Anonymous
Anonymous
>>3886872 are you a troll?
its the same pressure.
Anonymous
Quoted By:
Same pressure that the centre. Different pressures at the sides.
Anonymous
>>3886887 There's 1/100,000th the water on the right than the left.
There's more pressure on the right right.
Krakengineer !!5XY+x7grkpt
>>3886887 The force of water pressure on an object is proportional to the fluid column above it, the mass of the fluid column to the right is less. They are not the same.
Anonymous
Quoted By:
They have the same pressure even at the sides.
Anonymous
>>3886905 >>3886902 no1: the formula it P = g*rho*h
and look! g, rho and h are the same in both!
no2:
pressure is force per area,
force is mass * g, thus it is proportional to the area,
take the force and divide it by the area, now the area dependence is gone!
Anonymous
>>3886902 >>3886905 Swim 10' down in a small but deep pool, like one of those practice diving pools.
Swim 10' down in the ocean.
Same pressure, or different?
Answer: not exactly the same cause seawater weighs a bit more than pure water, but pretty darned close.
Krakengineer !!5XY+x7grkpt
>>3886924 force is mass * g, thus it is proportional to the areaThe mass of the column to the right is less. The area at the bottoms are the same in both cases. There is less pressure.
>>3886928 What makes this situation different is that the diameter of the column of water on the right expands at the bottom. The unit water pressure is distributed over a larger area. If we were dealing with two cylinders the water pressure would be equal regardless of the diameter.
Anonymous
Anonymous
>>3886948 Here is a hint for you.
Anonymous
Quoted By:
Hydrostatic Paradoxon in da house, bitches.
Anonymous
Quoted By:
>>3886948 >The mass of the column to the right is less. The area at the bottoms are the same in both cases. There is less pressure. where are you getting this? have you studied engineering or are you only impersonating one? the tube width does not make a difference. only the height of the water.
Anonymous
>>3886948 so you think this would work?
Teacup !obkpuSWh5M
Obviously the one on the right has more pressure because there's more water. A fucking housewife could get this right you ape.
Anonymous
Quoted By:
saw this the other day and put it on the board in the break room of the astronomy group, as much as it is uncomfortable to say, the pressure is the same.
Teacup !obkpuSWh5M
Krakengineer !!5XY+x7grkpt
>>3886962 P = F / A
The question does not give the area at the bottoms of the tanks, let's assume it's 1m^2.
Case 1
F =Mass*Acceleration = 1000 Kg * 9.8m/s/s = 9800 N
P = F / A = 9800 n / 1m^2 = 9800 Pa
Case 2
F = Mass * Acceleration = 0.01 kg * 9.8m/s/s = 0.01 * 9.8 = 0.098 n
P = F / A = 0.098 n / 1m^2 = 0.098 Pa
9800 Pa does not equal 0.098 Pa. What seams to be the problem?
Anonymous
If we replaced the upper columns of water with movable, watertight weights equal to the mass of the water, we would not have the same pressure. Therefore, the pressures in the two cylinders are different.
Anonymous
Anonymous
>>3887006 first case:
F = 1000 Kg * 9.8m/s/s = 9800 N
P = F / A = 9800 n / 1m^2 = 9800 Pa
second:
F = Mass * Acceleration = 0.01 kg * 9.8m/s/s = 0.01 * 9.8 = 0.098 N
P at diameter change: F / A = 0.098 N/0.00001m = 9800Pa
force on water in 1m^3 tube: 9800Pa*1m^3 = 9800Pa
9800Pa = 9800Pa
you fucking moron.
Anonymous
Anonymous
Quoted By:
>>3887006 >2 different objects >same area? Idk if troll or just stupid.
Tripfag at his finest..
Krakengineer !!5XY+x7grkpt
>>3887030 P at diameter change: F / A = 0.098 N/0.00001m = 9800Pa>> 0.00001m The diagram shows that the area where the rocks are stationed are the same diameter.
Anonymous
Quoted By:
>>3887047 thats why i used
>force on water in 1m^3 tube: 9800Pa*1m^3 = 9800Pa >1m^3 read the whole thing, that's the force btw, you need to divide it by 1m^3 again, (hint this gives 9800Pa)
Anonymous
Quoted By:
>>3887015 wrong. We would have the same pressure. You're stupid and a nigger.
Quoted By:
p = rho * g * h
Anonymous
Quoted By:
>>3886984 Until friction and turbulence eliminate the flow, yes, in some sense it works.
Anonymous
if the presure is not the same, this would happen,
Anonymous
Anonymous
>>3887061 This is exactly right. Anyone who still thinks it'd be the same is an idiot.
Anonymous
Quoted By:
>>3887061 actually, I should be nicer. This is 4chan, after all.
Anyone who still thinks the pressures are different should pay better attention.
Anonymous
Quoted By:
>>3887061 Amazing. While all the VIOLENT SIMIANS here are squabbling over equations and imaginary science, this prime specimen proposes that you can solve this problem with experimentation instead.
Now that, my primate friends- is real science. Take notes you lowly apes.
