buoyancy with sunk objects

[sv3]

so this topic seemed simple enough until i let my bloody brain loose on it and now I'm conjuring up all sorts of scenarios. Here's one that is likely easy but I just want confirmation: If an object is in water, let's say in a pool or the sea, and its density is greater than that of water, it will sink right? And if this is correct, how is there any buoyant force? I figure if an object sinks and is sitting at the bottom of a pool/sea, then there is no water underneath it, the pressures from the sides cancel, and it doesn't go anywhere. So is the buoyant force zero for sunk objects?

The equation "Vsub/V = pobject/pfluid" works well for floating scenarios but I am not sure I can apply it elsewhere? Can I? I'm just a little confused about the whole displaced fluid = mass versus displaced fluid = volume thing. I've been going about both scenarios the same way getting the right answers but afraid I don't get the bigger picture.

thanks
steve

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[Will Hunting]

so this topic seemed simple enough until i let my bloody brain loose on it and now I'm conjuring up all sorts of scenarios. Here's one that is likely easy but I just want confirmation: If an object is in water, let's say in a pool or the sea, and its density is greater than that of water, it will sink right? And if this is correct, how is there any buoyant force? I figure if an object sinks and is sitting at the bottom of a pool/sea, then there is no water underneath it, the pressures from the sides cancel, and it doesn't go anywhere. So is the buoyant force zero for sunk objects?

The equation "Vsub/V = pobject/pfluid" works well for floating scenarios but I am not sure I can apply it elsewhere? Can I? I'm just a little confused about the whole displaced fluid = mass versus displaced fluid = volume thing. I've been going about both scenarios the same way getting the right answers but afraid I don't get the bigger picture.

thanks
steve

I have to go, so I'll give a short answer. There is always a buoyant a force. the buoyant force is maximal when an object is completely submerged. It's just that the buoyant force isn't great enough to match the objects weight. So the net force is downward. You will do well on the MCAT. You have the right mindset. Keep it up.

 

[LaCasta]

There is still a buoyant force (pointing up), but the weight of the object (pointing) is greater than it. So net force is down.


A quick thing about the formula to remember is if the density of the ovject is 90% the density of water, then 90% of the object is submerged and only 10% is sticking out of the water.

Hope that helps

 

[Will Hunting]

I have to go, so I'll give a short answer. There is always a buoyant a force. the buoyant force is maximal when an object is completely submerged. It's just that the buoyant force isn't great enough to match the objects weight. So the net force is downward. You will do well on the MCAT. You have the right mindset. Keep it up.

Now, at the bottom. There is a normal force that acts to counter the weight so that it no longer moves. Perhaps this is what you were referring to. At the bottom it still displaces a volume so I would believe that there is still a buoyant force but it does get murky here. I'm happy to see that you're really questioning everything like the 40+ scorers. That's what separates the gamers from the goats.

 

[sv3]

Now, at the bottom. There is a normal force that acts to counter the weight so that it no longer moves. Perhaps this is what you were referring to. At the bottom it still displaces a volume so I would believe that there is still a buoyant force but it does get murky here. I'm happy to see that you're really questioning everything like the 40+ scorers. That's what separates the gamers from the goats.

is what i meant. like an object in direct contact with the bottom of the pool.....i figured if water isn't under it, then there's no buoyant force -just the normal force but it wouldn't move. But your point about fluid displaced is very true.....interesting. Another thing, if the net force is down due to the weight - buoyant forces then doesn't this mean there is a net acceleration down and eventually all sinking objects hit the bottom no matter how slowly they may go? (due to F=ma).

thanks for the kind words but thinking like this is driving is me mad when i see way more than there really is on some occasions - hopefully it all comes together for me in time for the writing.

As far as the formula, thanks for the explanation. I understand it but TPR says that the volume displaced = the volume of the object in the fluid (whether completely submerged or partially) and Ek gives me the "fluid displaced = mass if partially submerged and fluid displaced = volume if fully submerged". Perhaps they are both saying the same thing as I use the buoyant force formula for both types of problems. I just prefer the TPR way.

thanks very much for the replies. This is what it's come down to...hoping for replies on an internet board on saturday nights......my god.

steve

 

[Fort]

There is still a buoyant force when the object is at rest on the bottom. The argument is even though it may seem that there is no water between the object and the floor, there actually is. Thus, there is still a buoyant force, and it's equal to the weight of the fluid the object displaces.

I know it seems strange, but you have to imagine the fluid being able to squeeze in between the object and the floor.

 

[sv3]

i figured that was going to be the answer. It is hard to beleive but I am sure you are right. Thanks for the reply. At least if this shows up on the MCAT I will have a head start. Now onto electrostatics.......what joy!

steve

 

[BerkReviewTeach]

i figured that was going to be the answer. It is hard to beleive but I am sure you are right. Thanks for the reply. At least if this shows up on the MCAT I will have a head start. Now onto electrostatics.......what joy!

steve

Perhaps this might be a better way to think about your conundrum. Let's say the object was resting on the bottom of a tank. If you were to lift it from rest off of the bottom, would it be easier to lift than if it were in the air? The answer is yes, so there must be some force helping to lift the object.

It would be easier, because the water rushing in to fill the gap below the object would "push" the object up. What is pushing that water into that space is in essence the buoyant force. Now the thing that may help is figuring out what is causing the water to rush into the space below the object. The weight of the water above pushes down on the water below, forcing it to flow into a region of lower opposing force. That same force is always pushing down on the water at the bottom of the tank. The problem is that the force is not strong enough to push the water back into the space occupied by the object. But the fluid is always trying to push its way back into the space occupied by the water, which is what we refer to as the buoyant force.

I'm not sure that's a clear argument or example, but if you can twist your head around that thought over time, it may help.

 

[sv3]

that really helped me picture what is going on at the bottom of a tank/pool/etc. Thanks very much. Actually just came across such a question on my practice materials so it was good to see some of my mind-wandering wasn't a total waste.

cheers
steve

 

[BerkReviewTeach]

Actually just came across such a question on my practice materials so it was good to see some of my mind-wandering wasn't a total waste.

Thinking is never a waste when preparing for this exam.

 

[sv3]

i just have to get better and not going beyond and making things more difficult for myself. I haven't started my timed practice runs yet but want to be as effecient as possible.

thanks again for that s-i-c-k explanation

 

[Shadowplay]

I don't know if this will help, but the Normal force is the equal and opposite force of the object's force pushing down on the bottom of the tank (by definition). There is still a buoyant force, pushing up. Weight is pushing down. Net force is zero (the object is at rest), so:

W + Fb + N = 0

N = -(W+Fb)

W and Fb are in opposite directions. So:

|N| = |W| - |Fb|

The difference in weight, not already countered by the bouyancy, is equal to the normal force.