Other are treating the question as that boat was dropped in water rather than putting it very slowly, in that case yes it will but if you put it very slowly then it won't because of what I said in my previous reply.
Your more precise doesn't help the question. But to answer it no, there shouldn't be a change in torque after the boat from before (comparing steady state to steady state)
It’s the same as adding water to the tank. So long as the boat floats, the pressure is equal across the water. That’s what floating is
What's confusing me is this: without the boat each half of the container has equal amounts of water. With the boat, the left side has
half the water + weight of the boat - equal weight in water + some of the displaced water
and the right side has
half the water + some of the displaced water
But because the displaced water is in a flat layer, the right side must have more, because the boat takes up some space.
The weight of water moved to have the boat float is equal to the weight of the boat. Think of the boat as differently shaped water, but still the same mass. The boat takes up more space because it has a lower density to be able to float in the first place
The diagram is a bit off in my estimation, but no. Liquid only fills the space by the weight that is lost by the missing air, so it may as well be considered level.
>Liquid only fills the space by the weight that is lost by the missing air
???
2 years ago
Anonymous
so if you suck 1kg of air out, 1kg worth of liquid will fill the space.
2 years ago
Anonymous
i dont think thats right. if you suck out 29.92 inches of air from the right side, 29.92 inches of mercury takes its place, but mercury is obviously heavier than air, so the right side is now heavier.
2 years ago
Anonymous
>if you suck out 29.92 inches of air from the right side, 29.92 inches of mercury takes its place
only if the way you're measuring inches of air is how much the mercury rises. You can measure how much air is in an inch, and suck it out, and find that the mercury rises less than an inch. The reason for this is because mercury, being heavier than air, pulls on it and forms a partial vacuum. If this weren't the case, than after sucking out 29 inches of air, the rest of it would be at atmospheric pressure, the mercury having filled it's space, and if so, what force is keeping the mercury up?
2 years ago
Anonymous
i think i see what you mean now, but i think you meant to user units of force here
so if you suck 1kg of air out, 1kg worth of liquid will fill the space.
instead of units of mass. the weight of the air on the left side equals the weight of the mercury on the right, because theres an entire atmosphere sitting on top of the left side.
2 years ago
Anonymous
Yeah that sounds right!
and it's a good idea, to consider the atmosphere's pushing force rather than a vacuum's mythical "sucking force". Thanks
Forget pressure homosexual
Apply force on infinitesimal portion of the system = ρgdV and integrate over infinitesimal torque = ρgrdV
You find non-zero value.
Also pressure is not uniform gay, that's why things bend.
Your more precise doesn't help the question. But to answer it no, there shouldn't be a change in torque after the boat from before (comparing steady state to steady state)
Other are treating the question as that boat was dropped in water rather than putting it very slowly, in that case yes it will but if you put it very slowly then it won't because of what I said in my previous reply.
how do you contest that the center of mass is still directly above the fulcrum? are you suggesting the water is not at a uniform height?
This and basic distribution of water based on mass distributed to make the boat float. If the boat sunk and was at the bottom then definitely non-zero net torque added
oddly enough, something floating on another thing doesn't add any weight to it directly. Simply put, if the whole thing were more dense than the water it replaces, then it wouldn't float, and however much it floats above, is the extra weight. So if you 'replace' the water it's convex hull is taking up if it were level with the boat, the weight doesn't change. Since the water is already there, it does change, but there isn't a direction to the pressure.
The water level rises, but the weight doesn't change from the boat.
moron, weight doesn't add if the replaced water gets spilled somewhere else, aka leaves the system. Yes, cause the volume displaced has the same mass as the mass of the boat underneath the water.
If the water stays within, mass adds up.
Doesn't take a freaking genius to figure it out.
The center of mass is exactly above the tip with or without the boat, but there is a state in between before the system stabilizes after you add the boat or when it moves so I guess it would tip.
>it's otherwise not evident that the dimensions of the container are equivalent to prevent tipping without the boat.
Yes it is. You have autism. The question obviously isn't asking people to get out rulers and measure the container.
Having a boat in the container is identical to a situation where the "bulge" created by the boat below the surface level was filled with water. There would be no difference in weight, density, anything. So the container would not tip.
The weight of water moved to have the boat float is equal to the weight of the boat. Think of the boat as differently shaped water, but still the same mass. The boat takes up more space because it has a lower density to be able to float in the first place
Suppose we have just a normal container of water, nothing in it except water, everything normal. Now it's trivial to see that the container would not tip.
Now, what we are going to do is to "carve out" a hypothetical region of water (illustrated in picrel). By definition, the shape of our piece of water is the exact same shape that a boat would create if it was in the container in the same spot. Now, what we do next is to place that boat in that same place. According to our axioms, the boat has the same weight as the water that used to be there.
Now, gravity is pulling the boat downward with a certain force. Let's call that force X. However, our hypothetical region of water was also being pulled downward by gravity. Let's call the downward force of our water region Y. The region of water was defined to be the bulge created by the boat, so whatever force the boat exerts to the left side of the container, by definition, the hypothetical water region exerts the same force. Therefore X=Y so the container does not tip.
How would the water know that we took some water out and replaced it with a boat? After all, they have the same weight. The only difference is in material (one is made out of steel) and density (the boat takes more space).
Thank you, this got me about as close to getting it intuitively as I think I will get. I'm still struggling with the pseudo formulae of
[...]
[...]
What's confusing me is this: without the boat each half of the container has equal amounts of water. With the boat, the left side has
half the water + weight of the boat - equal weight in water + some of the displaced water
and the right side has
half the water + some of the displaced water
But because the displaced water is in a flat layer, the right side must have more, because the boat takes up some space.
Maybe an easy way to think of it is to think how pressure works. If you submerge your hand, basically a certain force is applied to every unit of surface area on your hand. So pressure is what prevents boats from sinking. But the same pressure that keeps the boat floating still exists in the water even without the boat. If the forces of pressure are equal with and without a boat, the container does not tip.
this experiment shouldn't be too hard to check. get a small container with water, balance it on a small fulcrum, then slowly put a boat in it near the edge.
Suppose we have just a normal container of water, nothing in it except water, everything normal. Now it's trivial to see that the container would not tip.
Now, what we are going to do is to "carve out" a hypothetical region of water (illustrated in picrel). By definition, the shape of our piece of water is the exact same shape that a boat would create if it was in the container in the same spot. Now, what we do next is to place that boat in that same place. According to our axioms, the boat has the same weight as the water that used to be there.
Now, gravity is pulling the boat downward with a certain force. Let's call that force X. However, our hypothetical region of water was also being pulled downward by gravity. Let's call the downward force of our water region Y. The region of water was defined to be the bulge created by the boat, so whatever force the boat exerts to the left side of the container, by definition, the hypothetical water region exerts the same force. Therefore X=Y so the container does not tip.
How would the water know that we took some water out and replaced it with a boat? After all, they have the same weight. The only difference is in material (one is made out of steel) and density (the boat takes more space).
You could think of the fact that the displaced area underwater created by the boat is related to the force of the boat. If you add some weight to a boat, it sinks and has to displace a bigger volume under the surface. If a boat weight 100 000 tons, the water must be lifting the boat upward by a force equal to 100 000 tons to prevent the boat from sinking. This lifting force is applied to that part of the boat which is underwater. But if you removed the boat and just replaced it with water, that region of water would also weight exactly 100 000 tons so that it also would not sink nor rise upward. Think of it as sort of a boat made out of water.
[...]
[...]
Thank you, this got me about as close to getting it intuitively as I think I will get. I'm still struggling with the pseudo formulae of [...] but I'll sit with it.
are you guys saying that there is more water to the right of the fulcrum than to the left of the fulcrum?
Yes. Put the container of OP's pic in a freezer, then take the boat out of the piece of ice. You can see that there is now sort of a concave bulge where the boat used to be, so that half of the block of ice has less ice than the other half.
It's impossible to put in the boat slowly enough for it not to not tip unless you put in the boat somewhere on the right and apply just enough force to counterbalance the tipping to the right until it stops moving left.
no it doesnt tip. good physicslet identifier, though.
[...]
>There are two external forces on the system containing both boat and the water tank.
yes. those two forces are gravity and the normal force between the fulcrum and the earth. thats it. these forces are collinear (the presence of the ship does not change the COM) so theres no moment: nothing to cause it to tip.
yes we understood archimedes principle the first seventy times it was posted here. what you neglect to mention is that this missing water is DISPLACED. you know what that means? it moves. it creates an instantaneous pressure gradient, i.e. a force field (newton's law... the accelerated water must also experience a force). so there must exist an instantaneous net torque somewhere in this tank, meaning it must tip.
here's the explanation in brainlet language for you: see that water with all the arrows on it that you drew? well, you have the boat where that water used to be. now you need to put that water back in the tank.
>what you neglect to mention is that this missing water is DISPLACED
this isn't actually something you have to consider unless you're talking about dumping the boat into the water with force. This is because 'replacing' and 'displacing' water, without that extreme force, is the same thing.
You can imagine slowly lowering the boat into the water, so that 'pressure gradient' is minimized, and then dumping out any rise in water level. There you have it: the non-displaced variant. Essentially, it doesn't matter how much water was in there before and after the boat is dropped, Archimedes principle still works.
>'pressure gradient' is minimized
it will never be zero, no matter how slowly you place the boat in the water. there will always exist a pressure gradient due to the displacement (acceleration) of water, no matter how small. as such, there must exist a nonzero force and thus a nonzero torque on the system. the tank tips.
2 years ago
Anonymous
[eqn]
lim_{t to infty} frac{Delta x}{t} = 0
[/eqn]
motion is negligible
2 years ago
Anonymous
lol, do the experiment then and post it here. prove my smug ass wrong.
2 years ago
Anonymous
brainlet: you realize you can simply draw a free body diagram and learn that there is no tipping moment in a matter of seconds, right? of course it would be nice to do the experiment for real, but I actually do not care enough.
prove you aren't a clinical shit-for-brain and draw me a free body diagram right now.
2 years ago
Anonymous
another anon here.
I have balanced a pot of water on a tiny bottle of rum extract. I find that when I drop a can of conola spray on the far side, it usually falls on the near side, as expected from 'displacement'.
When I slowly place the spray into the water, however, it does not fall. With the pot emptied, placing the conola spray slowly, in the same angle and arrangement it is placed into the water with the pot full of water, the pot tips over.
If you ask nicely, I may send pics, but this is a fun household experiment anybody can do.
2 years ago
Anonymous
Based empirical anon.
2 years ago
Anonymous
>balanced a pot of water on a tiny bottle of rum extract.
if you're clever enough to do this experiment, surely you recognize this isn't an ideal fulcrum right? pic rel is your setup. so it doesn't surprise me that you can slowly trickle water in, and it won't tip. you created forces, but the fulcrum is stable enough to withstand those forces.
consider a situation where you used a very, very, very long tank (10 meters long) on your little bottle of rum extract. i expect spraying will cause it to tip.
2 years ago
Anonymous
>surely you recognize this isn't an ideal fulcrum right?
you do realize physicists exaggerate their drawings right? it's to drive home the point that you have a stable fulcrum, as opposed to an ideal point-like fulcrum. here's the deal: in both cases there are forces (and hence, torques) introduced whether you drop your bottle or slowly place it. with an ideal fulcrum, even the tiniest force will cause it to tip. in your stable fulcrum situation, the setup can withstand tiny forces.
your experiment doesn't prove what you think it does.
