Hi Hasan; that is an interesting question.
If when you say the elevator has reached "equilibrium" you mean the elevator
has reached a constant speed then I think the block would be at the same level.
The block would only be subject to the force of gravity and therefore weigh the same
and therefore displace the same volume of water and therefore float at the same
level. That would be the way a normal elevator operates, i.e., from a stop the
elevator accelerates over a short period to bring the elevator to a constant
traveling speed.
On the other hand if you are saying the elevator acceleration is constant (the
elevator speed is continually increasing) or you are considering a point in time
during the short period when the acceleration is constant, the problem is more
complicated. During the acceleration period when the elevator is going upward then
the weight of the block would be increased because of the additional acceleration.
This additional block weight would lead to increased water displacement resulting
with the block floating lower in the beaker of water. With the elevator accelerating
downward the block would weigh less and float higher.
While I think I am correct in my comments my only hesitation is if Archimedes had
some qualifications (I do not recall that there were) about his principles of
displacement and bouncy if you were in an accelerating system.
Carlton Schroeder
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I assume you have gone over a few of the driving equations for buoyancy but I will
recap them just in case.
When something is floating in a liquid the force up which we will call Fb is equal
to the weight of the liquid that is displaced. In equation form that would be written.
Fb=V*density
Where V is the volume of the liquid displaced and density is the density of the
liquid.
When something is floating then it is in static equilibrium which means the downward
force of the object due to gravity is equal to the buoyant force up.
Force in its simplest terms is F=ma (thank Newton for that one)
Objects that are stationary have a gravitational field intensity of 9.81 m/s^2.
For this example let us call this downward force Fd. Fd=mg (when it is sitting
still)
O.K. So we have gone over some of the preliminary stuff. Now let us think what
happens in our theoretical scenario. We have a block that is happily floating
around and is in equilibrium meaning Fb=Fd
Now the elevator is moving up...let us think what does and does not change.
-The density of the water does not change
-The mass of the block does not change.
-The acceleration of the block is increased and now equal to g+a
-Since the acceleration is increased the Fd is increased because Fd=ma (effectively
the block now "weighs" more)
So now that Fd has increased and we know that Fb=Fd we know Fb has increased. Since
we know the density of the liquid is not changed the only way to increase Fb is to
increase the volume of displacement.
Since more water needs to be displaced the block is going to be more under water and
the block will be LOWER in the water then when it started.
When I think about problems like these I like to make them really extreme. Let us
take the same example but instead of an elevator let us place it in a rocket ship.
When you blast off in a rocket you might have 5 g's (your weight is increased by 5
times). So for the example if the block weighed 2 pounds before and now is
experiencing 5 times gravity it is going to weigh 10 pounds. If the block was
barely floating before it is definitely going to sink now cause it weighs 10
pounds!
Kevin Hardin
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Easy : SAME LEVEL.
The block is already equilibrated with its water-displacement by the 1.0
earth-gravity field it is in.
It displaces a volume of water with mass exactly equal to its own.
Changing to 1.2 gravities, (or even 10 g's) will not change the mass of either,
nor the ratio of the block's weight to the displaced water's weight.
The block also has the same inertia as the water it has displaced.
Gravity-weight and inertial-force are both given by mass.
The largest deviation from this picture is the surface-tension of the water,
and the forces exerted by the meniscus upon the wood block.
This is a rather small deviation.
Meniscus forces might not increase in exact proportion to the net apparent gravity.
Wood is generally wettable, so the meniscus would curve upwards, pulling the wood down.
The radius of curvature would get a bit smaller as apparent gravity increases,
and then the meniscus would pull down harder.
But I think this force-increase might be less than proportional to the net apparent gravity,
(perhaps proportional to the square-root instead)
and the wood might float higher by a fraction of a millimeter when the elevator accelerates upwards.
Or if the wood was hydrophobic from being coated with wax,
the meniscus would curve downwards and its force on the block would be upwards,
and that force would be a smaller fraction of the wood's weight when the local gravity increased,
so the hydrophobic block would sink very slightly as the elevator accelerates upwards.
Jim Swenson
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Hasan,
The block will float at the same level. From within the elevator, there
is no way to tell whether the elevator accelerates upward or "g" has
increased. Increasing "g" will not change the level at which the block
floats, so having the elevator accelerate will not change the elevator
either. The level is based on a ratio of densities.
Now look at it from the point of view of force and pressure. When the
elevator accelerates upward, force between the beaker and the water
increases. This in turn causes the pressure to increase. The increased
pressure will increase the buoyant force on the ball at the original
level. This makes the buoyant force greater than the actual weight of
the ball, providing the ball with a net upward force. This upward net
force then gives the ball an acceleration equal to that of the elevator.
The ball accelerates upward with the elevator without having to sink to
a greater depth.
Dr. Ken Mellendorf
Physics Instructor
Illinois Central College
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