Kettle, piece of wood, ball bearings

There are several stainless steel ball bearings on a piece of wood. The piece of wood is floating in a kettle filled with water all the way to the rim. One by one the ball bearings are pushed off the edge of the piece of wood so they fall into the water. What happens to the water level in the kettle?

Originally posted by talzamir
There are several stainless steel ball bearings on a piece of wood. The piece of wood is floating in a kettle filled with water all the way to the rim. One by one the ball bearings are pushed off the edge of the piece of wood so they fall into the water. What happens to the water level in the kettle?

Originally posted by joe shmo
I'm going to go with "nothing" (neglecting dynamic effects).

edit: well...maybe not

Originally posted by joe shmo
My latest

the change in height will be a function of the relative densities of the bearing and liquid

delta_h = V_b/A*((rho_b - rho_liq)/rho_liq)

A = cross sectional area of tank
V_b = volume of bearings
rho_liq = density of liquid
rho_b = density of bearing
delta_h = change in height

which says as long as the density of the bearing is greater than that of the liqiud delta_h will be positive.

Originally posted by LemonJello
That result does not seem right to me.

If on the wood, a bearing should displace an amount of water that has mass equal to the bearing's mass; on the other hand if sunken in the kettle, a bearing should displace an amount of water that has volume equal to the bearing's volume. So when rho_b > rho_liq this latter amount is smaller and delta_h should thus be negative, no?

In light of L.J. pointing out my flawed logic...I decided to give it one more go

delta_h = m_b/A*(1/rho_b - 1/rho_liq)


Indicating that if rho_b>rho_liq; delta_h<0

It seems to be consistent to me...but I've been wrong before. 😞...🙂

SOLUTION

The truth of the matter has been found in the above discussion, as has the reason for it. While the ball bearings are on the piece of wood, they displace enough water to equal their own mass; while underwater, enough water to equal their volume.

Let V_ballbearing be the volume of a ball bearing and rho_steel its density. Their mass is therefore V x rho_steel. We get

V_displaced_water_at_start x rho_water = V_ballbearing x rho_steel
V_displaced_water_at_end = V_ballbearing

V_displaced_water_at_start = V_displaced_water_at_end x rho_steel / rho_water

difference
= V_displaced_water_at_start - V_displaced_water_at_end
= V_displaced_water_at_end x (rho_steel / rho_water - 1)
= V_displaced_water_at_end x (rho_steel - rho_water) / rho_water

which is greater than zero, so the water is higher at the start, ergo, it goes down.

height difference = vol / area =

V_ball_bearing x (rho_steel - rho_water) / (rho_water x area).


Ok.. given that, a follow-up question; if the water is fairly hot, substances like salt, powdered sugar etc. dissolve fast, changing the density of the water without significantly changing the volume. What does that adding salt into the water do to the water level when the piece of wood is still floating in the water?

Not stating at which point of the process you are questioning I would have said down anyways because when the ball rolls off the log but is not in the water yet it isn't displacing anything but air. 😉

Originally posted by talzamir
SOLUTION

The truth of the matter has been found in the above discussion, as has the reason for it. While the ball bearings are on the piece of wood, they displace enough water to equal their own mass; while underwater, enough water to equal their volume.

Let V_ballbearing be the volume of a ball bearing and rho_steel its density. Their mass is therefo ...[text shortened]... salt into the water do to the water level when the piece of wood is still floating in the water?