>you do realize physicists exaggerate their drawings right?
you talk like a gay, or a woman.
2 years ago
Anonymous
>if you're clever enough to do this experiment, surely you recognize this isn't an ideal fulcrum right? pic rel is your setup. so it doesn't surprise me that you can slowly trickle water in, and it won't tip. you created forces, but the fulcrum is stable enough to withstand those forces.
this is true, and would cause a problem if the spray were truly weighing one side down as usual, and the fulcrum kept it stable regardless. But the fact that it tips either way when the pot is emptied is evidence to the contrary.
If you won't take an example from the real world(As there isn't any perfect fulcrum in real life), then what were you doing asking for an experiment?
2 years ago
Anonymous
>But the fact that it tips either way when the pot is emptied is evidence to the contrary.
well no. the case without water is trivial to analyze. as a matter of fact, it's a useful case to study. if you're dedicated enough, you can measure the minimum distance from the fulcrum that causes it to tip. i guarantee you this distance will be nonzero, and a bit larger than you expect. this is evidence your fulcrum is not ideal.
now fill the pot with water, and drop the canister in at different distances from the fulcrum. you'll see that the minimum distance that causes the pot to tip will be a bit larger than if it's empty. this is further evidence your fulcrum (with a pot filled with water) is again not ideal.
but most importantly, as you noticed, when the pot has water, and when you slowly submerge the canister inside the pot, the the pot+water+canister doesn't tip. again, i don't doubt you. i never claimed this was a large force. all i claimed was that this was a force (as a matter of fact, i claimed it was a small force). under the effect of an ideal fulcrum, this small force would cause the tank to tip.
because your fulcrum is not ideal, these small forces that occur when you submerge the canister will not cause the pot to tip. if, however, you were to use a longer pot, or a thinner fulcrum, you'll see it would be much, much more difficult to submerge the canister. under an ideal fulcrum, it'll be impossible to submerge the fulcrum without tipping the pot/tank.
2 years ago
Anonymous
>under an ideal fulcrum, it'll be impossible to submerge the fulcrum without tipping the pot/tank
I suppose this is true. The slower you go, the less the force, but any force will do it.
I still don't think that's what the picture was about, though. the OP just shows the boat in the water, presumably everything is still and settled, and nothing was dropped or changed.
2 years ago
Anonymous
>I still don't think that's what the picture was about
i disagree. i interpreted OPs scenario to be that the tank was empty, and then the boat was submerged. i think my interpretation is valid given what op showed (which was ambiguous). really, this all comes down to op not clarifying what the question actually was.
2 years ago
Anonymous
>i interpreted OPs scenario to be that the tank was empty, and then the boat was submerged
pure fricking autism. theres nothing to indicate this whatsoever.
2 years ago
Anonymous
you say autism like it's a bad thing. when did you first start browsing IQfy?
2 years ago
Anonymous
its been used here as an insult for a decade newfriend, like 2013 or something
this isn't to mention that you do legitimately have some mental deficiency for randomly inserting details to a prompt that aren't hinted at being relevant
2 years ago
Anonymous
>I still don't think that's what the picture was about
i disagree. i interpreted OPs scenario to be that the tank was empty, and then the boat was submerged. i think my interpretation is valid given what op showed (which was ambiguous). really, this all comes down to op not clarifying what the question actually was.
sorry, i'm drunk. firstly gotta correct the typo. >nder an ideal fulcrum, it'll be impossible to submerge the fulcrum without tipping the pot/tank
obviously you gotta submerge the cannister. >i interpreted OPs scenario to be that the tank was empty
i interpreted the scenario as the tank was filled with water, and then the boat was submerged.
2 years ago
Anonymous
A better way would be to have a long tank of water (like a fish tank) and have two scales. The tank would sit on the scales so that the opposite edges would have one scale underneath. You could use something like a food scale which is accurate up to one gram difference. Then you would place a boat very slowly directly on top of one scale and see if the numbers on boths scales change at the same time and stay equal.
2 years ago
Anonymous
clever experiment.
2 years ago
Anonymous
So just suppose you hold the tank in place while you put the boat in, and only let it free after a long time.
>Archimedes principle still works.
yes archimedes principles works. that's not why the tank tips though. the tank tips not because of the mass of the boat, but because of the displaced water which must accelerate
2 years ago
Anonymous
nobody is considering the case where the water is in motion you fricking autist. this is a hydroSTATIC problem.
2 years ago
Anonymous
before the boat is submerged, there are equal amounts of water to the right of the fulcrum and to the left of the fulcrum. agreed?
after the boat is placed in the tank, archimedes principle comes into play... and now the amount of water to the left of the fulcrum and to the right of the fulcrum are no longer equal. agreed? (otherwise the tank tips)
therefore, water must have MOVED from left to right. therefore, it's not a hydroSTATIC problem. dumbass.
2 years ago
Anonymous
you're fricking autisic if you think the transient motion of the water as the boat is placed is of concern.
imagine that the boat has been floating on the right hand side for a long time with the fulcrum pinned. then, a switch is activated and the pin is pulled and the fulcrum is allowed to rotate. does the thing tip? no. is there a torque causing it to tip? no.
2 years ago
Anonymous
you've changed the problem. obviously in a different problem the tank won't tip, lol. i've already explained it as best i can. if you can't understand that explanation, then idk what to tell you bro.
brainlet: you realize you can simply draw a free body diagram and learn that there is no tipping moment in a matter of seconds, right? of course it would be nice to do the experiment for real, but I actually do not care enough.
prove you aren't a clinical shit-for-brain and draw me a free body diagram right now.
free body diagram is a static picture. as i already explained, this isn't statics. it's dynamics.
2 years ago
Anonymous
>if you can't understand that explanation
I understand anon, it's just baffling that you think this is a dynamics problem. >free body diagram is a static picture. as i already explained, this isn't statics. it's dynamics.
schizophrenia
2 years ago
Anonymous
before: equal water left and right of fulcrum.
after: more water right of fulcrum than left of fulcrum
how the frick does that happen in a static picture? how does water not move for that to happen? do you even understand what archimede's principle is? the water is DISPLACED. how do you think water gets displaced in a static picture?
2 years ago
Anonymous
there is no """"before and after,"""" moron. we are GIVEN the state depicted in the OP as the initial state. does it tip? no it does not. it does not tip because there is no moment.
stop randomly inventing details to the problem that dont exist >do you even understand what archimede's principle is?
do YOU, homosexual? >the water is DISPLACED
yes >how do you think water gets displaced in a static picture?
by not being located inside the hull of the ship??? what the frick is this confusion
2 years ago
Anonymous
>there is no """"before and after,"""" moron. we are GIVEN the state depicted in the OP as the initial state >my assumptions are more valid than your assumptions
at least i had the intellectual honesty to admit from the beginning what my assumptions were when i gave my answer. you, however, refuse to admit that you're making an assumption.
2 years ago
Anonymous
We could also assume the tank is glued to the fulcrum, or the whole setup is floating in 0g, but that would be fricking autistic. The assumption that this is somehow a dynamics problem from a static image with nothing indicating motion is also fricking autistic.
2 years ago
Anonymous
you may not know this, but all images are static. are you trying to say that no static image can represent a static moment of a dynamic event? you're the one being autistic, staying fixated on a single interpretation of a literalist event.
On the left, everything is balanced. On the right, the water height is the same and the weight of the boat is pulling down to the side. There is a torque and it will tip.
Also, assuming that the boat did not exist at the side of the water for an infinite amount of time, any motion to the side will create an angular momentum about the fulcrim's tip which will also cause tipping.
Doesn't matter. The pressure on the bottom of the tank must be equal everywhere either way and the air pressure applies uniform force onto the water so nothing changes.
do you mean which would jump highest out of the water, or do you mean which, in the instant that this drawing describes, has the most upward force acting on it?
learn how buoyancy works for the love of christ the three on the right all experience exactly the same upward force because they are all presumably the same size.
LMAO FOOLED BY HIS OWN WORDS
yeah, exactly same force as much as they didn't leave the water.
But you forgot that force is proportional to the second derivative of motion, and initial velocity when they reach surface will not be the same (accounting for a perfectly homogenous fluid, without wavelets so that motion doesn't get tilted and stays upward).
OP used the word hardest, which is qualitative and doesn't precise in which way, so we assume he is asking which block will attain highest altitude. Or in a different scenerio where surface velocities = 0, which block needs the highest work to reach its respective height.
Answer is right block.
The above explanation answers the question.
>balanced a pot of water on a tiny bottle of rum extract.
if you're clever enough to do this experiment, surely you recognize this isn't an ideal fulcrum right? pic rel is your setup. so it doesn't surprise me that you can slowly trickle water in, and it won't tip. you created forces, but the fulcrum is stable enough to withstand those forces.
consider a situation where you used a very, very, very long tank (10 meters long) on your little bottle of rum extract. i expect spraying will cause it to tip.
you should be embarrassed by all of your posts ITT
2 years ago
Anonymous
not an argument.
2 years ago
Anonymous
>this cant be settled through experimentation because ideal point-fulcrums dont exist >i think a "tiny bottle" is like 6 inches wide
listen to how dumb you sound
2 years ago
Anonymous
you do realize physicists exaggerate their drawings right? it's to drive home the point that you have a stable fulcrum, as opposed to an ideal point-like fulcrum. here's the deal: in both cases there are forces (and hence, torques) introduced whether you drop your bottle or slowly place it. with an ideal fulcrum, even the tiniest force will cause it to tip. in your stable fulcrum situation, the setup can withstand tiny forces.
your experiment doesn't prove what you think it does.
2 years ago
Anonymous
im not the experimental anon >with an ideal fulcrum, even the tiniest force will cause it to tip.
no one here disagrees, so you can go ahead and leave now, but youre still moronic for thinking that was what OP was asking.
2 years ago
Anonymous
>no one here disagrees
you clearly haven't read the thread.
>surely you recognize this isn't an ideal fulcrum right?
[...] >you do realize physicists exaggerate their drawings right?
you talk like a gay, or a woman.
close. i talk like a scientist.
2 years ago
Anonymous
>you clearly haven't read the thread.
all i saw was some guy trying to explain to you why you shouldnt make up your own shit about the question.
2 years ago
Anonymous
>you shouldnt make up your own shit about the question.
take your own advice
im not the experimental anon >with an ideal fulcrum, even the tiniest force will cause it to tip.
no one here disagrees, so you can go ahead and leave now, but youre still moronic for thinking that was what OP was asking.
>youre still moronic for thinking that was what OP was asking.
unlike you, i clearly laid out my assumptions, which morons like you decried as the incorrect assumptions. i've already acknowledged that if your assumptions are correct, then the tank doesn't tip. dunno why you're incapable of doing the same. and if you are, idk what your problem is. do you think it's impossible your assumptions aren't the correct ones?
>OP used the word hardest, which is qualitative and doesn't precise in which way, so we assume he is asking which block will attain highest altitude.
being submerged lower would actually be a detriment to the height due to vortex shedding.
2 years ago
Anonymous
>[assumption 1] [assumption 2] [assumption 3]...
Look up my reply you will find them.
buoyancy = specific weight of water × volume
i dont see anything about depth here, do you????
youre only outing yourself as a pseud harder with this post
It's gotta tip, even though some anons are right under idealistic conditions of classical physics, the truth of the matter is pressure distributing is a system properties which emerges from particle interaction among other factors, so a perturbance bound to tip it is inevitable
>insert
what insertion? the boat is already in the water, this is the initial state
2 years ago
Anonymous
How the frick did it get in there? That's what I'm saying, a system like this has to have had a displacement which makes it tip in the real world, or it doesn't make any sense
>Best I can do
Thats not a free body diagram. are you seriously incapable of doing this?
