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The invisible lever in Archimedes’s gravity

  1. Mar 27, 2004 #1
    Say we have one weight floating in free space - vacuum and another weight in the same space on total distance D from the first. I’ll tell you the answer but any way my question for you is: “Is there a lever between the weights”? My answer is yes. Here is how I know it. First I have to find the center point of the invisible lever. I’ll make: P1 = (p1, 0, 0) the position of the first weight; P2 = (p2, 0, 0) the position of the second weight; M1 = (0, m1, 0) the mass of the first weight; M2 = (0, m2, 0) the mass of the second weight. I’ll place M1 with the beginning in the end of P2 and M2 with the beginning in the end of P1. Now I’ll draw the line passing thru the end of M1 and the end of M2. The intersection of that line with the line passing thru the P1 and P2 is the center point of the invisible lever. The distance from the center to P1 is D1. The distance from the center to P2 is D2. The law of the lever is:
    [tex]\frac{F_1}{F_2} = \frac{D_2}{D_1} = \frac{M_1}{M_2}[/tex]
    It’s because for every mass M in the lever there is equally proportional force F. The law of the lever is the simplest form of the Archimedes’s gravity.
    The more complex form is as follows:
    [tex] \frac{F_1}{D_2} = \frac {F_2}{D_1} = \sqrt{G^2 \frac{M_1}{D_1}\frac{M_2}{D_2}}[/tex]
    Here G is the force - mass ratio in the system and also
    [tex]G = \frac{F_1 + F_2}{M_1 + M_2}[/tex]
    If I make M1 = (m1, 0, 0) and M2 = (m2, 0, 0) the center point will still be the same.
     
  2. jcsd
  3. Mar 30, 2004 #2
    Taking the limit as the weight approaches zero and the same with the other weight, a double levels configuration is form.
     
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