Krakengineer !!5XY+x7grkpt
Quoted By:
>>3887061 You are confusing pressure with force.
Anonymous
Quoted By:
>>3887062 No. Disk on the right needs a smaller surface area for it to be analogous.
Anonymous
>>3887062 wrong, its not equivalent to the question
first consider the top part of both containers (the skinny bit on the right one and the same part on the fat one), pressure scales only with depth so at the bottom of both of these parts have the same pressure.
at the transition from the skinny part to the fat part the pressure at the boundary match so the top of the fat bit of the one on the right has the same pressure as the same bit on the left one. below this again pressure scales with depth, hence the pressure is the same.
Anonymous
Quoted By:
>gather 10000 straws >put them together to make a long chain >fill with water >put it on objects >break everything with shit loads of pressure>problem science?
Anonymous
Anonymous
Quoted By:
>>3887065 Congrats, you've created a perpetuum mobile.
Anonymous
Quoted By:
>swimming in the ocean >dive 1 feet >suddenly the pressure of the whole ocean is on you >explode
Anonymous
Quoted By:
>>3887088 add a turbine and free energy!
Anonymous
Quoted By:
forgetting that its 1 ml - not 1m (10^-3)^2 which would be dividing by 1 millionth therefore it is equal.
Anonymous
Quoted By:
>>3887088 thank you for simply explaining that boyle's self filling flask does not work.
You should put that mind to more productive use, like figuring out to make a bong out of a klein bottle.
Krakengineer !!5XY+x7grkpt
>>3887085 pressure scales only with depthThat is only when the area of the water column remains equal to the area the force is effecting, which is not the case here. MFW 95% of /sci/ falls for troll physics.
Anonymous
Quoted By:
>>3887103 im gonna assume you are a troll, no one can be this retarded.
>That is only when the area of the water column remains equal to the area the force is effecting, no, see
>>3887030 Anonymous
Quoted By:
the /g/ thread is much more productive.
Anonymous
Quoted By:
>>3887103 i described that for the top part and the bottom part of the right hand container the pressure scales with depth. at the boundary between the top and bottom parts the pressure must match. so the bottom part starts with the same pressure ans the bottom of the skinny part, so yes pressure scales with depth, the extra pressure comes from the top of the fat part of the container around the end of the skinny part.
try again this isn't troll physics.
Anonymous
Quoted By:
P = gamma(h) H is the same. Gamma is the same. The pressure is therefore the same.
Anonymous
>>3887030 this is only within the tubes. at the bottom the areas are equal.
Anonymous
>>3887128 thats why I used the area of 1m^3 in both calculations.
Anonymous
Quoted By:
BRB PATENT OFFICE
Anonymous
>>3887103 Krakengineer, what do you think the pressure is just above the widening of the cylinder? How about just below it?
Anonymous
>>3887134 if you're
>>3887006 then yes, you're right. but i was refering to
>>3887030 who did it wrong.
for the record. rho g h only applies if you say V=h*A which is not the case here. the bottom cross section is the A and if you do A*h you get the volume of a cylinder the size of the one on the left when in reality its much less. pressure = weight/area. the area of the bottoms is the same, the right one has left weight. figure it out
Anonymous
I know hardly any physics at all. Here is my attempt: There is a larger mass of water in the left tank compared to the right, meaning a larger force will be exerted on the same surface area (Mass*Acceleration of Gravity). Since the surface area is the same, the pressure on the object and the bottom of the tank on the left is larger than the pressure on the object and bottom of the tank on the right. Where did I go wrong?
Anonymous
>>3887156 i am
>>3887030 and i did it correctly, i used the smaller diameter up to the point where it became wider, then used the wider diameter of 1m^3, and get the and pressure.
>>3887006 is wrong
Anonymous
>>3887103 Incorrect. Pressure only scales with depth. Look it up. Consider the force from the weight of the water in both situations. The first one is gamma(A)(h) where A is the cross sectional area of the tube and h is the height. The Force induces a pressure at the interface whose magnitude is F/A or gamma(h). The same analysis can be done on the other tube, but notice how the area cancels. Therefore Pressure scales only with depth. P = gamma(h). You're wrong. Take it from an engineer.
>>3887163 That's wrong because the surface area is not the same.
>>3887179 The force on the object is the same, but you certainly did not prove it correctly. 9800PA * 1 m^3 = 9800 Pa? Seriously? Your method was wrong, your units are wrong, and your proof is wrong.
Anonymous
>>3887194 I figured that the equal and opposite force to the weight of the water must be applied along the bottom of the tank. Both tanks have equal surface area bottoms. Where did I go wrong this time?
Anonymous
>>3887194 >>3887194 ya, that last Pa should be N. congratulations! you are the first person to spot it so it seems the first person who even read it all the way through. have a doughnut.
Rainbow Dash !!Q/yRC4LcWxp
Quoted By:
>>3887061 That is what happens, though exaggerated.
Two words:
Capillary action
Anonymous
>>3887179 no you did it wrong.
force on water at transition is .098 newtons, you said it was the pressure. you wanted to say it was the force and then calculate the pressure with P=F/A with the new A at the bottom. what you did was calculate the pressure in the tube, say that that was the pressure in the bottom, and then reverse engineer the same answer as the left side.