Yeah I don't think I ever had to draw this shit, I'd probably do it wrong if I am precise but you get the general gist of it, you are applying only a macroscopical level model not even thinking about the dynamics over time to things which would have an effect on a microscopical one
2 years ago
Anonymous
>How the frick did it get in there?
???
It starts in there. The FBD is extremely simple. theres one force vertically down (gravity), and another force vertically up (normal force). They are equal in magitude and COLLINEAR. So there is no torque. Do you agree or disagree?
2 years ago
Anonymous
I agree it is a bullshit mathematical model impossible in reality which is stable by construction
2 years ago
Anonymous
>I agree it is a bullshit mathematical model impossible in reality which is stable by construction
Pin the fulcrum so that it doesnt move. Place the boat gently in the water. Wait until the water stops moving. Release the pin. Does it tip? No.
2 years ago
Anonymous
>Pin the fulcrum so that it doesnt move. Place the boat gently in the water.
why do you have to pin the fulcrum? you do realize you just agreed with people who say the tank tips, right?
2 years ago
Anonymous
i didn't, at no point did i agree that this was a dynamics problem
2 years ago
Anonymous
>why do you have to pin the fulcrum?
not him/her, but they're saying that to rule out transient effects from placing the boat in the water.
2 years ago
Anonymous
>transient effects from placing the boat in the water.
clarify
2 years ago
Anonymous
something about instantaneous pressure gradients. see
yes we understood archimedes principle the first seventy times it was posted here. what you neglect to mention is that this missing water is DISPLACED. you know what that means? it moves. it creates an instantaneous pressure gradient, i.e. a force field (newton's law... the accelerated water must also experience a force). so there must exist an instantaneous net torque somewhere in this tank, meaning it must tip.
here's the explanation in brainlet language for you: see that water with all the arrows on it that you drew? well, you have the boat where that water used to be. now you need to put that water back in the tank.
. idk man, not a physicist.
2 years ago
Anonymous
>drop the boat in the water >immediately, the water is pushed from the hull and sloshes in the container >this is transient dynamics >after a few minutes, the water comes to a halt and the boat is floating still >this is hydrostatic
homie please please please for the love of god just open a book on freshman mechanics
2 years ago
Anonymous
>a hydrodynamics problem is hydrostatic after it stops moving >therefore the tank doesn't tip because it's a hydrostatic problem
...
2 years ago
Anonymous
the problem is PRESENTED as a hydrostatic one; the ship is ALREADY in the water and this is no motion indicated. not only is there no motion indicated, there is NO TORQUE that could ever possibly put it into motion!
2 years ago
Anonymous
sigh. i cannot believe i have to say this, but i will. you're making an assumption that >the tank was filled with water >the boat is in the tank >the fulcrum was fixed as this was occurring >all until the arrangement in op pic rel is achieved >ergo it's hydrostatic and the tank doesn't tip
meanwhile, i make the assumption >the tank of water is balanced on the fulcrum >the boat is then placed in the tank of water >ergo it's a hydrodynamics problem and the tank tips
why are you getting to pissy about my assumptions being faulty, while refusing to admit that you yourself are making assumptions? you're literally incapable of recognizing you're making assumptions. i'm not even saying you're wrong (like you are to me). i'm saying that under different assumptions, different things occur. the fact you're unable to recognize the assumptions i make says more about you than it does about me.
2 years ago
Anonymous
>meanwhile, i make the assumption
that's not what is depicted though
2 years ago
Anonymous
OP here, I was asking about the hydrostatic state.
2 years ago
Anonymous
assuming you are OP (see what i'm doing? i'm clearly laying out my assumptions), then you should more clearly communicate the problem next time.
2 years ago
Anonymous
The diagram clearly communicated the underlying question. You would have to be insane to believe that someone was asking if you could tip a balance by pushing on one side.
2 years ago
Anonymous
i have to be insane to assume realism? frick off.
2 years ago
Anonymous
Do you agree there are precisely two external forces or no? Do you agree the forces are collinear or no?
2 years ago
Anonymous
>initial value problems are impossible to solve because you actually need to know the values before t=0 otherwise it doesn't make sense
how is that possible? ignoring the displaced water, the mass of water+boat should be uniformly distributed. now, once you consider the displaced water, there's more on the right side, hence a greater mass on the right side.
>the mass of water+boat should be uniformly distributed
what? are you conflating mass with pressure? yes, theres more water on the right. there's less water on the left, and also a ship on the left. there is exactly equal mass on either side. the COM remains on the vertical line passing through the fulcrum.
>and yet there is equal mass on either side of the fulcrum, and the center of mass is directly above the fulcrum.
If the boat weighed as much as water, it wouldn't float idiot.
Center of mass will never be balanced in the middle unless the boat is also in the middle.
the boat indeed weighs ***exactly*** as much as the
weight of the water that would occupy its hull.
holy frick lmao how are you this dense getting filtered by 2500 year old physics like damn moron
you claim to be a cs graduate yet you don't know how archimedes pressure works? the submerged part of the boat endures an upwards push from the water below it because the boat is indeed less dense than water. however it is only submerged partially because in its current state, the displaced water exerts the same force upwards as the boat's weight downwards, that's why it doesn't move. in its current state the boat isn't accelerating so there are no forces or all the forces cancel each other out, so you might as well ignore the boat entirely
2 years ago
Anonymous
>so there are no forces or all the forces cancel each other out
So you genuinely believe that gravity magically gets buffered by the water and ignores the relative position of the boats mass because it's displacing water?
2 years ago
Anonymous
The column of water directly underneath the boat contains exactly a boat weight worth less water than the mirror column on the opposite side.
2 years ago
Anonymous
You're silly, buoyancy isn't about columns like pressure it's about displaced volumes, here the water volume displaced by the boat's hull. You learn this in high school is everyone here a kid or what
2 years ago
Anonymous
>You learn this in high school is everyone here a kid or what
Not that anon, but for me it's the opposite. I've been out of school for 15 years and since I haven't needed this knowledge even once that entire time in my day to day life I've naturally just forgotten it. Use it or lose it, and like 80% of what you learn in school goes unused.
2 years ago
Anonymous
...it was an explanation of that principle. Do you have trouble following conversations generally, or do you just make being a c**t a habit for fun?
2 years ago
Anonymous
I have trouble following conversations I'm a bit of a sperg. although I don't see how using columns is in any way relevant to buoyancy, since it implies that the force is dependant on depth in any way.
2 years ago
Anonymous
Gravity is a body force that acts on every molecule in the system at once. This makes it tricky to visualize, fortunately it can ALWAYS be replaced (in the static case) by an equivalent force equal to the total weight of the system, and acting at the center of weight or center of mass: [math] text{CoM}=intmathbf{r}dm / int dm [/math]
So think carefully: does the presence of the ship change the COM at all?
2 years ago
Anonymous
vvvvv
[...]
The column of water directly underneath the boat contains exactly a boat weight worth less water than the mirror column on the opposite side.
The mass-density gradient across the surface of the tub will have a noticeable dip where the boat is. There is less water under the boat due to displacement, ergo there is a higher density on the other side. It is not balanced.
2 years ago
Anonymous
Explain my diagram then
[...]
[...]
I went out of my way to draw you an accurate picture of an example. The total mass of water is conserved and the submerged block is half as dense as water which means it sticks halfway out of the water.
Bonus points if you can figure out where the displaced water went.
2 years ago
Anonymous
your box is lighter than water
2 years ago
Anonymous
it literally weighs the same as the water.it is less dense than water (like all boats ever built), and it weighs the same as the water (just like all boats)
2 years ago
Anonymous
>The mass-density gradient
wrong. the "mass-density gradient" is uniform. all surfaces of equal pressure are parallel to the free surface.
2 years ago
Anonymous
No its literally not, the density in the region of the boat and the water under it is not the same as the density from the other water around it.
2 years ago
Anonymous
the boat weighs the same, and its hull the same volume of water that would otherwise be there. so the weight/volume=density is the same.
2 years ago
Anonymous
The column of mass pressing on the bottom of the tank has the exact same mass per area as at every other point.
your box is lighter than water
It isn't. The weight of the water + the weight of the box in that region is the same as the weight of the water everywhere else.
2 years ago
Anonymous
vvvvv
[...]
[...]
I went out of my way to draw you an accurate picture of an example. The total mass of water is conserved and the submerged block is half as dense as water which means it sticks halfway out of the water.
Bonus points if you can figure out where the displaced water went.
2 years ago
Anonymous
If gravity isn't being cancelled out, why isn't the boat sinking, dumbass. Does newton's second law ring a bell or are you actually that behind
how is that possible? ignoring the displaced water, the mass of water+boat should be uniformly distributed. now, once you consider the displaced water, there's more on the right side, hence a greater mass on the right side.
I was confounded by the same thought process, this was me
[...]
[...]
What's confusing me is this: without the boat each half of the container has equal amounts of water. With the boat, the left side has
half the water + weight of the boat - equal weight in water + some of the displaced water
and the right side has
half the water + some of the displaced water
But because the displaced water is in a flat layer, the right side must have more, because the boat takes up some space.
What I've realised is that the displaced water raises the water level, so the boat lifts, effectively making the displaced water an even layer underneath the boat. I can't draw for shit, but imagine doing this: >container full to the brim, add the boat and collect the displaced spill >take a picture from the side of the layout >raise the walls of the container, add the spill back >take a picture of the new layout
The boat will float to the same height in the water, so you can overlay the first picture on top of the second one, just slightly higher up. The only difference is a flat layer of water at the bottom (and the higher walls).
Ok, so it sounds like some of the displaced water will be under the boat, and the amount that's under is equal to the amount required to balance the container. Basically, the equilibrium condition of having equal pressure across the container still holds.
>Ok, so it sounds like some of the displaced water will be under the boat
genuinely what the frick did he mean by this
2 years ago
Anonymous
i posted this pic before:
it will tip to the right you brainlets
>t. computer science grad
. in that pic, none of the displaced water is under the boat, so it seems like there's more displaced water to the right (hence tipping to the right). however, it seems the water will still try to reach pressure equilibrium, meaning some of the displaced water will also be under the boat.
sigh. i cannot believe i have to say this, but i will. you're making an assumption that >the tank was filled with water >the boat is in the tank >the fulcrum was fixed as this was occurring >all until the arrangement in op pic rel is achieved >ergo it's hydrostatic and the tank doesn't tip
meanwhile, i make the assumption >the tank of water is balanced on the fulcrum >the boat is then placed in the tank of water >ergo it's a hydrodynamics problem and the tank tips
why are you getting to pissy about my assumptions being faulty, while refusing to admit that you yourself are making assumptions? you're literally incapable of recognizing you're making assumptions. i'm not even saying you're wrong (like you are to me). i'm saying that under different assumptions, different things occur. the fact you're unable to recognize the assumptions i make says more about you than it does about me.
I'm assuming the initial state is one in which there is no motion. that's an extremely normal assumption to make.
>I'm assuming the initial state is one in which there is no motion. that's an extremely normal assumption to make
by your reasoning, this is an initial state with no motion.
pardon? it's a normal assumption to make. this a perfectly reasonable static situation. just, ugh, draw the free body diagram.
