Buoyant Force on Suspended Object
Date: Spring 2012
An object is hung from the ceiling and immersed in a beaker of water. When it is placed on a mass balance, what would the reading be if the combined weight of the water and beaker is W, the weight of the object immersed is X and the upthrust that the object experiences is U?
Whether hanging from the ceiling, immersed in the water, or whatever, the weight of the object will be the weight of the object.
There is nothing special about it being in water, air or whatever.
For your dilemma, we need to re-define the parameters.
VO = Volume of the Object in cubic inches, feet, or whatever you wish to choose.
WO = Weight of the Object in pounds
WW = Weight of the water the VO displaces in pounds.
If WO is less than WW, the object will weigh the same, but it will float. It will float at the waterline of the object where VO = WW
If WO = WW, the object will be in neutral buoyancy, whether on the surface (at which point the top of the object will be just at the surface of the water) or at a depth. It just won’t sink lower or rise higher.
If WO is more than WW, the object will sink.
When a submarine wants to dive to a depth,
It first adds enough weight (in this case in the form of water in onboard ballast tanks) to make WO greater than WW.
To level off at depth, the submarine has to adjust WO to WW by using compressed air to blow the ballast water out of the submarine’s ballast tanks.
To surface, the submarine has to adjust its WO to be less than WW.
Hope this helps.
You are putting the cart before the horse. Before trying to develop a formula, you need to develop a mental “picture” of what is happening. Then the formula will become obvious. Archemedes' principle states that, under the influence of gravity, an object is buoyed up by a force that is equal to the weight of the volume of the fluid displaced by the object. So, for example, if the object is in a vacuum there is no buoyant force since “the weight of the volume of the displaced” is zero, because a vacuum has no mass no matter what its size. The attached web site provides more detail than can be presented in a medium like NEWTON. See http://hyperphysics.phy-astr.gsu.edu/hbase/pbuoy.html . This provides several examples you can examine. Do not be in a big hurry to derive a “formula”. Rather think about the examples on the web site.
Let us do a thought experiment. If you raise the scale and beaker up so that the rock rests on the bottom of the beaker, you get the combined weight of beaker, water and rock.
Now, move the beaker down until the rock is off the bottom and being held by the string. What has changed? The srting is now holding the rock so that its weight is "lost" to the scale.
Draw a free body diagram of the entire system, ceiling, string, rock, water, beaker and scale and it should become clear.
Hope this helps.
Think of it in terms of Newton’s third law. The suspended object must be pushing down on the water with a downward force of magnitude U.
Richard E. Barrans Jr., Ph.D., M.Ed.
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Update: June 2012