Anonymous
Pressure is force per unit area, faggots. P = F/A F = volume * Density V = A * Height P = A*H*D/A P = H*D FUCKING HINT IF YOU'RE RETARDED: THE PRESSURE DOES NOT DEPEND ON HORIZONTAL AREA OF THE STRAW OF THE WATER. SAME PRESSURES.
Anonymous
Anonymous
>>3887215 too bad,
>>3887203 already got the doughnut.
Anonymous
>>3887194 you are wrong sir. so much for your engineering skills
rho g h comes from P=F/A, F=ma. m=rhoV and V=hA.
this is true sometimes but in OPs example V≠hA because A is the A of the bottom and if you take h* that A you get a volume equal to the tube on the left, not the right.
instead you should use P=F/A or pressure = weight divided by area. the area is the bottom where the rock is, so they're equal, and obviously the left has more weight.
Anonymous
Quoted By:
>>3887179 >>3887194 >>3887217 please keep fighting the good fight i need sleep.
Anonymous
People who think the pressure is less in the setup on the right: what do you think is going on at the interface of the two tubes? Just above the interface, the pressure is the same in either case. There can be no finite jump in pressure across the interface. Consider a very thin cylindrical control volume straddling the interface (axis parallel to the symmetry axis). Its mass can be made arbitrarily small, the only way it can be in equilibrium is if the pressure is continuous across the interface: the pressure in the wide part is the same in both cases. The pressure at the bottom is the same in both cases.
Rainbow Dash !!Q/yRC4LcWxp
To whomever thinks the pressures are the same. If you step into a bathtub and I put a really long straw into it, do you really think I could crush you with 1 cup of water? Actually think this through, just because you know the equations doesn't mean you have to throw basic reasoning out the window.
Anonymous
>>3887239 yes, it will crush you, assuming the bath is sealed airtight.
Anonymous
Anonymous
>>3887239 If the bathtub is sealed and the straw is 10m high, then yes. you can crush me with just water. It is possible, because each of the walls of the tub would need to push against the straw per unit area.
Pressures are the same:
P = D * g * H
pressure, density, gravity, height.
Anonymous
Quoted By:
These ones always trick people. Same goes for tension problems.
Anonymous
>>3887217 see
>>3887235 V≠hA
>>3887224 i wasnt comparing your units im saying your method was wrong. you wanted to use the force and instead you used the pressure. you didnt mistype it, you actually did it wrong.
>>3887006 did it right
Rainbow Dash !!Q/yRC4LcWxp
>>3887250 >>3887250 No it won't.
You are an ignorant little human who doesn't understand anything. Put a transparent straw in a cup of water, the water will actually flow up the straw above the water level.
LEARN 2 CAPILLARY ACTION
Anonymous
>>3887201 Someone answer me.
Anonymous
Quoted By:
>>3887253 >>3887252 >>3887250 NO
thats not how science works. positioning of 1lb of water does not change the fact that it is 1lb of water.
please god read
>>3887235 >>3887256 and stop saying rho g h
YOU CANNOT USE RHO G H HERE AS THE AREA AT THE BOTTOM IS DIFFERENT THAN THE AREA AT THE TOP, THIS V≠hA
OK?
Krakengineer !!5XY+x7grkpt
>>3887145 At the point just before the cylinder on the right widens the pressures are equal.
P1 = F1 / A1 = 9800 n / 1m^2 = 9800 Pa
P2 = F2 / A2 = 0.098 N/0.00001m^2 = 9800Pa
After the tube on the right widens the area increases but the force of gravity from the water above remains the same, so the pressure drops.
P2 = F2 / A2 = 0.098 n / 1m^2 = 0.098 Pa
I can't believe all of /sci/ is being trolled because they don't realize their pressure depth formulas don't account for situations like this. The force of gravity on the two water columns is not equal, but the area the force acts on AT THE BOTTOM is equal, so the pressure is not equal.
Saying the pressure is the same in both cases is like saying 100 men will be pushed down on by a shared load of 10kg as much as by a shared load of 10000kg.
>>3887260 >>3887256 ok you know what, fine, keep on being ignorant. im done, i dont care anymore, two more idiots among millions isn't going to make a difference. il be in a interesting thread if you need me.
Anonymous
Quoted By:
>>3887271 you had it right in the beginning, the other guy was wrong. the right one has less pressure.
Rainbow Dash !!Q/yRC4LcWxp
Quoted By:
>>3887278 You simply don't understand how the implementation of RhoGH works.
Anonymous
>>3887276 Then please refer to this post:
>>3887238 There can't be a sudden drop in pressure as the tube widens in a steady state configuration.
Anonymous
emm. all the people who say the pressures are different you may want to read up on hydraulics. the point being that you can get a bigger force out of a fluid that the one you put in. force isn't conserved pressure is. the force from the small tube is supplied to only a small area and hence produces a large pressure. if the forces were different hydraulic machinery wouldn't work. top pic related.
