Fluids (buoyant force) question

[daftypatty]

Hi, so this is my question: if an object with a lighter density than water is placed into water, its buoyant force (Fb) would be greater than its (W), so it floats up. However, once the object reaches the top of the fluid, what is to make it stop floating up? I.e., isn't the Fb still greater than W?

And also, I was just wondering about the equation of buoyant force, Fb = density of fluid x volume of displaced fluid x g ; is the volume of displaced fluid (and therefore the buoyant force) only dependent on the volume of the object put into the fluid & not its mass? Just needed a clarification.

Thanks!

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[desertrat12]

I think there are 2 answers to your first question, 1st: why does it not keep floating up once it reaches the top of the fluid? Well, because it would be impossible for an object to float out of the liquid it was in. That would turn in to levitation. Beyond the scope of the MCAT. I think what you meant, however, is what makes the object - once it reaches the tope of the fluid - remain partially submerged and not continue until the very top of the fluid? If this is the case, the answer lies in what is the buoyant force. Like you said, the buoyant force is equal to the density of the fluid x volume of displaced fluid x g. Density of fluid (mass/volume) is multiplied by the volume of the fluid to leave us with the mass of the displaced fluid. This is then multiplied by g to leave us with Newtons, which makes sense because it is a force. The answer to your second question can help us answer your first question. You are right in that the buoyant force only depends on the volume of the object placed in the fluid. Consider an air craft carrier, super heavy, it manages to float, but if you were able to compact that entire weight into a smaller volume, it would have a much harder time floating. This is because the volume of water displaced is much less. So, the buoyant force = the weight of the water displaced. So, when an object is fully submerged it is displacing an "objects worth of the liquid". If the buoyant force is greater than the weight of the object then it will begin to rise. Because buoyant force does not depend on depth it will continue rising up through the liquid. Things get a little different once we reach the top of the liquid. Once it pops out of the liquid it is no longer displacing the same amount of liquid it once was. Therefore the weight of the displaced water is less and the buoyant force is therefore less. The object will float in equilibrium at the point where buoyant force is equal to the weight.

So, the buoyant force doesn't depend on the weight of the object BUT whether or not the buoyant force will cause the object to rise in the liquid does depend on the weight of the object.

Correct me if I am wrong, but this is how I like to think about it.

 

[daftypatty]

Ah, thanks so much for the detailed response.

"If the buoyant force is greater than the weight of the object then it will begin to rise. Because buoyant force does not depend on depth it will continue rising up through the liquid. Things get a little different once we reach the top of the liquid. Once it pops out of the liquid it is no longer displacing the same amount of liquid it once was. Therefore the weight of the displaced water is less and the buoyant force is therefore less. The object will float in equilibrium at the point where buoyant force is equal to the weight."

So, in a theoretical example, if we were to try submerging a realtively low density object into water, it would start floating up once we let go of it (b/c the Fb > W), but then once it reaches to the top of the fluid, the Fb actually decreases b/c the volume of fluid that was initially displaced becomes lower (so the weight of the displaced fluid decreases). This allows the W(object) and Fb to come to equilibrium?

 

[desertrat12]

You got it. Archimedes principle states that an object displaces an amount of fluid exactly equal to the volume of whatever fraction of the object is being submerged. Also, keep in mind that Apparent weight = Actual weight - buoyant force. When buoyant force is larger than actual weight, the apparent weight of the object is negative so it rises. Once it pops out, archimedes principle tells us that the amount of fluid displaced will then be equal only to the fraction of the object that is being submerged. As the amount of water displaced decreases, so does buoyant force. As this decreases our Apparent weight gets closer and closer to zero until we hit zero. At this point the object is in equilibrium.

 

[daftypatty]

Thanks a lot!!