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/sci/ - Science & Math

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3886863 No.3886863 [Reply] [Original]

90% of first year students get this wrong, can you solve it?

>> No.3886872

Different pressures.

>> No.3886887

>>3886872
are you a troll?

its the same pressure.

>> No.3886900

Same pressure that the centre. Different pressures at the sides.

>> No.3886902

>>3886887
There's 1/100,000th the water on the right than the left.

There's more pressure on the right right.

>> No.3886905

>>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.

>> No.3886910

They have the same pressure even at the sides.

>> No.3886924

>>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!

>> No.3886928

>>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.

>> No.3886948

>>3886924 force is mass * g, thus it is proportional to the area

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.

>>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.

>> No.3886949

depends

>> No.3886962
File: 923 B, 168x41, 51c1e96f5b03dceb3635b9e62170d57a[1].png [View same] [iqdb] [saucenao] [google]
3886962

>>3886948
Here is a hint for you.

>> No.3886969

Hydrostatic Paradoxon in da house, bitches.

>> No.3886973

>>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.

>> No.3886984
File: 59 KB, 320x282, Boyle'sSelfFlowingFlask.png [View same] [iqdb] [saucenao] [google]
3886984

>>3886948
so you think this would work?

>> No.3886992

Obviously the one on the right has more pressure because there's more water.

A fucking housewife could get this right you ape.

>> No.3886993

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.

>> No.3886997

>>3886992
*left

>> No.3887006

>>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?

>> No.3887015

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.

>> No.3887025
File: 5 KB, 581x421, sss.png [View same] [iqdb] [saucenao] [google]
3887025

sup?

>> No.3887030

>>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.

>> No.3887032

>>3887025
Aw shit nigga,

>> No.3887035

>>3887006
>2 different objects
>same area?

Idk if troll or just stupid.
Tripfag at his finest..

>> No.3887047

>>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.

>> No.3887052

>>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)

>> No.3887053

>>3887015
wrong. We would have the same pressure. You're stupid and a nigger.

>> No.3887056

p = rho * g * h

>> No.3887057

>>3886984
Until friction and turbulence eliminate the flow, yes, in some sense it works.

>> No.3887061
File: 8 KB, 581x421, 1318372083593.png [View same] [iqdb] [saucenao] [google]
3887061

if the presure is not the same, this would happen,

>> No.3887062
File: 24 KB, 1245x421, 1318372083593.png [View same] [iqdb] [saucenao] [google]
3887062

>>3887025

DUN DUN DUN

>> No.3887065

>>3887061
This is exactly right. Anyone who still thinks it'd be the same is an idiot.

>> No.3887066

>>3887061
actually, I should be nicer. This is 4chan, after all.
Anyone who still thinks the pressures are different should pay better attention.

>> No.3887074

>>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.

>> No.3887075

>>3887061
You are confusing pressure with force.

>> No.3887081

>>3887062
No. Disk on the right needs a smaller surface area for it to be analogous.

>> No.3887085

>>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.

>> No.3887087
File: 38 KB, 400x400, 1307576907917.jpg [View same] [iqdb] [saucenao] [google]
3887087

>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?

>> No.3887088
File: 7 KB, 581x421, 1318372083593.png [View same] [iqdb] [saucenao] [google]
3887088

the this will happen!

>> No.3887090

>>3887065
Congrats, you've created a perpetuum mobile.

>> No.3887091
File: 354 KB, 300x300, 1304060612231.gif [View same] [iqdb] [saucenao] [google]
3887091

>swimming in the ocean
>dive 1 feet
>suddenly the pressure of the whole ocean is on you
>explode

>> No.3887092
File: 9 KB, 581x421, 1318373078513.png [View same] [iqdb] [saucenao] [google]
3887092

>>3887088
add a turbine and free energy!

>> No.3887093

forgetting that its 1 ml - not 1m (10^-3)^2 which would be dividing by 1 millionth therefore it is equal.

>> No.3887097

>>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.