Rainbow Dash !!Q/yRC4LcWxp
Quoted By:
>>3887293 Straw in water.
Here's a pic of what happens
Anonymous
>>3887201 Think of it as two separate parts. The bottom part is the wider part of the cylinder with the object in it. Now consider the pressure acting on the bottom part from the top part. The force is of course greater from the larger upper cylinder than the force from the smaller one. But remember, we are concerned with force on the object, not just the bottom section. The pressure (since this fluid is incompressible), transmits the force to all surfaces that the fluid touches (Pascal's principle). Therefore we need the pressure at the tube interface. The pressure is simply the force per unit area. So let's figure out what the force is. The force can be represented by gamma(h)(A) where gamma is the specific weight of the fluid (look it up for more info), h is the height of the tube, and A is the cross sectional area of the tube. Notice, however, that the pressure is the force per unit area. So 'A' cancels. Therefore the pressure depends only on depth, as does the force on the object.
>>3887235 Think again, buddy. The area concerned is not where the rock is. It's the cross sectional area of the upper section. If you really need convincing, enforce static equilibrium on the container. You clearly don't grasp basic principles of fluid statics.
>>3887239 You actually would be crushed if the "straw" was a tube where capillary forces are negligible and the bathtub was closed off (only containing water).
Anonymous
Quoted By:
77 posts? wtf /sci/? simple answer: sump pumps are rated by feet of head pressure is the same regardless of pipe size
Rainbow Dash !!Q/yRC4LcWxp
>>3887300 In hydraulics, the force comes from the piston, not the liquid itself.... The liquid simply transfers the force...
WHY IS EVERYONE ON HERE SO RETARDED????
Anonymous
>>3887300 you're right but only in the tubes. once the tube widens at the bottom it no longer works that way because the forces havent changed but the area has. thus the pressure drops
Anonymous
>>3887324 everyones retarded but me!!!
Rainbow Dash !!Q/yRC4LcWxp
>>3887329 From what I read krakengineer seems to have it right too.
Anonymous
>>3887324 HA! You deleted your post! What's it like being wrong, you little bitch?
Rainbow Dash !!Q/yRC4LcWxp
>>3887334 I deleted it because it had a typo, I reposted it right away
>>3887245 Anonymous
>>3887331 o please, krakengineer hasn't been correct since he started coming on sci, weather it be thus or QM or anything. i dont know if he is a very good troll or just retarded, but if he said something, its save to assume the opposite.
Anonymous
>>3887328 wrong the small tube has a very small area and a small force and hence a large pressure. pascals principle says that pressure is transferred to all surfaces of the fluid including the top of the fat bit of the container. hence the whole area of the fat tube is pressing down with the same pressure so the small area of the small tube drops out and the situations are the same.
Anonymous
>>3887339 >Who the hell cares about a typo >You're still wrong My question still stands: what's it like being wrong?
Rainbow Dash !!Q/yRC4LcWxp
>>3887341 Look, the pressure would be the same if the tube stayed the same length width. The width opened up creating more area for the water to be pushed down upon. Thus capillary action would have a drastic effect here.
He is most definitely correct in this case.
Anonymous
Quoted By:
>>3887345 FUCKING THANK YOU. Enforce static equilibrium on the container. You'll see that I'm correct. (I'm these posts, BTW).
>>3887320 >>3887194 Anonymous
>>3887320 actually you're wrong "buddy". the question was whats the pressure on the rock not whats the pressure in the tubes. up until the change in diameter the pressure is the same but the area is larger where the rock is and thus the pressure goes down.
anyone who thinks a cup of water can crush you just because its stacked up high on itself is retarded. its force is the same. 1lb = 1lb no mattter what the configuration. 1lb on a needle head is alot more pressure than 1lb on a sidewalk tile but putting 1lb on a needle head and then saying that that pressure can crush you from all sides when you touch it to a sealed bath tub is retarded. the pressure spreads back out.
Rainbow Dash !!Q/yRC4LcWxp
Quoted By:
>>3887346 Allow me to recap your argument.
Anon: You deleted your post you admitted your wrong
Rainbow Dash: It's still up, just corrected.
Anon: HA you didn't answer my question
I stand by my point you are retarded, and I'm not wrong. You simply don't understand P = F/A
Anonymous
>>3887351 Lol capillary action. Change it to 10m diameter and 1m diameter. Surface tension and capillary effects are negligible.
Anonymous
Quoted By:
Guy who said he knows hardly any physics here. I'm starting to think the term "pressure" is being used to describe 2 different things here.
Anonymous
Quoted By:
Alright, you fucking niggers. Give me a few minutes. I'm writing a proof that will end this argument once and for all.
Anonymous
http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html first box shows how the small tube would transmit a force grater than its mass to the fluid below.
Rainbow Dash !!Q/yRC4LcWxp
Quoted By:
>>3887364 The determination would be the diameter change. As you can see there is a significant change on the picture on the right, and no change in the picture on the left. If you kept the tube really thin throughout (so that there isn't that tank), the pressure would be the same in both pictures.
