buoyant force on a sumberged body is a product of pressure, which is a product of the height of the water column. When the steel ball is submerged, the displaced water causes the depth to increase, but this does not chance the mass of the water supported by the scale.
Giant mechanical spider
Carson Kelly
Easton Rivera
You can just imagine the equivalent problem of a perfectly balanced scale with two beakers with equal amounts of water.
On the left pan of the scale you put a pingpong ball on a string.
Near the right side of the scale, but not touching anything, is a steel ball suspended by a stand.
It's pretty obvious that the additional weight of the pingpong ball and string on the left pan unbalances things and the scale tips to the left.
Zachary Nelson
>a product of the height of the water column
No it's not, if it was then it would increase as you got further down, and there would be no seabed because all the sand and stones would float upwards to the depth their weight equalled their buoyancy.
Justin Johnson
>equivalent problem
>equivalent
You need to prove this, you can't just assert it.
Matthew Clark
Hadn't considered that. I'm revising mine to say that the right side will drop.
The steel ball will apply less force on the string supporting it. The remaining force will be applied to the beaker of water making it weigh more than the other side.
The ping pong ball still has no net force on it. The ball is trying to float by pushing the water down, but the string prevents this. It's two equal forces cancelling each other out.
Thus the right side has more force which is equal to the amount of force that the steel ball's string is no longer supporting due to being submerged in water. This outweighs any force the ping pong ball's mass could possibly apply.