>> No.3887103
File: 85 KB, 480x600, 10-son-i-am-disappoint.gif [View same] [iqdb] [saucenao] [google]
3887103

>>3887085 pressure scales only with depth

That 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.

>> No.3887121

>>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

>> No.3887122

the /g/ thread is much more productive.

>> No.3887123

>>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.

>> No.3887124

P = gamma(h)

H is the same. Gamma is the same. The pressure is therefore the same.

>> No.3887128

>>3887030
this is only within the tubes. at the bottom the areas are equal.

>> No.3887134

>>3887128
thats why I used the area of 1m^3 in both calculations.

>> No.3887143
File: 22 KB, 779x421, 1318372083593.png [View same] [iqdb] [saucenao] [google]
3887143

BRB PATENT OFFICE

>> No.3887145

>>3887103
Krakengineer, what do you think the pressure is just above the widening of the cylinder? How about just below it?

>> No.3887156

>>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

>> No.3887163

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?

>> No.3887179

>>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

>> No.3887194

>>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.

>> No.3887201

>>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?

>> No.3887203

>>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.

>> No.3887205

>>3887061
That is what happens, though exaggerated.

Two words:
Capillary action

>> No.3887215

>>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.

>> No.3887217

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.

>> No.3887223

>>3887194
[x] Hydrotold!

>> No.3887224

>>3887215
too bad, >>3887203
already got the doughnut.

>> No.3887235

>>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.

>> No.3887237

>>3887179
>>3887194
>>3887217
please keep fighting the good fight i need sleep.

>> No.3887239 [DELETED] 

To whomever things 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.

>> No.3887238

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.

>> No.3887245

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.

>> No.3887250

>>3887239
yes, it will crush you, assuming the bath is sealed airtight.

>> No.3887252

>>3887245
yes it will.

>> No.3887253

>>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.

>> No.3887254

These ones always trick people. Same goes for tension problems.

>> No.3887256

>>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

>> No.3887260

>>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

>> No.3887271

>>3887201
Someone answer me.

>> No.3887274

>>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?

>> No.3887276

>>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.

>> No.3887278

>>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.

>> No.3887280

>>3887271
you had it right in the beginning, the other guy was wrong. the right one has less pressure.

>> No.3887291

>>3887278
You simply don't understand how the implementation of RhoGH works.

>> No.3887293

>>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.

>> No.3887300
File: 157 KB, 1087x2276, Hydraulic_Force_Torque_275px.png [View same] [iqdb] [saucenao] [google]
3887300

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.

>> No.3887314
File: 1 KB, 266x325, cap.gif [View same] [iqdb] [saucenao] [google]
3887314

>>3887293
Straw in water.

Here's a pic of what happens

>> No.3887320

>>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).

>> No.3887323

77 posts?
wtf /sci/?

simple answer:
sump pumps are rated by feet of head
pressure is the same regardless of pipe size

>> No.3887324

>>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????

>> No.3887328

>>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

>> No.3887329

>>3887324
everyones retarded but me!!!

>> No.3887331

>>3887329
From what I read krakengineer seems to have it right too.

>> No.3887334

>>3887324

HA! You deleted your post! What's it like being wrong, you little bitch?

>> No.3887339

>>3887334
I deleted it because it had a typo, I reposted it right away >>3887245

>> No.3887341

>>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.

>> No.3887345

>>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.

>> No.3887346

>>3887339

>Who the hell cares about a typo
>You're still wrong

My question still stands: what's it like being wrong?

>> No.3887351

>>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.

>> No.3887354

>>3887345

FUCKING THANK YOU. Enforce static equilibrium on the container. You'll see that I'm correct. (I'm these posts, BTW).

>>3887320
>>3887194

>> No.3887355

>>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.

>> No.3887361

>>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

>> No.3887364

>>3887351
Lol capillary action. Change it to 10m diameter and 1m diameter. Surface tension and capillary effects are negligible.

>> No.3887365

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.

