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Name: Hasan
Status: student
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A beaker with water and a floating wooden block is placed on the floor of a stationary elevator. The elevator now begins to accelerate upward at constant rate. When equilibrium is reached during the acceleration phase, will the block float HIGHER LOWER SAME LEVEL above the water?

The equilibrium position of a floating object is independent of the force of gravity. The downward force of gravity (Fg=m*g) is balanced by the upward buoyant force (Fb=Vs*rho*g) where Fg is the force of gravity, m is the mass of the floating object, g is the acceleration due to gravity, Fb is the buoyant force, Vs is the volume submerged, and rho is the density of the fluid. The gravity terms cancel when the forces are in equilibrium. If the elevator accelerates parallel to the gravitational field, the acceleration is simply additive with gravity, and applies uniformly to both the water (Fb) and the wood (Fg).

There are lots of 'theoretical' complicating factors like compressibility of the fluid, variance in density of the wood or of the water, changing of position/orientation of the wood (if non-spherical), and even surface tension that could cause different results, but I'm assuming you are choosing to neglect these.

Hope this helps,

Burr Zimmerman

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


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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