Anonymous
Quoted By:
>>3887355 >up until the change in diameter the pressure is the same but the area is larger where the rock is and thus the pressure goes down. now boys and girls what is Pascal's principle, "pressure is transmitted undiminished in a static enclosed fluid", hence the force does not go down just because the area where the pressure is applied is not the whole top.
Rainbow Dash !!Q/yRC4LcWxp
>>3887380 because of a mechanical advantage, it would have to push for a longer distance. the distance is the same in this case, as there is nothing being pushed down or up. Just the mass of the water at a constant height.
Anonymous
>>3887355 Pascal's principle, you fucking nigger. Look it up. I'm right and you're wrong. Fucking children.
Anonymous
Quoted By:
>>3887006 Let's assume this is correct.
Pressure in the right tube is 0.098Pa.
Obviously the cross-sectional area of the thin tube is less than 1m^2, lets call it 0.1m^2.
So the force exerted upwards on the column of water by the tank is 0.098Pa * 0.1m^2 = 0.0098N.
Column of water exerts 0.098N down, tank exerts 0.0098N up. Why doesn't the water churn?
Anonymous
>>3887412 there is nothing about distance its all static like the question, nothing is moving. it shows pressure is transmitted to all surfaces in the fluid and because the tube has the same pressure as the tank (smaller force but smaller area) hence that same pressure is transmitted to the object. it's the same.
Rainbow Dash !!Q/yRC4LcWxp
Quoted By:
>>3887447 Are you kidding?
Do you think you can push down with 1 pound of force for 1 meter, and move a 2000 pound object 1 meter up because of a change in diameter?
Anonymous
>>3887417 that doesnt even fucking apply here. would you please read the goddamn equations:
>>3887235 the force that the water on the tube exerts on the water in the base is less than that of the left side. less force means less pressure. the two bases are not connected.
in the small tube there is high pressure. when the area is increased there is no increase in force so the pressure must go down. fucking start from the beginning. rho g h cannot be used in this situation and the pressures therefore are not the same.
pressure = weight/area
the weight of the water in the right tube is less
the area of the bases (or of the rock) are the same
explain how the pressure would be equal.
Anonymous
>>3886863 Ofc it's the one on the left, the one on the left affect only part of the rock (the centre in this case) and the lefty one entire rock, therefore there's more pressure on the rock on the left.
Anonymous
>>3887485 > Doesn't know about Pascal's principle > 2011 Rainbow Dash !!Q/yRC4LcWxp
>>3887495 OK THOUGHT EXPERIEMENT.
I take a 10 meter long straw to the ocean filled with water.
When I place the straw in the ocean, does the pressure in the ocean suddenly shoot up to over 9000 atm and explode the ocean?
YOU GUYS ARE FUCKING RETARDED GO KILL YOURSELF IF YOU THINK YES.
Anonymous
Quoted By:
>>3887485 This is a statics problem, meaning that nothing is moving. So the force exerted by the column of water on the tank must equal the force exerted by the tank on the column.
There is indeed less water and less force exerted by the column on the right, but the force is exerted over a smaller area.
So the pressures are equal.
Anonymous
>>3887523 Are you twelve or what?
The ocean is not a closed container.
Anonymous
>>3887495 You're an idiot, pascal's principle implies that the area of water covers in the same percentage the test object as the one on the left. If the rock on the left was smaller, around the size of the tube's diameter or smaller, then it applies, otherwise the water in the tube affect that rock partially.
Anonymous
>>3887523 > Implying the ocean is an enclosed static fluid > Implying the water level in the column doesn't drop due to gravity > Implying Pascal's principle is in question Anonymous
P=F/A>Force exerted on area >Objects are the same >same area >less water in the right tank >water weighs less >less force >smaller pressure in the right one It's that simple.
Rainbow Dash !!Q/yRC4LcWxp
Quoted By:
>>3887540 Ok so naturally the ocean would rise 10 meters so that it wouldn't increase in pressure then?
Rainbow Dash !!Q/yRC4LcWxp
>>3887553 OK a frozen lake.
Anonymous
>>3887555 >same area the area is just the are of the tube not the tank below.
Anonymous
>>3887572 if the ice could take the pressure and was air tight yes.
Anonymous
>>3887545 >If the rock on the left was smaller, around the size of the tube's diameter or smaller, then it applies wtf does the rock have to do with this?
Anonymous
Quoted By:
>>3887574 same area meaning both objects have the same area as the actual question was if there is the same pressure on the object.
Rainbow Dash !!Q/yRC4LcWxp
>>3887579 So a straw 20 meters high, holding only 10 ml of water, drilled into a frozen lake, would explode the ice of said frozen lake?
Anonymous
Quoted By:
>>3887574 And i forgot. With object i meant the fucking rock.
Anonymous
My first reaction is that the pressure is equal, but can I ask a retarded question? Why is pressure at the bottom of the ocean much greater than at the top?
Anonymous
this is my last post. if you cant understand it after this, you're beyond my help. i replaced the tubes of water with weights of equal size. the masses are the same, the answer should be the same.
Anonymous
Quoted By:
>>3887596 As i said, P=F/A.
to make it more clearly, the force of the water directly above the area counts.
