say "hey look I just invented a pivot" and we have to work out the physics of the pivot
    when we only have torques and forces

    like you said - we need to work out the answer by modelling the pivot as forces - not as a pivot

    the pivot could be something with very large mass or very large coefficient of static friction

    either way is the same - it won't budge - but we can't consider it a pivot
     
  20. Jul 30, 2010 #19

    kuruman

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    My suggestion is to replace the pivot with its effect on the rectangle, namely an additional force. You can find what it is by the method I described to you. So don't use a pivot to find that force, use the idea of "how much force must be exerted at point P located 0.5 m in from the end opposite to the one where F is applied such that point P is at rest while the rest of the rectangle rotates about it." Don't call it a pivot if you don't want to, but whatever it is, it must simulate the motion of the rectangle which requires that point P behave like a pivot, i.e. it is at rest while the rest of the body rotates about it. To do that, a force must be applied at P that you can find as I indicated to you.
     
  21. Jul 30, 2010 #20
    and this is where the problem lays

    I have spent days working out solutions to this and got nowhere

    it seems to come down to that weirdness whereby:

    force X applied at CoM - result = linear force N
    force X applied away from CoM - result = linear force N + rotational force N (huh?)

    i can't even decide what the force against the pivot is because:

    the linear force is directed to the right
    the rotational force is directed to the left

    what is an "equal and opposite torque" ?

    and how to relate these when they are no longer equal?

    F = linear force + torque / d
    F = a / (a+b) N + b / (a+b) N / d

    there are some alternative approaches along the lines of:

    Ax = Ay * |x-p| / |c-p|
    alpha = Ax - Ay / |x-c|
    etc. which might lead to a solution

    but I cannot determine a force diagram that explains how a pivot works!
    and that is the "hole" in the physics books I have
     
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