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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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8y ago
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8y ago

By putting a mass on the other side.

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Q: How do you balance a see saw with one mass on one side?
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Why is the centroid the balancing point of a triangle?

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.


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Related questions

How do you use algebra in a see saw?

The see-saw needs to balance so the force on wither side of the pivot needs to be equal. The turning force is calculated by distance from pivot (in metres) by force. The side with the largest turning force will be lower than the other. If they are equal then the see saw will balance.


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Why is the centroid the balancing point of a triangle?

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.


When children play on a see-saw the long board acts as a?

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Counter tension is when gymnasts perform a balance which involves two or more of them pulling away from each other (or a piece of apparatus), where the weight is not even, whereas counter balance is when a move is balanced (like a sea-saw) or like when a gymnast performs an arabesque, the front and back of their body counter-balance one another.


When children play on a see-saw the long board they sit on acts like a what?

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