2 years ago
Anonymous
even supposing these cars
>I'm assuming the initial state is one in which there is no motion. that's an extremely normal assumption to make
by your reasoning, this is an initial state with no motion.
have no initial velocity, this still isn't a position of static equilibrium. that's fricking obvious when considering the forces at play. on the other hand, this
https://i.imgur.com/Y3RHf3n.png
To put it more precisely, is there a torque on the container of water.
IS a position of static equilibrium
2 years ago
Anonymous
>this still isn't a position of static equilibrium
draw the free body diagram. it's clearly static equilibrium. it's a static image, after all. no assumptions about initial velocity necessary.
2 years ago
Anonymous
>it's clearly static equilibrium
it isn't, there is a net torque on at least one of those cars.
2 years ago
Anonymous
huh? it's a static image homey. draw the free body diagram. can you do that? it's basic physics.
2 years ago
Anonymous
d is clearly nonzero and so there is a net torque and there is no equilibrium. can you do the same?
Finally a good thread. An interesting problem that doesn't have an obvious solution. When i saw OP i was eager to dive into the thread to see discussions and respectful arguments about what the solution is with unique insight and explanations to go with it.
Too bad the replies are filled with narcissistic outbursts, insults and dick measuring contests.
>An interesting problem that doesn't have an obvious solution
but the splution does indeed arrive pretty quickly if you draw a free body diagram, apply the principle of buoyancy, and consider the only 2 forces at play (and the location at which they act)
The Titanic had similar dimensions, albeit completely ignoring the shape (not a cylinder, etc.)
The real issue would be keeping the Burj from tipping over.
it's shouldn't be too difficult to build a mockup and test yourself, btw. just replace the fulcrum with something more stable. get two identical, decent scales and place a long rectangular container of water with the scales supporting at either end. place a floating model of a boat closer to one end, and observe the scales show no bias (they report the same increase in weight).
The easy way to solve this is by imagining a more extreme case.
The boat is touching the side wall of the tank. Will it tip?
And from there simply ask yourself if there's any friction between the wall and the boat. There isn't. So it won't tip. And since you know the boat won't tip the container in the center and it won't tip on the very edge, you can infer it won't tip the container on any arbitrary point between the two edges where the boat is buoyant.
And you don't need to know a damn thing about fluids to get to this except that they float shit.
Yes there is a torque. There is more pressure underneath the boat than next to it. The fluid pressure depends on specific weight which depends on the amount of mass. The boat will have more pressure underneath it than the pressure of the fluid it can displace
>There is more pressure underneath the boat than next to it.
no there isn't >The fluid pressure depends on specific weight which depends on the amount of mass
the fluid pressure depends only on its depth below the free surface. the surfaces of equal pressure are all parallel to the free surface. specifically, [math] p_text{gage}=gamma z [/math]. this is baby stuff. >The boat will have more pressure underneath it than the pressure of the fluid it can displace
actual schizobabble
this isn't something that happens in reality, anon
2 years ago
Anonymous
maybe not in your reality. i can feel someone swimming atop me.
2 years ago
Anonymous
so this pressure should be measurable, right? how come nobody has ever measured this before?
2 years ago
Anonymous
they have. your ignorance about their existence doesn't preclude their existence.
2 years ago
Anonymous
>they have
show me right away. It hope it's obvious I'm not talking about the sensation of feeling moving water from a nearby swimmer.
2 years ago
Anonymous
>I'm not talking about the sensation of feeling moving water from a nearby swimmer.
that's exactly what you're talking about. you may not know it, but it is.
2 years ago
Anonymous
you are sitting at the bottom of a pool, 6 feet under. your friend is floating above you. neither of you are moving, the water is calm. can you feel him above you?
kinder-level physics
2 years ago
Anonymous
that's a different question than the one you originally posed. let's stay with one question, yeah?
2 years ago
Anonymous
You're denser than I assumed. > let's stay with one question, yeah?
yes.
you are sitting at the bottom of a pool, 6 feet under. your friend is floating above you. neither of you are moving, the water is calm. can you feel him above you?
kinder-level physics
this is the one question I want answered.
2 years ago
Anonymous
if that's the one question you want answered, why did you originally ask something different? >when you're swimming in a pool and your friend swims over you when you're a few feet underwater, can you feel the pressure immediately triple?
i can't assign a definite value, but the pressure will obviously change due to water moving. or do you think moving water doesn't induce changes in pressure? if so, you may be on to solving one of the millennium prize problems, because navier stokes has it all wrong. after i pointed out your critical misunderstanding, you try to shift to a different question to save face. just admit you made a fricking mistake dude. we all do at times.
2 years ago
Anonymous
>i can't assign a definite value
why not? I can
2 years ago
Anonymous
bullshit.
2 years ago
Anonymous
(6 feet)*(62.4 pounds/cu.ft)/(144 in^2/ft^2) = 2.6 psi (gauge) or something like a fifth of a bar if you are a metricgay.
is "pressure is proportional to depth" seriously not something you've ever learned before?
2 years ago
Anonymous
and in a dynamic situation when the water is moving? you forgot your own question, numbnuts. your friend swims over you. he's moving. that changes the pressure.
2 years ago
Anonymous
sure I'll compute it. give me the specific parameters.
2 years ago
Anonymous
you're the one who asked the question. i won't hold your hand through it. have fun
2 years ago
Anonymous
>different question than the one you originally posed
The start of the reply chain:
Yes there is a torque. There is more pressure underneath the boat than next to it. The fluid pressure depends on specific weight which depends on the amount of mass. The boat will have more pressure underneath it than the pressure of the fluid it can displace
>Yes there is a torque. There is more pressure underneath the boat than next to it.
It was you who made a moronic statement and then started being artistically pedantic when the other anon challenged that notion. You clung to him making a not 100% analogous example as if you pretended to miss the point.
I don't think you are a moron. I think you are just a dishonest butthole who realised the mistake and tried to spin it around as if it wasn't you but the other anon.
2 years ago
Anonymous
>It was you who made a moronic statement
which statement do you think was me, moron?
whatever helps you sleep at night, champ. congrats, you can solve a problem in your physics 1 book, lol. what are you trying to prove? that you don't know how to actually apply them to real world situations?
water rushes into the boat. the concentration of mass on the boat's side increases. the center of mass will slowly shift to that side, quicker and quicker as water fills the hull, and then suddenly jumps even farther toward the end of the container when the boat collides with the floor.
Except that's false. Leaving hydrodynamics of the water aside, the tank will only tip once the boat sinks and touches the bottom. At no point during the sinking or flooding with there be any torque in the system.
2 years ago
Anonymous
you're right, anon. I thought about it again and container is indeed stable until the boat touches the floor. The COM remains above the fulcrum the entire time.
2 years ago
Anonymous
Yeah, in fact, you could submerge any object of any size, shape and density and fix it to some outside frame to keep it in place in the tank and it will still not tip. In essence, the only thing that matters is the pressure exerted on the tank walls which will always be symmetrical in steady state.
2 years ago
Anonymous
good post
2 years ago
Anonymous
That suggests an even more counterintuitive situation - a person pushing an inflated ball down into the water on one side.
2 years ago
Anonymous
Correct. Trying to push a balloon underwater on one side, for example, doesn't put any torque on the container.
2 years ago
Anonymous
To be perfectly clear to those who are truly mentally deficient: I am ignoring all friction and all transient motion of the water.
2 years ago
Anonymous
It is more counterintuitive but it is true. It's not really the weight exerting the force onto the water but the buoyant force. If you submerge a 1 L big beach ball, the ball will displace 1 L (1 kg) of water and the buoyant force exerted on the beach ball (and your hand) is exactly equal to the weight of displaced water which is 1 kg.
So no matter what you put into the tank or how you hold it, there will never be any torque as long as you don't touch the walls of the container.
2 years ago
Anonymous
>there will never be any torque as long as you don't touch the walls of the container.
it's almost as if...this is literally a defining characteristic of what is meant by "fluid" or something
damn I love fluid mechanics bros
2 years ago
Anonymous
Well, you can just say that the pressure exerted on the walls only depends on the height of the water column and the pressure of the water is what exerts a direct force on the tank. So the actual shape the water takes inside the tank doesn't matter as long as it's in the state of equilibrium.
The boat doesn't take up the entire left side and the water will be evenly displaced throughout the entire tank, not exclusively on the right side. Therefore, it's imbalanced.
why do you assume the boat weighs more than the extra water opposite side? do you realize they weigh exactly the same?
2 years ago
Anonymous
Because some of the water displaced is still left of the pivot, it didn't all teleport to the right of the pivot.
2 years ago
Anonymous
The water displaced was evenly distributed in the tank. Equally on both sides. The missing mass on the left is supplied by the mass of the boat itself.
The boat displaces the same mass of water as its own mass. So x kg of water disappears (it's evenly distributed so it might as well be removed) and x kg of boat is put in its place.
No, the removed mass is completely irrelevant. It only changes the height of the water column. You could physically scoop that water displace by the hull out and nothing would change.
That'd be great if the boat was the size of the entire left side of the container, but it's not and that image misrepresents the space the boat occupies.
Doesn't matter what the size of the boat is. The boat displaces the same amount of mass of water as its own mass. And the displaced water is of no consequence. It doesn't distribute "mostly on the right side" or anything. It equally raises the water level in the whole tank.
>It doesn't distribute "mostly on the right side" or anything. It equally raises the water level in the whole tank.
Correct, and half of the whole tank is on the left side. There is a space between the boat and the pivot where the water level has raised, so SOME of the displaced mass is on the left and SOME of the displaced mass is on the right. But the left ALSO has a boat with mass equal to the WHOLE displacement. So it's no longer balanced because the boat's equivalent displaced mass is not entirely located to the right of the pivot.
You clearly need a picture to visualise this. Gimme a second.
2 years ago
Anonymous
>So it's no longer balanced because the boat's equivalent displaced mass is not entirely located to the right of the pivot.
yeah, it's spread perfectly evenly everywhere.
I made a picture of my own. For simplicity, I split the tank into 4 segments. The boat adds 3 units of mass and displaces 3 units of water. The displaced water is then evenly distributed across the remaining sections. The right side is only +2 because 1 of the displaced water units remains left of the pivot. Therefore, not enough water is displaced to the right side of the pivot to counterbalance the left side.
2 years ago
Anonymous
>The displaced water is then evenly distributed across the remaining sections.
WRONG. it's evenly distributed across ALL sections, including the section to the left of the boat, and including the section that is the boat.
2 years ago
Anonymous
No, the water is displaced out from underneath the boat. You can't displace water into the space the boat occupies.
2 years ago
Anonymous
Alright. I gave it my best. I will never successfully explain this to you, I guess.
2 years ago
Anonymous
You can actually. Some of the displaced water ends up under the boat because the water level in the whole tank raised.
2 years ago
Anonymous
But that would require an increase in water density below the boat, but with nothing to contain it the water would spread out and not remain more dense beneath the boat. Then we're back to my illustration with the even displacement around the boat.
2 years ago
Anonymous
Let's think this through.
Here we have four sections. Let's say that the boat weighs three kilos. Now, what happens to the weight of Section 1? By the weight of the boat we will add 3 kilos. Then, by the Archimedes' principle we will subtract 3 kilos of displaced water. Now those 3 kilos of water must be put back to the tank in one way or another.