>> No.3887375

Alright, you fucking niggers. Give me a few minutes. I'm writing a proof that will end this argument once and for all.

>> No.3887380

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.

>> No.3887389

>>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.

>> No.3887404

>>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.

>> No.3887412

>>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.

>> No.3887417

>>3887355

Pascal's principle, you fucking nigger. Look it up. I'm right and you're wrong. Fucking children.

>> No.3887419

>>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?

>> No.3887447

>>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.

>> No.3887477

>>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?

>> No.3887485

>>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.

>> No.3887488

>>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.

>> No.3887495

>>3887485
> Doesn't know about Pascal's principle
> 2011

>> No.3887523

>>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.

>> No.3887532

>>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.

>> No.3887540

>>3887523
Are you twelve or what?

The ocean is not a closed container.

>> No.3887545

>>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.

>> No.3887553

>>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

>> No.3887555

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.

>> No.3887557

>>3887540
Ok so naturally the ocean would rise 10 meters so that it wouldn't increase in pressure then?

>> No.3887572

>>3887553
OK a frozen lake.

>> No.3887574

>>3887555
>same area
the area is just the are of the tube not the tank below.

>> No.3887579

>>3887572
if the ice could take the pressure and was air tight yes.

>> No.3887582

>>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?

>> No.3887587

>>3887574
same area meaning both objects have the same area as the actual question was if there is the same pressure on the object.

>> No.3887589

>>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?

>> No.3887592

>>3887574
And i forgot. With object i meant the fucking rock.

>> No.3887596

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?

>> No.3887603
File: 11 KB, 552x648, 1318369942350.png [View same] [iqdb] [saucenao] [google]
3887603

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.

>> No.3887605

>>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.

>> No.3887609
File: 28 KB, 1308x746, Untitled.png [View same] [iqdb] [saucenao] [google]
3887609

hey rho*g*h fag

explain this

>> No.3887613

>>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.

>> No.3887616

>>3887609
>>3887609
thats caused by capalary action you dumbass, not the presure difference.

>> No.3887618

>>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

>> No.3887622
File: 11 KB, 514x531, dumbasses.png [View same] [iqdb] [saucenao] [google]
3887622

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

>> No.3887625

>>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)

>> No.3887626

>>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.

>> No.3887627

>>3887006
7/10 pretty good

>> No.3887637

>>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.

>> No.3887640

>>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

>> No.3887649

>>3887640

Nope. He did it correctly. Pascal's principle.

>> No.3887653

>>3887622
It's only considers the area directly underneath the tube, not the entire are of the bottom part.

>> No.3887654

>>3887637
I feel bad for you. I'm glad I'm not as retarded as you.

>> No.3887658

>>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.

>> No.3887659

>>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.

>> No.3887660

>>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.

>> No.3887663

>>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?

>> No.3887667

>>3887663

Yes. Learn the difference between pressure and force.

>> No.3887670
File: 37 KB, 1486x1017, Untitled.png [View same] [iqdb] [saucenao] [google]
3887670

more complete version of the picture so you can see what I'm saying

>> No.3887677

>>3887663
Have your mom step on your foot with a snowshoe, you'll barely feel it.

Now have her do it with high heels.

>> No.3887683

>>3887654
>a personal attack due to lack of an argument.

>> No.3887688

>>3887654
Unfortunately you are much retardeder.

>> No.3887689

>>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

>> No.3887694

>>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.

>> No.3887697

>>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.

>> No.3887702

>>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.

>> No.3887704

>>3887694
If the water is in a full, sealed container and the hole is airtight when the needle is inserted, then yes you can.

>> No.3887709

>>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.

>> No.3887715

>>3887704
You just when full blown autistic.

>> No.3887718

>>3887702

Contact a reputable fluids engineer and state that conclusion. Just be prepared to be laughed at. The area is not the same.

>> No.3887722

>>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.

>> No.3887723

>>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.

