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Fine but forces are vectors->

  1. Mar 18, 2004 #1
    This post is in relation with some previous post about center of mass. The mass-center of several masses could be found this way:
    [tex]R=\frac{\sum R_iM_i}{\sum M_i}[/tex]
    Here the masses are considered equal in every direction i.e. they are each any vector with M_i intensity. How do we find the center of force knowing that forces are vectors with particular direction?
    A? A? A?
  2. jcsd
  3. Mar 18, 2004 #2


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    There's two parts:

    [tex]F_{net}=\sum \vec{F}_i[/tex]
    [tex]\tau_{net}=\sum \vec{F}_i \times \vec{R}_i[/tex]
    [tex]\times[/tex] is the vector (cross) product of vectors.

    The location that you choose to calculate the net torque from on a rigid body is arbitrary. There is an offsetting change in moment of inertia so that the resulting motion will be the same.
    Last edited: Mar 22, 2004
  4. Mar 18, 2004 #3


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    NateTG's post is completely correct but I am puzzled by the original post:
    "Here the masses are considered equal in every direction i.e. they are each any vector with M_i intensity."

    What??? Masses have no direction, they are not vectors. Force vectors, and vectors in general, have nothing to do with "center of mass".
  5. Mar 19, 2004 #4
    I'll tell you right away:
    "The 1st equation will not work in Archimedes'-lever work space"
    -The forces deeply depend on the position in which they act (take one lever in balnce and you'll see what I mean)
    -The closer you get to the origin (the center) of the system the force is larger by intensity.

    Two particles/objects that perfectly balance each other must:
    -have parallel and oposite distances from the center
    -have parallel and oposite forces in those positions
    -have intensities for the forces and distances that meet this condition F1D1=F2D2;

    Due to the formal equality between the charge, the mass and the geometrical position which are all punctuations their potentials (electrical, gravitational and geometrical-"the force") must also be formally equal meaning the law of lever has same form electrically and gravitationally like it has geometrically.

    As for the 2nd equation I can tell you, after long dedication to the law of lever, that F x D = E x P where:
    F is force
    D is distance
    E is energy
    P is energy potential (different from potential energy)
    all with respect to the center. In other words the product of any thing about one particle and its potential must be same regardless of the type of the quality it applies to.

  6. Mar 22, 2004 #5

    Chi Meson

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    It looks like we all need to be informed of your new difinitions of vectors, forces, cross-products as well as this new, unitless vector quantity you call "energy potential."
  7. Mar 22, 2004 #6
  8. Mar 22, 2004 #7


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    If you have an object that has two parralel opposite forces acting on it, and the two forces are not in line, then they will exert a net torque on the object and it will spin.

    Considering that a lever is a rigid body with (at least) three forces on it - one for each arm and one for the fulcrum - and you're claiming that there are only two I won't be responding anymore.

    It looks like this thread should be moved to theory development.
  9. Mar 23, 2004 #8

    just like if their forces were same by direction and the centre was fixed (like in the lever) there will be no motion at all. you should know that equality of the forces can only be verified when you make the distances same. (Oposite distances and oposite forces) is very much same as (same distances and same forces).
    Right, then the lever is not consisted of two objects but of three where the ground is the third one you are not considering.
  10. Mar 27, 2004 #9
    Lets not quarrel - Lets fight

    Say we have two weights holding on a rigid bar; the bar holding on fulcrum; the fulcrum holding on the ground. We then have a lever. Having the weights performing no motion is trivial case of lever in equilibrium state. In general the lever is in equilibrium state when only its center point is performing no motion. That’s why Archimedes says: “Give me a place to stand and REST my lever on and I’ll break every Newton’s bone”. Because there is no part of the lever performing motion there is no part of the lever subjected to force. Now push the left weight down the right weight will go up. This means that the force - input was distributed on both weights so that the left weight’s force is pointing down while the right weight’s force is pointing up. The lever’s center point still remains immovable preserving the lever’s equilibrium state. So, how are the forces and the distances related?

    I want to accent this: The weights before their shifting might have had both forces pointing down but, consequentially their center point must carry some force also pointing down which usually is different from the pure sum of the two forces (we must shift the weights at the center point to estimate the force carried by the center). Now this lever is not in equilibrium state because its center point carries nonzero force either or the lever is in equilibrium and this force is balanced with some third weight (in our case the ground). Having slightly bigger mass the center point of the system consisted of the two weights and the ground (the global lever) must be closer to the ground which enables the center point of the lever consisted of the two weights only (the local lever) to have noticeable smaller force than the force carried by the ground cause its more distant from the "global" center. You better draw it.
  11. Mar 27, 2004 #10
    The force - mass ratio for each weight consisting the lever must be same. That's simply how the lever works. The mass must also be considered as vector collinear with the force if there's no charge in the system.
  12. Mar 27, 2004 #11

    Doc Al

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    Archimedes is rolling in his grave...

    I presume you are describing a first class lever, in which the fulcrum is between the effort force and load force.
    Equilibrium of a lever occurs when the net torque is zero and the net force is zero. Motion is irrelevant.
    Oh really? Then what is holding those weights (and the bar) up?
  13. Mar 27, 2004 #12
    Nothing is holding any thing up.
    Because of no motion there is no force.
    Because of no force there is no mass.
    There is no need of holding any thing up.
    There's only zero - mass all around.

    When you push one weight you input force and mass.
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