if seesaw is balanced under its own weight with no added mass on it you cannot balance on one side. If it is unbalanced under its own weight u can add mass to balance on one side with mass depending on distance to pivot
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This is sometimes referred to as center of mass, as well. If you could take the vector sum of all the torques produced by all of the 'point-masses' to this particular point they will net to zero. For simplicity, consider a weightless see-saw. On the right side of the see-saw is a person weighing 100 pounds, who is 6 feet away from the pivot point. This produces (100 pounds force) x (6 feet ) = 600 foot-pounds force of torque, in the clockwise direction. On the other side of the see saw is a 200 pound person at 3 feet away from the pivot. This will produce (200 pounds force) x (3 feet ) = 600 foot-pounds force of torque, in the counter-clockwise direction. So the see-saw is balanced. Now for each 'point-mass' (this is an infintessimaly small area with an associated small mass), a certain distance away from the centroid, there would be an equal mass the same distance away on the other side, but a vertex is farther away than the opposite side, so there will be two points, each at an angle from the centroid to a point which are a shorter distance, but add to balance the farther points. This is kind-of hard to explain without pictures.
It is the displacement of teeth on the saw, i.e., to bend the teeth of the saw alternately to either side of the blade. ( then I noticed this is in a computer section, sorry).
One way... saw.
I just bought one today. Dealer said it probably was late 50s or 60s model. Paid 250. Was listed for 269.00. Just saw where the same gun was estimated to be worth 300 as a trade in item.
7/4 or 1.75 this is correct i saw one seventeenth hence 4 1/4