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Problem from principia

  1. Nov 14, 2011 #1
  2. jcsd
  3. Nov 16, 2011 #2
    I believe it's a consequence of the fact that he starts out with an initial condition of balance where A balances P when KOL is a straight line. We can assume weight P is less than weight A because weight P is suspended from a longer arm. Arm OL > KO, and yet the wheel balances. Weight P must, therefore, be less than weight A.

    When he draws the new radius OD, which gives us the new point, point D, and breaks force DA into its components, component DC, by itself, still balances weight P. This is because, despite being a smaller force than DA, it is now pulling at a more favorable angle (perpendicularly to) radius OD. That perpendicular force component must represent the magnitude of the force that has been in equilibrium with force LP (which also pulls perpendicular to its respective radius) all along.

    Since the forces are in equilibrium, we can take triangle DAC to be a diagram of the magnitude of the forces DA and DC and their proportions to each other, as manifested in the proportions of their literal lengths to each other. The exact proportion of weight A to weight P has, at the same time, also been revealed. Weight P has been, all along, that much less than weight A as line DC is shorter than line DA. P is to A in weight, as DC is to DA in length. If you imagine gravity to be acting along line DC (on the left side of the wheel only), then the only weight that will balance P is a weight equal to P. Weight A placed at that point and pulling in that direction at g would upset the equilibrium.

    Maybe this will help give perspective, too: If you consider the right angle ODC, and imagine it is a separate component that can be rotated around point O by itself, imagine rotating it up so that line OD lies right on top of line OK. Line OD is longer and will stick out to the left farther. Now imagine everything is fused back together in this new position. What weight would we place at point C to balance weight P? It could only be a weight equal to weight P. Arm OD is equal in length to arm OL, so only equal weights (pulling perpendicularly, of course) will balance.

    Weight A, though, in the drawing as it actually is presented, is suspended from a shorter arm. Therefore, to balance the wheel, it has to be greater than weight P. Breaking force DA into its components reveals the proportion of how much greater it has to be when pulling straight down on the radius KO.

    Newton, himself, "deduces" (proves? demonstrates?) all this by pointing out that triangle DAC is similar to triangle KOD ("similar" meaning they have all the same proportions and same angles, despite being different in size). Line KO is to line DO as line DC is to line DA. Since line KO is to line DO in a certain proportion it is also to line OL in that same proportion because line OL is equal in length to line DO.

    In this complicated way he's stated what we knew at the start, that weight P must be somewhat less than weight A. But he's sharpened that up by showing that weight P is to line KO as weight A is to line OL. Weight P is exactly as much less in weight as line KO is in length. If line KO is 1/8 less in length than line OL, then weight P must be 1/8 less in weight than weight A.
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