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Adding forces Again

  1. Jul 29, 2010 #1

    rew

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    I've searched the forum, and found several topics about adding vectors. This one is not as easy... :-)

    Suppose I have an object. There are two forces acting upon it. One is acting on a point at -5,0 and the other at +5, 0. The one at -5,0 has magnitude sqrt(17), pointing up (positive y) 4 units and towards the origin 1 unit. The one at +5,0 is pointing -1,4 also sqrt(17) N.

    Now I can add the vectors, and I end up with an 0,8 vector of resultant force. However, where does this resultant force act? In this case it will be on the Y-axis, but where?

    When the forces are parallel I can find the resulting point: say 1N (0,1) at -1,0 and 3N (0,3) at 3,0 will result in 4N (0,4) in the origin. But how do I find the resulting point if the forces are not parallel ?

    Now, back to my original situation. If I add an 0,-8 force in the origin, the total force will be zero. Total moment as well.

    However if that force is tilted, say in the -1,-4 direction, do I get a moment around the origin? I think I do. I think I could model this resulting moment by finding a point where the resulting 8N force acts.

    Can someone explain how I can find that point?
     
  2. jcsd
  3. Jul 29, 2010 #2

    ZapperZ

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  4. Jul 29, 2010 #3

    rew

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    Ehmm no. I can add vectors just fine.

    Let me try to paraprase my question.

    Suppose I have an object (if you want, the object can be defined as -6<x<6;-1<y<1, a 12x2 rectangle). If I push on it on the left side (force vector (0,4)), it will start to move forward, but also turn clockwise. If I push on it on the right side (again (0,4)), it will also move forward, but start to rotate counterclockwise.

    So where a force acts upon an object makes a difference.

    So given the two vector forces (1, 4) and (-1, 4) I can add them together: (0, 8) Vector addition.

    What we're doing here, is we have multiple forces and "simplify" things by calculating a resulting force.

    However, besides a magnitude, a resulting force also has an resulting point where it affects the object.

    If the forces (0,4) and (0,4) affect an object at -5,0 and 5,0 the resulting force will be (0,8) affecting the object at the origin ( 0,0 ).

    However, when those forces affecting the object at -5,0 and 5,0 are not (0,4) but (1, 4) and (-1, 4) I suspect that the point where the resulting force (again (0,8)) can be thought to affect the object is not at the origin, but somewhere on the positive Y axis.

    I want to be able to calculate that point.
     
    Last edited: Jul 29, 2010
  5. Jul 29, 2010 #4

    K^2

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    If each Fi is applied at location ri, then the total torque on the object is given by ΣrixFi. Then your total force, ΣFi can be applied to any point r, such that rxΣFi = ΣrixFi. You will find that in 2D, there is an entire line of such points. Usually, you take r to be perpendicular to ΣFi, which gives you a unique effective arm.

    x denotes cross product, if it's not clear. In 2D, it is a scalar quantity, and can be found thus.

    axb = axby-aybx
     
  6. Jul 30, 2010 #5

    rew

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    Thanks! In the practical problem I'm trying to solve, I think this results in "doesn't work". So I'm happy I can now solve the physics involved, but not happy that the physics doesn't solve my problem. :-)
    (As you can see from the smiley, for me the theoretical "problem solved" outweighs the practical "have to find another solution")
     
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