Putting the water back raises the boat and the surface of water equally which adds 3*(1/4) kilos to Section 1. So for Section 1 we have +3 (boat) -3 (Archimedes) +3*(1/4) (putting the water back) = 3*(1/4). One quarter of three kilos of weight were added to each section which means that the forces stay balanced and the tank does not tip. So overall...
So some of the displaced water goes into the section from which it was displaced from. You can get an idea why this works if you think about the fact that submerging a floating object into a tank simultaneously raises the surface level as you are submerging it. So if you did this in real life you would simultaneously be displacing water and adding 1/4 of that water underneath the boat.
2 years ago
Anonymous
No, the water is displaced out from underneath the boat. You can't displace water into the space the boat occupies.
But that would require an increase in water density below the boat, but with nothing to contain it the water would spread out and not remain more dense beneath the boat. Then we're back to my illustration with the even displacement around the boat.
I went out of my way to draw you an accurate picture of an example. The total mass of water is conserved and the submerged block is half as dense as water which means it sticks halfway out of the water.
Bonus points if you can figure out where the displaced water went.
2 years ago
Anonymous
Oops, mistake. In the right case, the left text should say mass on left instead of right.
2 years ago
Anonymous
Huh... that actually does make sense. I concede, it wouldn't tip.
2 years ago
Anonymous
>argues an incorrect point for several posts >gets convinced by an effort-post >admits they were wrong >learns something
very based
>So it's no longer balanced because the boat's equivalent displaced mass is not entirely located to the right of the pivot.
yeah, it's spread perfectly evenly everywhere.
imagine the tank with no boat. everything is perfectly symmetric, right?
now imagine the boat floating statically at one end. the system is the same everywhere, except at the boat. but the boat weight exactly as much as that chunk of water would weight anyway, so it doesn't effect the total distribution of mass in the system. the distribution of mass in the system IS ALL THAT MATTERS in determining if it tips, because gravity is the only external force besides the normal at the base.
While the mass is the same, it is no longer evenly distributed left and right of the fulcrum. On the left, there should be more torque from having mass above the surface of the water and hence at a large radius.
No. The water on the right side becomes heavier to counterbalance the weight of the boat until they reach an equilibrium. It's basic newtonian mechanics.
I HAVE THE SOLUTION
The boat is a sailboat
Therefore this simulation accepts wind (else the sailboat would be absurd)
-> It WILL tip because of the wind.
No it won't tip, pressure distributes uniformly.
Other are treating the question as that boat was dropped in water rather than putting it very slowly, in that case yes it will but if you put it very slowly then it won't because of what I said in my previous reply.
What's confusing me is this: without the boat each half of the container has equal amounts of water. With the boat, the left side has
half the water + weight of the boat - equal weight in water + some of the displaced water
and the right side has
half the water + some of the displaced water
But because the displaced water is in a flat layer, the right side must have more, because the boat takes up some space.
the boat displaces an amount of water equal to its mass
That's what I meant to say by
>weight of the boat - equal weight in water
so the left side simplifies to half the water + some of the displaced water
The weight of water moved to have the boat float is equal to the weight of the boat. Think of the boat as differently shaped water, but still the same mass. The boat takes up more space because it has a lower density to be able to float in the first place
>pressure distributes uniformly
inspired this
The diagram is a bit off in my estimation, but no. Liquid only fills the space by the weight that is lost by the missing air, so it may as well be considered level.
>Liquid only fills the space by the weight that is lost by the missing air
???
so if you suck 1kg of air out, 1kg worth of liquid will fill the space.
i dont think thats right. if you suck out 29.92 inches of air from the right side, 29.92 inches of mercury takes its place, but mercury is obviously heavier than air, so the right side is now heavier.
>if you suck out 29.92 inches of air from the right side, 29.92 inches of mercury takes its place
only if the way you're measuring inches of air is how much the mercury rises. You can measure how much air is in an inch, and suck it out, and find that the mercury rises less than an inch. The reason for this is because mercury, being heavier than air, pulls on it and forms a partial vacuum. If this weren't the case, than after sucking out 29 inches of air, the rest of it would be at atmospheric pressure, the mercury having filled it's space, and if so, what force is keeping the mercury up?
i think i see what you mean now, but i think you meant to user units of force here
instead of units of mass. the weight of the air on the left side equals the weight of the mercury on the right, because theres an entire atmosphere sitting on top of the left side.
Yeah that sounds right!
and it's a good idea, to consider the atmosphere's pushing force rather than a vacuum's mythical "sucking force". Thanks
Forget pressure homosexual
Apply force on infinitesimal portion of the system = ρgdV and integrate over infinitesimal torque = ρgrdV
You find non-zero value.
Also pressure is not uniform gay, that's why things bend.
>Also pressure is not uniform gay, that's why things bend
Your mom bends over without pressure though
>Also pressure is not uniform gay, that's why things bend.
Max moron
Yes
Your more precise doesn't help the question. But to answer it no, there shouldn't be a change in torque after the boat from before (comparing steady state to steady state)
how do you contest that the center of mass is still directly above the fulcrum? are you suggesting the water is not at a uniform height?
>are you suggesting the water is not at a uniform height?
yeah, the water is lower underneath the boat
This and basic distribution of water based on mass distributed to make the boat float. If the boat sunk and was at the bottom then definitely non-zero net torque added
just to be clear: are you saying there is less water to the left of the fulcrum than on the right?
yeah. look at the picture. pretty obvious.
so the water must have redistributed in some way. during this redistribution, how can you say the net torque is zero at all times?
well i didnt say that. i dont know much about fluid dynamics.
if there's a net torque at any time during this redistribution, the tank tips over.
here are my posts
i didnt say the container wouldnt tip
Owned, holy shit
The boat was not added, OPs image shows the boat is already there from the start.
You can't read. Follow the reply chain, it literally starts with a post saying "comparing steady state to steady state"
oddly enough, something floating on another thing doesn't add any weight to it directly. Simply put, if the whole thing were more dense than the water it replaces, then it wouldn't float, and however much it floats above, is the extra weight. So if you 'replace' the water it's convex hull is taking up if it were level with the boat, the weight doesn't change. Since the water is already there, it does change, but there isn't a direction to the pressure.
The water level rises, but the weight doesn't change from the boat.
moron, weight doesn't add if the replaced water gets spilled somewhere else, aka leaves the system. Yes, cause the volume displaced has the same mass as the mass of the boat underneath the water.
If the water stays within, mass adds up.
Doesn't take a freaking genius to figure it out.
The center of mass is exactly above the tip with or without the boat, but there is a state in between before the system stabilizes after you add the boat or when it moves so I guess it would tip.
assuming this is the before picture, then yes obviously it will tip
Wrong.
made your question better defined
yea you don't need this. the question is about balance on a fulcrum bisecting the bottom of the container, you don't need the before picture at all.
Yes you do because it's otherwise not evident that the dimensions of the container are equivalent to prevent tipping without the boat.
>it's otherwise not evident that the dimensions of the container are equivalent to prevent tipping without the boat.
Yes it is. You have autism. The question obviously isn't asking people to get out rulers and measure the container.
Question was inspired by this thing.
looks like something straight out of an Escher drawing
No it does not look like that, brainlet.
yes it does
strange image to fake
Is this safe? I'd worry the consequences of this breaking.
it's an obvious shop. water bridges do exist, though.
Perhaps the same could be said of all bridges.
That's the Veluwemeer aqueduct, here's a more representative photo of its size.
Is the boat moving?
It’s the same as adding water to the tank. So long as the boat floats, the pressure is equal across the water. That’s what floating is
Having a boat in the container is identical to a situation where the "bulge" created by the boat below the surface level was filled with water. There would be no difference in weight, density, anything. So the container would not tip.
Thank you, this got me about as close to getting it intuitively as I think I will get. I'm still struggling with the pseudo formulae of
but I'll sit with it.
Maybe an easy way to think of it is to think how pressure works. If you submerge your hand, basically a certain force is applied to every unit of surface area on your hand. So pressure is what prevents boats from sinking. But the same pressure that keeps the boat floating still exists in the water even without the boat. If the forces of pressure are equal with and without a boat, the container does not tip.
this experiment shouldn't be too hard to check. get a small container with water, balance it on a small fulcrum, then slowly put a boat in it near the edge.
>Will it tip?
If the system is black or israeli you can guarantee it will not tip
Suppose we have just a normal container of water, nothing in it except water, everything normal. Now it's trivial to see that the container would not tip.
Now, what we are going to do is to "carve out" a hypothetical region of water (illustrated in picrel). By definition, the shape of our piece of water is the exact same shape that a boat would create if it was in the container in the same spot. Now, what we do next is to place that boat in that same place. According to our axioms, the boat has the same weight as the water that used to be there.
Now, gravity is pulling the boat downward with a certain force. Let's call that force X. However, our hypothetical region of water was also being pulled downward by gravity. Let's call the downward force of our water region Y. The region of water was defined to be the bulge created by the boat, so whatever force the boat exerts to the left side of the container, by definition, the hypothetical water region exerts the same force. Therefore X=Y so the container does not tip.
How would the water know that we took some water out and replaced it with a boat? After all, they have the same weight. The only difference is in material (one is made out of steel) and density (the boat takes more space).
You could think of the fact that the displaced area underwater created by the boat is related to the force of the boat. If you add some weight to a boat, it sinks and has to displace a bigger volume under the surface. If a boat weight 100 000 tons, the water must be lifting the boat upward by a force equal to 100 000 tons to prevent the boat from sinking. This lifting force is applied to that part of the boat which is underwater. But if you removed the boat and just replaced it with water, that region of water would also weight exactly 100 000 tons so that it also would not sink nor rise upward. Think of it as sort of a boat made out of water.
are you guys saying that there is more water to the right of the fulcrum than to the left of the fulcrum?
Yes. Put the container of OP's pic in a freezer, then take the boat out of the piece of ice. You can see that there is now sort of a concave bulge where the boat used to be, so that half of the block of ice has less ice than the other half.
It's impossible to put in the boat slowly enough for it not to not tip unless you put in the boat somewhere on the right and apply just enough force to counterbalance the tipping to the right until it stops moving left.
Dude frick off there's water all over my kitchen floor now
no it doesnt tip. good physicslet identifier, though.
>There are two external forces on the system containing both boat and the water tank.
yes. those two forces are gravity and the normal force between the fulcrum and the earth. thats it. these forces are collinear (the presence of the ship does not change the COM) so theres no moment: nothing to cause it to tip.
show a video and prove it.
yes we understood archimedes principle the first seventy times it was posted here. what you neglect to mention is that this missing water is DISPLACED. you know what that means? it moves. it creates an instantaneous pressure gradient, i.e. a force field (newton's law... the accelerated water must also experience a force). so there must exist an instantaneous net torque somewhere in this tank, meaning it must tip.
here's the explanation in brainlet language for you: see that water with all the arrows on it that you drew? well, you have the boat where that water used to be. now you need to put that water back in the tank.
The poster above your post seems like he didn't understand yet. I tried to make it really simple
Yeah, the weight of the boat pulling down over there is the same as the water that was pulling down there when the boat was in the middle.
tooker please think clearly: what are the external forces on the system? is there a moment? please draw a FBD, I know you're smarter than this
>what you neglect to mention is that this missing water is DISPLACED
this isn't actually something you have to consider unless you're talking about dumping the boat into the water with force. This is because 'replacing' and 'displacing' water, without that extreme force, is the same thing.