If you dive deeper, the force is lager as there is a greater pile of water, causing pressure to increase.
Anonymous
hey rho*g*h fag explain this
Anonymous
Quoted By:
>>3887589 no the ice would just lift at the edges, or crack. no exploding necessary. in the real world the water would run out the straw as the lake leaked.
Anonymous
>>3887609 >>3887609 thats caused by capalary action you dumbass, not the presure difference.
Anonymous
Quoted By:
>>3887589 not him but i suppose you will likely be pouring
water for a while into the straw before the straw begins to
fill up, and then either water will leak out
of the ice or the straw will rupture due to
any increase in pressure
Anonymous
Alright, excuse the poor drawing. Water isn't moving, so Fc1 = Ft1 and Fc2 = Ft2. The force from the columns is equal to the weights of the water: Fc1 = m1 * g = 9800N Fc2 = m2 * g = 0.098N The heights of the columns are equal. Since V=HA (only talking about the columns here) A2 = A1 * (V2 / V1) = 0.00001 * A1. Now, the forces exerted by the tanks on the columns are determined by the pressure and the areas of the interfaces: Ft1 = P1 * A1 Ft2 = P2 * A2 Substituting we have: 9800N = P1 * A1 0.098N = P2 * A1 * 0.00001 So: P1 = 9800N / A1 P2 = 9800N / A1 QE fucking D
Anonymous
>>3887582 Then the object, whatever it is.
You either have to enlarge the tube on the left to cover entire plan are of the rick or minimize the rock. The tube on the left affects only the centre of the rock, not the entire rock, or not equally at least (the farther away from the centre, the smaller the pressure)
Anonymous
Quoted By:
>>3887485 >He thinks pressure is a vector quantity >He thinks a change in force is equivalent to a change in pressure >He thinks Pascal's principle does not apply Pressure is a scalar quantity.
A change in force does not equate to a change in pressure.
Why would you think that the bottom of the container is the relevant area?
You can't just assume that the force due to the fluid is transmitted directly to the object. If that were the case, you could make the object any size you like, and the force would be the same. Don't you think the windows on submersibles would be really large if the force exerted on them was not dependent on their surface area? Pascal's principle absolutely applies.
Also, your "equations" give no insight whatsoever. (You also forgot to incorporate the acceleration due to gravity). This is all down to your refusal to accept that Pascal's principle applies. You'd be laughed out of an engineering department for thinking otherwise.
>>3887488 No.
>>3887545 Think again nigger.
http://en.wikipedia.org/wiki/Pascal's_law >>3887603 If the area ratio is correct, then yes.
Anonymous
Quoted By:
>>3887006 7/10 pretty good
Anonymous
>>3887603 but as shown in your diagram the masses take up different area's, the smaller area means the smaller weight presses down with the same force per unit area (pressure). by pascals principle this pressure is transferred thought the fluid so both have the same pressure.
honestly take this to your physics teacher, or someone who does physics.
Anonymous
>>3887622 you did it wrong. all you found was that the pressure at the very bottom of the tubes is the same. for the pressure on the rock, A1=A2
Anonymous
Quoted By:
>>3887640 Nope. He did it correctly. Pascal's principle.
Anonymous
>>3887622 It's only considers the area directly underneath the tube, not the entire are of the bottom part.
Rainbow Dash !!Q/yRC4LcWxp
>>3887637 I feel bad for you. I'm glad I'm not as retarded as you.
Anonymous
Quoted By:
>>3887625 if the pressure was different under the tube than beside that, there would be a pressure difference and hence a flow, so by your argument this would cause a current to flow.
Anonymous
Quoted By:
>>3887640 If the pressure in the tank is different than the pressure in the column then the water would be moving.
Unless you think that the force exerted by the tank on the top glass somehow holds up the column of water.
Anonymous
>>3887616 Yes, capillary action is one of the many scenarios in which delta P = rho *g *h doesn't apply.
Maybe if you could derive this formula from a more complete equation, like Navier-Stokes. Then you would understand that rho*g*h is an extremely limited equation with many assumptions. It should never be used by people as stupid as yourself.
Anonymous
>>3887637 so if the hole was super super small and i stuck a needle in there you think i could make the pressure higher than a 1000kg weight (over a ton) with just my hand?
Anonymous
>>3887663 Yes. Learn the difference between pressure and force.
Anonymous
Quoted By:
more complete version of the picture so you can see what I'm saying
Anonymous
Quoted By:
>>3887663 Have your mom step on your foot with a snowshoe, you'll barely feel it.
Now have her do it with high heels.
Anonymous
>>3887654 >a personal attack due to lack of an argument. Anonymous
>>3887654 Unfortunately you are much retardeder.
Anonymous
Quoted By:
>>3887603 not the same at all, the point people seem to be missing is it's a closed container.
the pressure must be the same, the force does not, so where does the extra force go?
it's the pressure acting up on the top of the bottom section
Anonymous
>>3887667 pressure is force divided by area you think that the pressure of the needle on the water is the same thing as the pressure of the water? LOL?
hey guys if you put 5lbs on a needle which is .01inches^2 you can have a 500psi glass of water.