You can imagine slowly lowering the boat into the water, so that 'pressure gradient' is minimized, and then dumping out any rise in water level. There you have it: the non-displaced variant. Essentially, it doesn't matter how much water was in there before and after the boat is dropped, Archimedes principle still works.
>'pressure gradient' is minimized
it will never be zero, no matter how slowly you place the boat in the water. there will always exist a pressure gradient due to the displacement (acceleration) of water, no matter how small. as such, there must exist a nonzero force and thus a nonzero torque on the system. the tank tips.
[eqn]
lim_{t to infty} frac{Delta x}{t} = 0
[/eqn]
motion is negligible
lol, do the experiment then and post it here. prove my smug ass wrong.
brainlet: you realize you can simply draw a free body diagram and learn that there is no tipping moment in a matter of seconds, right? of course it would be nice to do the experiment for real, but I actually do not care enough.
prove you aren't a clinical shit-for-brain and draw me a free body diagram right now.
another anon here.
I have balanced a pot of water on a tiny bottle of rum extract. I find that when I drop a can of conola spray on the far side, it usually falls on the near side, as expected from 'displacement'.
When I slowly place the spray into the water, however, it does not fall. With the pot emptied, placing the conola spray slowly, in the same angle and arrangement it is placed into the water with the pot full of water, the pot tips over.
If you ask nicely, I may send pics, but this is a fun household experiment anybody can do.
Based empirical anon.
>balanced a pot of water on a tiny bottle of rum extract.
if you're clever enough to do this experiment, surely you recognize this isn't an ideal fulcrum right? pic rel is your setup. so it doesn't surprise me that you can slowly trickle water in, and it won't tip. you created forces, but the fulcrum is stable enough to withstand those forces.
consider a situation where you used a very, very, very long tank (10 meters long) on your little bottle of rum extract. i expect spraying will cause it to tip.
>surely you recognize this isn't an ideal fulcrum right?
>you do realize physicists exaggerate their drawings right?
you talk like a gay, or a woman.
>if you're clever enough to do this experiment, surely you recognize this isn't an ideal fulcrum right? pic rel is your setup. so it doesn't surprise me that you can slowly trickle water in, and it won't tip. you created forces, but the fulcrum is stable enough to withstand those forces.
this is true, and would cause a problem if the spray were truly weighing one side down as usual, and the fulcrum kept it stable regardless. But the fact that it tips either way when the pot is emptied is evidence to the contrary.
If you won't take an example from the real world(As there isn't any perfect fulcrum in real life), then what were you doing asking for an experiment?
>But the fact that it tips either way when the pot is emptied is evidence to the contrary.
well no. the case without water is trivial to analyze. as a matter of fact, it's a useful case to study. if you're dedicated enough, you can measure the minimum distance from the fulcrum that causes it to tip. i guarantee you this distance will be nonzero, and a bit larger than you expect. this is evidence your fulcrum is not ideal.
now fill the pot with water, and drop the canister in at different distances from the fulcrum. you'll see that the minimum distance that causes the pot to tip will be a bit larger than if it's empty. this is further evidence your fulcrum (with a pot filled with water) is again not ideal.
but most importantly, as you noticed, when the pot has water, and when you slowly submerge the canister inside the pot, the the pot+water+canister doesn't tip. again, i don't doubt you. i never claimed this was a large force. all i claimed was that this was a force (as a matter of fact, i claimed it was a small force). under the effect of an ideal fulcrum, this small force would cause the tank to tip.
because your fulcrum is not ideal, these small forces that occur when you submerge the canister will not cause the pot to tip. if, however, you were to use a longer pot, or a thinner fulcrum, you'll see it would be much, much more difficult to submerge the canister. under an ideal fulcrum, it'll be impossible to submerge the fulcrum without tipping the pot/tank.
>under an ideal fulcrum, it'll be impossible to submerge the fulcrum without tipping the pot/tank
I suppose this is true. The slower you go, the less the force, but any force will do it.
I still don't think that's what the picture was about, though. the OP just shows the boat in the water, presumably everything is still and settled, and nothing was dropped or changed.
>I still don't think that's what the picture was about
i disagree. i interpreted OPs scenario to be that the tank was empty, and then the boat was submerged. i think my interpretation is valid given what op showed (which was ambiguous). really, this all comes down to op not clarifying what the question actually was.
>i interpreted OPs scenario to be that the tank was empty, and then the boat was submerged
pure fricking autism. theres nothing to indicate this whatsoever.
you say autism like it's a bad thing. when did you first start browsing IQfy?
its been used here as an insult for a decade newfriend, like 2013 or something
this isn't to mention that you do legitimately have some mental deficiency for randomly inserting details to a prompt that aren't hinted at being relevant
sorry, i'm drunk. firstly gotta correct the typo.
>nder an ideal fulcrum, it'll be impossible to submerge the fulcrum without tipping the pot/tank
obviously you gotta submerge the cannister.
>i interpreted OPs scenario to be that the tank was empty
i interpreted the scenario as the tank was filled with water, and then the boat was submerged.
A better way would be to have a long tank of water (like a fish tank) and have two scales. The tank would sit on the scales so that the opposite edges would have one scale underneath. You could use something like a food scale which is accurate up to one gram difference. Then you would place a boat very slowly directly on top of one scale and see if the numbers on boths scales change at the same time and stay equal.
clever experiment.
So just suppose you hold the tank in place while you put the boat in, and only let it free after a long time.
>Archimedes principle still works.
yes archimedes principles works. that's not why the tank tips though. the tank tips not because of the mass of the boat, but because of the displaced water which must accelerate
nobody is considering the case where the water is in motion you fricking autist. this is a hydroSTATIC problem.
before the boat is submerged, there are equal amounts of water to the right of the fulcrum and to the left of the fulcrum. agreed?
after the boat is placed in the tank, archimedes principle comes into play... and now the amount of water to the left of the fulcrum and to the right of the fulcrum are no longer equal. agreed? (otherwise the tank tips)
therefore, water must have MOVED from left to right. therefore, it's not a hydroSTATIC problem. dumbass.
you're fricking autisic if you think the transient motion of the water as the boat is placed is of concern.
imagine that the boat has been floating on the right hand side for a long time with the fulcrum pinned. then, a switch is activated and the pin is pulled and the fulcrum is allowed to rotate. does the thing tip? no. is there a torque causing it to tip? no.
you've changed the problem. obviously in a different problem the tank won't tip, lol. i've already explained it as best i can. if you can't understand that explanation, then idk what to tell you bro.
free body diagram is a static picture. as i already explained, this isn't statics. it's dynamics.
>if you can't understand that explanation
I understand anon, it's just baffling that you think this is a dynamics problem.
>free body diagram is a static picture. as i already explained, this isn't statics. it's dynamics.
schizophrenia
before: equal water left and right of fulcrum.
after: more water right of fulcrum than left of fulcrum
how the frick does that happen in a static picture? how does water not move for that to happen? do you even understand what archimede's principle is? the water is DISPLACED. how do you think water gets displaced in a static picture?
there is no """"before and after,"""" moron. we are GIVEN the state depicted in the OP as the initial state. does it tip? no it does not. it does not tip because there is no moment.
stop randomly inventing details to the problem that dont exist
>do you even understand what archimede's principle is?
do YOU, homosexual?
>the water is DISPLACED
yes
>how do you think water gets displaced in a static picture?
by not being located inside the hull of the ship??? what the frick is this confusion
>there is no """"before and after,"""" moron. we are GIVEN the state depicted in the OP as the initial state
>my assumptions are more valid than your assumptions
at least i had the intellectual honesty to admit from the beginning what my assumptions were when i gave my answer. you, however, refuse to admit that you're making an assumption.
We could also assume the tank is glued to the fulcrum, or the whole setup is floating in 0g, but that would be fricking autistic. The assumption that this is somehow a dynamics problem from a static image with nothing indicating motion is also fricking autistic.
you may not know this, but all images are static. are you trying to say that no static image can represent a static moment of a dynamic event? you're the one being autistic, staying fixated on a single interpretation of a literalist event.
On the left, everything is balanced. On the right, the water height is the same and the weight of the boat is pulling down to the side. There is a torque and it will tip.
Also, assuming that the boat did not exist at the side of the water for an infinite amount of time, any motion to the side will create an angular momentum about the fulcrim's tip which will also cause tipping.
If you take air buoyancy in account, will it still not tip?
Doesn't matter. The pressure on the bottom of the tank must be equal everywhere either way and the air pressure applies uniform force onto the water so nothing changes.
There is no information, so the answer is ist will either tip or not. It depends of the level of simplification you would allow to apply in this model
It's not stable, so it will tip.
which wooden block get pushed up the hardest? my guess is the first one from the left
do you mean which would jump highest out of the water, or do you mean which, in the instant that this drawing describes, has the most upward force acting on it?
all the same, except for left obviously
are you saying the net force acting on each box is the same?
He should be saying that because it's true
learn how buoyancy works for the love of christ the three on the right all experience exactly the same upward force because they are all presumably the same size.
LMAO FOOLED BY HIS OWN WORDS
yeah, exactly same force as much as they didn't leave the water.
But you forgot that force is proportional to the second derivative of motion, and initial velocity when they reach surface will not be the same (accounting for a perfectly homogenous fluid, without wavelets so that motion doesn't get tilted and stays upward).
OP used the word hardest, which is qualitative and doesn't precise in which way, so we assume he is asking which block will attain highest altitude. Or in a different scenerio where surface velocities = 0, which block needs the highest work to reach its respective height.
Answer is right block.
The above explanation answers the question.
you should be embarrassed by this post
you should be embarrassed by all of your posts ITT
not an argument.
>this cant be settled through experimentation because ideal point-fulcrums dont exist
>i think a "tiny bottle" is like 6 inches wide
listen to how dumb you sound
you do realize physicists exaggerate their drawings right? it's to drive home the point that you have a stable fulcrum, as opposed to an ideal point-like fulcrum. here's the deal: in both cases there are forces (and hence, torques) introduced whether you drop your bottle or slowly place it. with an ideal fulcrum, even the tiniest force will cause it to tip. in your stable fulcrum situation, the setup can withstand tiny forces.
your experiment doesn't prove what you think it does.
im not the experimental anon
>with an ideal fulcrum, even the tiniest force will cause it to tip.
no one here disagrees, so you can go ahead and leave now, but youre still moronic for thinking that was what OP was asking.
>no one here disagrees
you clearly haven't read the thread.
close. i talk like a scientist.
>you clearly haven't read the thread.
all i saw was some guy trying to explain to you why you shouldnt make up your own shit about the question.
>you shouldnt make up your own shit about the question.
take your own advice
>youre still moronic for thinking that was what OP was asking.
unlike you, i clearly laid out my assumptions, which morons like you decried as the incorrect assumptions. i've already acknowledged that if your assumptions are correct, then the tank doesn't tip. dunno why you're incapable of doing the same. and if you are, idk what your problem is. do you think it's impossible your assumptions aren't the correct ones?
>OP used the word hardest, which is qualitative and doesn't precise in which way, so we assume he is asking which block will attain highest altitude.
being submerged lower would actually be a detriment to the height due to vortex shedding.
>[assumption 1] [assumption 2] [assumption 3]...