Anonymous
Quoted By:
>>3887653 Actually, take that back.
The pressure would be higher at the centre than on the example on the left, but smaller on the sides, the overall pressure, however, would be the same.
My bad.
Rainbow Dash !!Q/yRC4LcWxp
>>3887688 >>3887683 Whoever did this
>>3887603 Did it perfectly.
the fluid is going to distribute the weight equally over the same area. Weight one is much larger than weight 2, so the pressure is going to have a massive difference.
The fact you guys can't see it, makes you beyond help.
It's not personal attacks because I don't have an argument, it's personal attacks because you are too stupid to comprehend it.
Anonymous
>>3887694 If the water is in a full, sealed container and the hole is airtight when the needle is inserted, then yes you can.
Anonymous
>>3887660 Capillary action is not relevant for the original problem. Secondly, the Navier-Stokes equations simplify down to the pressure gradient and the specific weight. This actually simplifies to gamma(h) = P.
>>3887702 Nope. You're beyond help. The forces do not act over the same area.
Rainbow Dash !!Q/yRC4LcWxp
>>3887704 You just when full blown autistic.
Anonymous
Quoted By:
>>3887702 Contact a reputable fluids engineer and state that conclusion. Just be prepared to be laughed at. The area is not the same.
Rainbow Dash !!Q/yRC4LcWxp
>>3887709 The irony here, is that what you are saying is a violation of pascal's law, while you claim everyone else is the one who's violating it.
Anonymous
>>3887702 Gonna try this again.
The weights aren't moving, so there must be a force equal to their weight exerted by the tank on them.
Force is distributed evenly across the inside surface of the container.
There is much less area under the lighter weight (the area is proportional to the weight).
So the pressure has to be the same in order for there to be enough force to hold the weight up.
Anonymous
Can someone just post a video demonstrating this and let this thread die
Anonymous
Anonymous
Quoted By:
>>3887722 Pascal's law states that the pressure in a sealed fluid is transmitted through the fluid. If the area over which the weights were acting was the same, you'd be correct. The area over which the forces act, however, is not the same. Please go into detail about how this violates Pascal's principle.
Anonymous
>>3887715 jesus christ dash me and you seem to be the only ones who understand this thing.
seriously. 500psi glass of water from a 500psi needle. NOT HOW PRESSURE WORKS
Anonymous
Anonymous
Quoted By:
>>3887702 >the fluid is going to distribute the weight equally over the same area. well there's yer problem. quote me one principle or law or anything else that says this.
on the other hand we have pascals principle which says and pressure applied to a liquid is transmitted throughout.
Anonymous
>>3887709 The fact that rho*g*h does not apply in increasingly small containers because of capillary effects is relevant.
Also it applies now.
Rainbow Dash !!Q/yRC4LcWxp
>>3887723 The area under the weight is not what is causing the pressure on the rock. I think you are trying to say something like the rock has to be less dense then the water to float. but you could just have a mechanical piston pushing down at exactly 9800N on one side, and thinner one pushing .0098 N on the other. The actual size of the piston doesn't matter though, because the water will distribute the force evenly regardless thanks to PASCALS LAW.
Anonymous
Quoted By:
>>3887746 The original problem never specified the inclusion of capillary effects. Why don't we throw in electromagnetic forces on the water molecules or deflections of the container due to pressure that alters the volume while we're at it?
Anonymous
Anonymous
Quoted By:
>>3887752 You're definitely trolling. The original problem has nothing to do with pistons. The size of the piston (for the other picture) DOES matter.
Why don't you state Pascal's principle and explain it to avoid any confusion? I'd love to see what you cook up.
Anonymous
>>3887731 >>3887731 >>3887731 >>3887731 >>3887731 >>3887731 >>3887731 >>3887731 >>3887731 >>3887731 >>3887731 >>3887731 >>3887731 i could fucking kiss you thank you thank you thank you finely i can get some sleep.
so Rainbow Dash what the fuck were you saying about me being retarded.
Anonymous
Quoted By:
>>3887752 I'm trying to say nothing about density.
The rock is irrelevant. The pressure felt by the rock will be equal to the water pressure, so it's easier just to talk about the water pressure.
It doesn't matter if we're talking about a piston or a weight or whatever. If the thing exerting the force doesn't move, then there must be an equal force pushing back on it.
If the big weight is pushing down with 1000N on a 1000m^2, the water is pushing back with 1000N of force on a 1000m^2 area. 1 Pascal.
If the small weight is pushing down with 1N of force on a 1m^2 area, the water is pushing up with 1N of force on a 1m^2 area. 1 Pascal.
Anonymous
>>3887731 The problem is that he implies that the pressure is constant from one end to another on both containers.
It's not.
Anonymous
pressure distribution on the rock
Anonymous
Quoted By:
>>3887793 No, the pressure distribution is exactly the same. Might be a good idea to check out that lecture posted earlier.
Anonymous
Quoted By:
>>3887792 >>3887793 18:00 Watch for a few minutes...
Anonymous
Quoted By:
>>3887792 >MIT lectures says im wrong on a video checked by countless physicists >he must be wrong Anonymous
Anonymous
>>3887768 you're really listening to those jews at MIT?
Anonymous
Quoted By:
>>3887824 >bad at physics >blame the jews Anonymous
Quoted By:
Sure is taking Rainbow Dash a long time to watch the relevant two minutes of that video.... =P
Ricers gonna !RiceOCNvto
I did some math and found the capilary action of 10ml of water at 10m in length is 7.847533632286995515695067264574m So you would have nearly 8 meters of air on the right side So it would in fact be less here is the math in pic Radius of the tube is 0.0001784m Someone correct me if this is wrong So with that being said the pressure would be like 1/4 depending on the base size ect
Anonymous
If you had two tubes, one at each end, does that mean that the pressure would be double of that on the left example?
Anonymous
>>3887840 What? The original problem statement doesn't give enough information to determine the radius of the tube.
Anonymous
>>3887768 He didn't take into account capillary action. He was teaching classical physics. l2 apply it or don't learn it at all.
Anonymous
>>3887846 The water would fall out of the tube coming from the bottom end.
Anonymous
Quoted By:
>>3887848 Disregard that, I suck cocks.
Ricers gonna !RiceOCNvto
Quoted By:
>>3887848 how does it not?
10ml is a volume
10m is a height
its a tube so its round
Here explains the effects of capillary action on pressure.
http://en.wikipedia.org/wiki/Young%E2%80%93Laplace_equation Anonymous
Quoted By:
>>3887660 arent some of the assumptions
incompressible fluid
and reynolds <2200 flow
Anonymous
Quoted By:
>>3887840 i'm sure your calculation is sound and yes capillary action would be significant but i think the original task was ignoring that and most of the wrong people (Rainbow Dash and krakenengineer) were saying the force would be less because they didn't understand pascal's principle.
Anonymous
>>3887848 it does, we know height and the volume.
Interesting point. so in this case capilary action is revelant.
Anonymous
Anonymous
>>3887855 I might be dumb or tires or both, but I can't make sense out of your comment.
Ricers gonna !RiceOCNvto
Quoted By:
>>3887865 very relevant
assuming the tubes are glass the pressure is around 1/4 of the other sides 980,000 pascals depending on the size of the base the rock is in
Anonymous
>>3887869 Definitely tires.
He said one tube at each end.
A cylinder has two ends, a top and a bottom.
Anonymous
Quoted By:
>>3887852 he didn't specify the dimensions in his case but the point stands the argument people were making was bollocks.
Anonymous
Quoted By:
Of course you make the capillary contribution arbitrarily small by scaling up the problem. The people who were arguing for unequal pressure were wrong even if you do take capillary into account (and there's not enough info to do that).
Ricers gonna !RiceOCNvto
SO THE FINAL ANSWER IS THE LEFT SIDE HAS MORE PRESSURE This is based on capillary action not retard physics of hurp derp more water if it was not 10ml water and maybe 1 liter the pressures would be very close but based on what is stated in the picture the pressure would be quite a bit different
Anonymous
I think the biggest problem most people have with this problem is differentiating force with pressure. The left tube has more mass and will weigh more on a scale than the one on the right but they have the same pressure because they're at the same depth.
Anonymous
>>3887875 Yeah, by that I meant one on the right and one on the left.
Soz for the poor drawing.
Anonymous
Quoted By:
>>3887910 And that question I asking everyone who says the pressure on the OP question is the same in both examples.
Anonymous
Quoted By:
>>3887910 Why would it matter? Twice as much water column weight, twice as much interface area, once as much pressure.
Anonymous
>>3887905 So a tube of single molecules of water stacked directly on top of each other will have the same pressure as the left tube of water?
Anonymous
Quoted By:
>>3887910 *Ignoring cappilary action* that scenario gives the exact same pressure as the other two.
Anonymous
Quoted By:
>>3887905 They don't have the same pressure.
Anonymous
>>3887892 what about this?
Anonymous
Quoted By:
>>3887920 Pressure is a continuum quantity, it doesn't apply on that scale (same as density, temperature...)
Anonymous
Quoted By:
>>3887920 Our assumptions only hold in completely filled, sealed containers.
These kinds of things break down at the atomic scale.
Anonymous
Quoted By:
>>3887920 yup. They'll have totally different mass though and thus a totally different amount of force on the rocks. But there will be the same amount of pressure at the same depth.
To help, imagine the rocks are a single molecule in size. There's now roughly the same number of molecules stacked on top of each rock. The sizes of the tubes don't matter with pressure, but the depth of the tubes does.
Anonymous
Quoted By:
The problem some people have here is that they think that pressure = force. The left one emits more force but same pressure (pressure = per unit area)
Ricers gonna !RiceOCNvto
Quoted By:
>>3887931 the pressure is pushing outward
Surface tension is what is pulling it up the sides of the tube and pushing it outward
The pressure is not pushing directly downward as your implying it would
see this
http://en.wikipedia.org/wiki/Young%E2%80%93Laplace_equation Anonymous