Look up my reply you will find them.
what? did you mean to reply to someone else?
buoyancy = specific weight of water × volume
i dont see anything about depth here, do you????
youre only outing yourself as a pseud harder with this post
The one on the right experiences the most upward force because the water is denser towards the bottom
>Boat on one side
>Will it tip
the boat weighs exactly as much as the water it is displacing
so the water level will rise in the container, but it will not put the scale off balance
It's gotta tip, even though some anons are right under idealistic conditions of classical physics, the truth of the matter is pressure distributing is a system properties which emerges from particle interaction among other factors, so a perturbance bound to tip it is inevitable
>It's gotta tip
so there's a torque on the container, yes? where does the torque come from, anon? compute this please
draw a force body diagram immediately
Best I can do
When you insert the boat it will cause the water molecules to displace in a chain reaction before they settle on a stable identical pressure state
>Best I can do
Thats not a free body diagram. are you seriously incapable of doing this?
>insert
what insertion? the boat is already in the water, this is the initial state
How the frick did it get in there? That's what I'm saying, a system like this has to have had a displacement which makes it tip in the real world, or it doesn't make any sense
Yeah I don't think I ever had to draw this shit, I'd probably do it wrong if I am precise but you get the general gist of it, you are applying only a macroscopical level model not even thinking about the dynamics over time to things which would have an effect on a microscopical one
>How the frick did it get in there?
???
It starts in there. The FBD is extremely simple. theres one force vertically down (gravity), and another force vertically up (normal force). They are equal in magitude and COLLINEAR. So there is no torque. Do you agree or disagree?
I agree it is a bullshit mathematical model impossible in reality which is stable by construction
>I agree it is a bullshit mathematical model impossible in reality which is stable by construction
Pin the fulcrum so that it doesnt move. Place the boat gently in the water. Wait until the water stops moving. Release the pin. Does it tip? No.
>Pin the fulcrum so that it doesnt move. Place the boat gently in the water.
why do you have to pin the fulcrum? you do realize you just agreed with people who say the tank tips, right?
i didn't, at no point did i agree that this was a dynamics problem
>why do you have to pin the fulcrum?
not him/her, but they're saying that to rule out transient effects from placing the boat in the water.
>transient effects from placing the boat in the water.
clarify
something about instantaneous pressure gradients. see
. idk man, not a physicist.
>drop the boat in the water
>immediately, the water is pushed from the hull and sloshes in the container
>this is transient dynamics
>after a few minutes, the water comes to a halt and the boat is floating still
>this is hydrostatic
homie please please please for the love of god just open a book on freshman mechanics
>a hydrodynamics problem is hydrostatic after it stops moving
>therefore the tank doesn't tip because it's a hydrostatic problem
...
the problem is PRESENTED as a hydrostatic one; the ship is ALREADY in the water and this is no motion indicated. not only is there no motion indicated, there is NO TORQUE that could ever possibly put it into motion!
sigh. i cannot believe i have to say this, but i will. you're making an assumption that
>the tank was filled with water
>the boat is in the tank
>the fulcrum was fixed as this was occurring
>all until the arrangement in op pic rel is achieved
>ergo it's hydrostatic and the tank doesn't tip
meanwhile, i make the assumption
>the tank of water is balanced on the fulcrum
>the boat is then placed in the tank of water
>ergo it's a hydrodynamics problem and the tank tips
why are you getting to pissy about my assumptions being faulty, while refusing to admit that you yourself are making assumptions? you're literally incapable of recognizing you're making assumptions. i'm not even saying you're wrong (like you are to me). i'm saying that under different assumptions, different things occur. the fact you're unable to recognize the assumptions i make says more about you than it does about me.
>meanwhile, i make the assumption
that's not what is depicted though
OP here, I was asking about the hydrostatic state.
assuming you are OP (see what i'm doing? i'm clearly laying out my assumptions), then you should more clearly communicate the problem next time.
The diagram clearly communicated the underlying question. You would have to be insane to believe that someone was asking if you could tip a balance by pushing on one side.
i have to be insane to assume realism? frick off.
Do you agree there are precisely two external forces or no? Do you agree the forces are collinear or no?
>initial value problems are impossible to solve because you actually need to know the values before t=0 otherwise it doesn't make sense
it will tip to the right you brainlets
>t. computer science grad
and yet there is equal mass on either side of the fulcrum, and the center of mass is directly above the fulcrum.
how is that possible? ignoring the displaced water, the mass of water+boat should be uniformly distributed. now, once you consider the displaced water, there's more on the right side, hence a greater mass on the right side.
>the mass of water+boat should be uniformly distributed
what? are you conflating mass with pressure? yes, theres more water on the right. there's less water on the left, and also a ship on the left. there is exactly equal mass on either side. the COM remains on the vertical line passing through the fulcrum.
>and yet there is equal mass on either side of the fulcrum, and the center of mass is directly above the fulcrum.
If the boat weighed as much as water, it wouldn't float idiot.
Center of mass will never be balanced in the middle unless the boat is also in the middle.
the boat indeed weighs ***exactly*** as much as the
weight of the water that would occupy its hull.
holy frick lmao how are you this dense getting filtered by 2500 year old physics like damn moron
you claim to be a cs graduate yet you don't know how archimedes pressure works? the submerged part of the boat endures an upwards push from the water below it because the boat is indeed less dense than water. however it is only submerged partially because in its current state, the displaced water exerts the same force upwards as the boat's weight downwards, that's why it doesn't move. in its current state the boat isn't accelerating so there are no forces or all the forces cancel each other out, so you might as well ignore the boat entirely
>so there are no forces or all the forces cancel each other out
So you genuinely believe that gravity magically gets buffered by the water and ignores the relative position of the boats mass because it's displacing water?
The column of water directly underneath the boat contains exactly a boat weight worth less water than the mirror column on the opposite side.
You're silly, buoyancy isn't about columns like pressure it's about displaced volumes, here the water volume displaced by the boat's hull. You learn this in high school is everyone here a kid or what
>You learn this in high school is everyone here a kid or what
Not that anon, but for me it's the opposite. I've been out of school for 15 years and since I haven't needed this knowledge even once that entire time in my day to day life I've naturally just forgotten it. Use it or lose it, and like 80% of what you learn in school goes unused.
...it was an explanation of that principle. Do you have trouble following conversations generally, or do you just make being a c**t a habit for fun?
I have trouble following conversations I'm a bit of a sperg. although I don't see how using columns is in any way relevant to buoyancy, since it implies that the force is dependant on depth in any way.
Gravity is a body force that acts on every molecule in the system at once. This makes it tricky to visualize, fortunately it can ALWAYS be replaced (in the static case) by an equivalent force equal to the total weight of the system, and acting at the center of weight or center of mass: [math] text{CoM}=intmathbf{r}dm / int dm [/math]
So think carefully: does the presence of the ship change the COM at all?
The mass-density gradient across the surface of the tub will have a noticeable dip where the boat is. There is less water under the boat due to displacement, ergo there is a higher density on the other side. It is not balanced.
Explain my diagram then
your box is lighter than water
it literally weighs the same as the water.it is less dense than water (like all boats ever built), and it weighs the same as the water (just like all boats)
>The mass-density gradient
wrong. the "mass-density gradient" is uniform. all surfaces of equal pressure are parallel to the free surface.
No its literally not, the density in the region of the boat and the water under it is not the same as the density from the other water around it.
the boat weighs the same, and its hull the same volume of water that would otherwise be there. so the weight/volume=density is the same.
The column of mass pressing on the bottom of the tank has the exact same mass per area as at every other point.
It isn't. The weight of the water + the weight of the box in that region is the same as the weight of the water everywhere else.
vvvvv
If gravity isn't being cancelled out, why isn't the boat sinking, dumbass. Does newton's second law ring a bell or are you actually that behind
I was confounded by the same thought process, this was me
What I've realised is that the displaced water raises the water level, so the boat lifts, effectively making the displaced water an even layer underneath the boat. I can't draw for shit, but imagine doing this:
>container full to the brim, add the boat and collect the displaced spill
>take a picture from the side of the layout
>raise the walls of the container, add the spill back
>take a picture of the new layout
The boat will float to the same height in the water, so you can overlay the first picture on top of the second one, just slightly higher up. The only difference is a flat layer of water at the bottom (and the higher walls).
Ok, so it sounds like some of the displaced water will be under the boat, and the amount that's under is equal to the amount required to balance the container. Basically, the equilibrium condition of having equal pressure across the container still holds.
>Ok, so it sounds like some of the displaced water will be under the boat
genuinely what the frick did he mean by this
i posted this pic before:
. in that pic, none of the displaced water is under the boat, so it seems like there's more displaced water to the right (hence tipping to the right). however, it seems the water will still try to reach pressure equilibrium, meaning some of the displaced water will also be under the boat.
There is more water on the right side but more boats on the left side. The both sides have equal mass.
The boat's weight isn't the same weight of the displaced water?
yes
I'm assuming the initial state is one in which there is no motion. that's an extremely normal assumption to make.
>I'm assuming the initial state is one in which there is no motion. that's an extremely normal assumption to make
by your reasoning, this is an initial state with no motion.
>le no physics
pardon? it's a normal assumption to make. this a perfectly reasonable static situation. just, ugh, draw the free body diagram.
even supposing these cars
have no initial velocity, this still isn't a position of static equilibrium. that's fricking obvious when considering the forces at play. on the other hand, this
IS a position of static equilibrium
>this still isn't a position of static equilibrium
draw the free body diagram. it's clearly static equilibrium. it's a static image, after all. no assumptions about initial velocity necessary.
>it's clearly static equilibrium
it isn't, there is a net torque on at least one of those cars.
huh? it's a static image homey. draw the free body diagram. can you do that? it's basic physics.
d is clearly nonzero and so there is a net torque and there is no equilibrium. can you do the same?
it's balanced by the silver car.
The boat applies a force even without the finger you silly. Only if the boat touches the bottom it will tip
Yes, have you ever tried balancing anything? This looks like trying to balace a 40 gallon tub on a traffic cone.
It would tip even without the boat
Finally a good thread. An interesting problem that doesn't have an obvious solution. When i saw OP i was eager to dive into the thread to see discussions and respectful arguments about what the solution is with unique insight and explanations to go with it.
Too bad the replies are filled with narcissistic outbursts, insults and dick measuring contests.
>An interesting problem that doesn't have an obvious solution
but the splution does indeed arrive pretty quickly if you draw a free body diagram, apply the principle of buoyancy, and consider the only 2 forces at play (and the location at which they act)
give me a foothold and i will hold the world
How much water must the boat displace to carry the burj khalifa?
Apparently it weighs 500 000 tonnes, so a cylinder 20m deep and 200m wide would do it. I can't decide if that seems plausible.
The Titanic had similar dimensions, albeit completely ignoring the shape (not a cylinder, etc.)
The real issue would be keeping the Burj from tipping over.
What's the answer. This thing is driving me insane because I forgot most of my physics class and no one makes a compelling argument.
it doesn't tip, anon.
it's shouldn't be too difficult to build a mockup and test yourself, btw. just replace the fulcrum with something more stable. get two identical, decent scales and place a long rectangular container of water with the scales supporting at either end. place a floating model of a boat closer to one end, and observe the scales show no bias (they report the same increase in weight).
fluid mechanics is such a filter class. most of you will not make it. m0whv
The easy way to solve this is by imagining a more extreme case.
The boat is touching the side wall of the tank. Will it tip?
And from there simply ask yourself if there's any friction between the wall and the boat. There isn't. So it won't tip. And since you know the boat won't tip the container in the center and it won't tip on the very edge, you can infer it won't tip the container on any arbitrary point between the two edges where the boat is buoyant.
And you don't need to know a damn thing about fluids to get to this except that they float shit.
If there is contact AND friction between the boat and the container, it absolutely will tip.
no tipping
Yes there is a torque. There is more pressure underneath the boat than next to it. The fluid pressure depends on specific weight which depends on the amount of mass. The boat will have more pressure underneath it than the pressure of the fluid it can displace
>There is more pressure underneath the boat than next to it.
no there isn't
>The fluid pressure depends on specific weight which depends on the amount of mass
the fluid pressure depends only on its depth below the free surface. the surfaces of equal pressure are all parallel to the free surface. specifically, [math] p_text{gage}=gamma z [/math]. this is baby stuff.
>The boat will have more pressure underneath it than the pressure of the fluid it can displace
actual schizobabble
Very interesting arguments. It appears I learned nothing in that class.
when an airplane flies over your head, are you immediately fricking crushed by the tremendous weight of an airliner?
when you're swimming in a pool and your friend swims over you when you're a few feet underwater, can you feel the pressure immediately triple?
can't you?
this isn't something that happens in reality, anon
maybe not in your reality. i can feel someone swimming atop me.
so this pressure should be measurable, right? how come nobody has ever measured this before?
they have. your ignorance about their existence doesn't preclude their existence.
>they have
show me right away. It hope it's obvious I'm not talking about the sensation of feeling moving water from a nearby swimmer.
>I'm not talking about the sensation of feeling moving water from a nearby swimmer.
that's exactly what you're talking about. you may not know it, but it is.
you are sitting at the bottom of a pool, 6 feet under. your friend is floating above you. neither of you are moving, the water is calm. can you feel him above you?
kinder-level physics
that's a different question than the one you originally posed. let's stay with one question, yeah?
You're denser than I assumed.
> let's stay with one question, yeah?
yes.
this is the one question I want answered.
if that's the one question you want answered, why did you originally ask something different?
>when you're swimming in a pool and your friend swims over you when you're a few feet underwater, can you feel the pressure immediately triple?
i can't assign a definite value, but the pressure will obviously change due to water moving. or do you think moving water doesn't induce changes in pressure? if so, you may be on to solving one of the millennium prize problems, because navier stokes has it all wrong. after i pointed out your critical misunderstanding, you try to shift to a different question to save face. just admit you made a fricking mistake dude. we all do at times.
>i can't assign a definite value
why not? I can
bullshit.
(6 feet)*(62.4 pounds/cu.ft)/(144 in^2/ft^2) = 2.6 psi (gauge) or something like a fifth of a bar if you are a metricgay.
is "pressure is proportional to depth" seriously not something you've ever learned before?
and in a dynamic situation when the water is moving? you forgot your own question, numbnuts. your friend swims over you. he's moving. that changes the pressure.
sure I'll compute it. give me the specific parameters.
you're the one who asked the question. i won't hold your hand through it. have fun
>different question than the one you originally posed
The start of the reply chain:
>Yes there is a torque. There is more pressure underneath the boat than next to it.
It was you who made a moronic statement and then started being artistically pedantic when the other anon challenged that notion. You clung to him making a not 100% analogous example as if you pretended to miss the point.
I don't think you are a moron. I think you are just a dishonest butthole who realised the mistake and tried to spin it around as if it wasn't you but the other anon.
>It was you who made a moronic statement
which statement do you think was me, moron?
"acktually I'm gonna need more information than that, teach... all hydrostatic problems are ACKTUALLY dynamics problems afterall"
this might surprise you... but real life problems are actually quite a bit more complicated than problems in the back in your physics 1 book. moron.
and you can't solve either
whatever helps you sleep at night, champ. congrats, you can solve a problem in your physics 1 book, lol. what are you trying to prove? that you don't know how to actually apply them to real world situations?
That is one dense motherfricking block.
Uhhh, I get 31.65kg.
ACKTUALLY NO BRO
No. Displaces the same weight of water to the other side.
Yeah because balance what the frick
in completely idealized circumstances where there are no waves created by the boat or anything, no
it will tip if you poke a hole in the boat and let it sink.
When though?
water rushes into the boat. the concentration of mass on the boat's side increases. the center of mass will slowly shift to that side, quicker and quicker as water fills the hull, and then suddenly jumps even farther toward the end of the container when the boat collides with the floor.
Except that's false. Leaving hydrodynamics of the water aside, the tank will only tip once the boat sinks and touches the bottom. At no point during the sinking or flooding with there be any torque in the system.
you're right, anon. I thought about it again and container is indeed stable until the boat touches the floor. The COM remains above the fulcrum the entire time.
Yeah, in fact, you could submerge any object of any size, shape and density and fix it to some outside frame to keep it in place in the tank and it will still not tip. In essence, the only thing that matters is the pressure exerted on the tank walls which will always be symmetrical in steady state.
good post
That suggests an even more counterintuitive situation - a person pushing an inflated ball down into the water on one side.
Correct. Trying to push a balloon underwater on one side, for example, doesn't put any torque on the container.
To be perfectly clear to those who are truly mentally deficient: I am ignoring all friction and all transient motion of the water.
It is more counterintuitive but it is true. It's not really the weight exerting the force onto the water but the buoyant force. If you submerge a 1 L big beach ball, the ball will displace 1 L (1 kg) of water and the buoyant force exerted on the beach ball (and your hand) is exactly equal to the weight of displaced water which is 1 kg.
So no matter what you put into the tank or how you hold it, there will never be any torque as long as you don't touch the walls of the container.
>there will never be any torque as long as you don't touch the walls of the container.
it's almost as if...this is literally a defining characteristic of what is meant by "fluid" or something
damn I love fluid mechanics bros
Well, you can just say that the pressure exerted on the walls only depends on the height of the water column and the pressure of the water is what exerts a direct force on the tank. So the actual shape the water takes inside the tank doesn't matter as long as it's in the state of equilibrium.
The boat doesn't take up the entire left side and the water will be evenly displaced throughout the entire tank, not exclusively on the right side. Therefore, it's imbalanced.
>Therefore, it's imbalanced.
where is the COM located, anon?
Somewhere between the boat and the pivot.
why do you assume the boat weighs more than the extra water opposite side? do you realize they weigh exactly the same?
Because some of the water displaced is still left of the pivot, it didn't all teleport to the right of the pivot.
The water displaced was evenly distributed in the tank. Equally on both sides. The missing mass on the left is supplied by the mass of the boat itself.
The boat displaces the same mass of water as its own mass. So x kg of water disappears (it's evenly distributed so it might as well be removed) and x kg of boat is put in its place.
Not all the mass moved to the right of the pivot, at least some of it remained to the left of the pivot and is contributing to an imbalance of mass.
No, the removed mass is completely irrelevant. It only changes the height of the water column. You could physically scoop that water displace by the hull out and nothing would change.
That'd be great if the boat was the size of the entire left side of the container, but it's not and that image misrepresents the space the boat occupies.
Doesn't matter what the size of the boat is. The boat displaces the same amount of mass of water as its own mass. And the displaced water is of no consequence. It doesn't distribute "mostly on the right side" or anything. It equally raises the water level in the whole tank.
>It doesn't distribute "mostly on the right side" or anything. It equally raises the water level in the whole tank.
Correct, and half of the whole tank is on the left side. There is a space between the boat and the pivot where the water level has raised, so SOME of the displaced mass is on the left and SOME of the displaced mass is on the right. But the left ALSO has a boat with mass equal to the WHOLE displacement. So it's no longer balanced because the boat's equivalent displaced mass is not entirely located to the right of the pivot.
You clearly need a picture to visualise this. Gimme a second.
I made a picture of my own. For simplicity, I split the tank into 4 segments. The boat adds 3 units of mass and displaces 3 units of water. The displaced water is then evenly distributed across the remaining sections. The right side is only +2 because 1 of the displaced water units remains left of the pivot. Therefore, not enough water is displaced to the right side of the pivot to counterbalance the left side.
>The displaced water is then evenly distributed across the remaining sections.
WRONG. it's evenly distributed across ALL sections, including the section to the left of the boat, and including the section that is the boat.
No, the water is displaced out from underneath the boat. You can't displace water into the space the boat occupies.
Alright. I gave it my best. I will never successfully explain this to you, I guess.
You can actually. Some of the displaced water ends up under the boat because the water level in the whole tank raised.
But that would require an increase in water density below the boat, but with nothing to contain it the water would spread out and not remain more dense beneath the boat. Then we're back to my illustration with the even displacement around the boat.
Let's think this through.
Here we have four sections. Let's say that the boat weighs three kilos. Now, what happens to the weight of Section 1? By the weight of the boat we will add 3 kilos. Then, by the Archimedes' principle we will subtract 3 kilos of displaced water. Now those 3 kilos of water must be put back to the tank in one way or another.
Putting the water back raises the boat and the surface of water equally which adds 3*(1/4) kilos to Section 1. So for Section 1 we have +3 (boat) -3 (Archimedes) +3*(1/4) (putting the water back) = 3*(1/4). One quarter of three kilos of weight were added to each section which means that the forces stay balanced and the tank does not tip. So overall...
Section 1 = + 3 - 3 + 3*(1/4) = 3*(1/4)
Section 2 = + 3*(1/4)
Section 3 = + 3*(1/4)
Section 4 = + 3*(1/4)
So some of the displaced water goes into the section from which it was displaced from. You can get an idea why this works if you think about the fact that submerging a floating object into a tank simultaneously raises the surface level as you are submerging it. So if you did this in real life you would simultaneously be displacing water and adding 1/4 of that water underneath the boat.
I went out of my way to draw you an accurate picture of an example. The total mass of water is conserved and the submerged block is half as dense as water which means it sticks halfway out of the water.
Bonus points if you can figure out where the displaced water went.
Oops, mistake. In the right case, the left text should say mass on left instead of right.
Huh... that actually does make sense. I concede, it wouldn't tip.
>argues an incorrect point for several posts
>gets convinced by an effort-post
>admits they were wrong
>learns something
very based
This thread is psyops by US navy recruiters
>So it's no longer balanced because the boat's equivalent displaced mass is not entirely located to the right of the pivot.
yeah, it's spread perfectly evenly everywhere.
imagine the tank with no boat. everything is perfectly symmetric, right?
now imagine the boat floating statically at one end. the system is the same everywhere, except at the boat. but the boat weight exactly as much as that chunk of water would weight anyway, so it doesn't effect the total distribution of mass in the system. the distribution of mass in the system IS ALL THAT MATTERS in determining if it tips, because gravity is the only external force besides the normal at the base.
it will tip because part of the boat is above water
Which beaker has higher pressure at the bottom?
AAAAAAAAAAAA
yes
is the mass of the boat over the volume displaced equal to the density of water?
While the mass is the same, it is no longer evenly distributed left and right of the fulcrum. On the left, there should be more torque from having mass above the surface of the water and hence at a large radius.
No. The water on the right side becomes heavier to counterbalance the weight of the boat until they reach an equilibrium. It's basic newtonian mechanics.
If it floats it stays, if it sinks it tips.
The picture is too vague of a description and can be used interpreted in multiple ways. This isn’t science, or math, this is philosophy.
I HAVE THE SOLUTION
The boat is a sailboat
Therefore this simulation accepts wind (else the sailboat would be absurd)
-> It WILL tip because of the wind.
If Some of the ships volume isn't displacing water than I imagine that would count as extra mass on one side.
Why wouldn't it be displacing water?
so many people ITT figuring out how buoyancy works for the first time.