>> No.3887725

Can someone just post a video demonstrating this and let this thread die

>> No.3887731
File: 293 KB, 688x475, Capture.png [View same] [iqdb] [saucenao] [google]
3887731

>>3887725
http://academicearth.org/lectures/pressure-in-static-fluid
skip to minute 20.

>> No.3887736

>>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.

>> No.3887739

>>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

>> No.3887743
File: 1.59 MB, 1440x852, minecraft-trollface.png [View same] [iqdb] [saucenao] [google]
3887743

>>3887739

Nice try. 0/10.

>> No.3887744

>>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.

>> No.3887746
File: 38 KB, 1471x757, Untitled.png [View same] [iqdb] [saucenao] [google]
3887746

>>3887709


The fact that rho*g*h does not apply in increasingly small containers because of capillary effects is relevant.

Also it applies now.

>> No.3887752

>>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.

>> No.3887753

>>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?

>> No.3887756
File: 2 KB, 126x118, 1317777982266.jpg [View same] [iqdb] [saucenao] [google]
3887756

>>3887731
>>3887731
>>3887731
>>3887731
>>3887731

Fucking THIS!

>> No.3887767

>>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.

>> No.3887768

>>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.

>> No.3887791

>>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.

>> No.3887792

>>3887731
The problem is that he implies that the pressure is constant from one end to another on both containers.
It's not.

>> No.3887793
File: 14 KB, 732x378, derp.png [View same] [iqdb] [saucenao] [google]
3887793

pressure distribution on the rock

>> No.3887807

>>3887793
No, the pressure distribution is exactly the same. Might be a good idea to check out that lecture posted earlier.

>> No.3887809

>>3887792
>>3887793
18:00 Watch for a few minutes...

>> No.3887812

>>3887792
>MIT lectures says im wrong on a video checked by countless physicists
>he must be wrong

>> No.3887821
File: 16 KB, 732x378, 1318380452943.png [View same] [iqdb] [saucenao] [google]
3887821

>>3887793
No.

>> No.3887824

>>3887768
you're really listening to those jews at MIT?

>> No.3887834

>>3887824
>bad at physics
>blame the jews

>> No.3887838

Sure is taking Rainbow Dash a long time to watch the relevant two minutes of that video....


=P

>> No.3887840
File: 762 B, 108x45, 208f9d14f34c6d28ef6486e22c792609.png [View same] [iqdb] [saucenao] [google]
3887840

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

>> No.3887846

If you had two tubes, one at each end, does that mean that the pressure would be double of that on the left example?

>> No.3887848

>>3887840
What? The original problem statement doesn't give enough information to determine the radius of the tube.

>> No.3887852
File: 31 KB, 1097x827, Untitled.png [View same] [iqdb] [saucenao] [google]
3887852

>>3887768


He didn't take into account capillary action. He was teaching classical physics. l2 apply it or don't learn it at all.

>> No.3887855

>>3887846
The water would fall out of the tube coming from the bottom end.

>> No.3887863

>>3887660
arent some of the assumptions
incompressible fluid
and reynolds <2200 flow

>> No.3887864

>>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.

>> No.3887860

>>3887848
Disregard that, I suck cocks.

>> No.3887862

>>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

>> No.3887865

>>3887848
it does, we know height and the volume.

Interesting point. so in this case capilary action is revelant.

>> No.3887868

archival?

>> No.3887869

>>3887855
I might be dumb or tires or both, but I can't make sense out of your comment.

>> No.3887872

>>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

>> No.3887875

>>3887869
Definitely tires.

He said one tube at each end.

A cylinder has two ends, a top and a bottom.

>> No.3887877

>>3887852
he didn't specify the dimensions in his case but the point stands the argument people were making was bollocks.

>> No.3887889

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).

>> No.3887892

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

>> No.3887905

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.

>> No.3887910
File: 18 KB, 552x648, 1318369942350.png [View same] [iqdb] [saucenao] [google]
3887910

>>3887875
Yeah, by that I meant one on the right and one on the left.
Soz for the poor drawing.

>> No.3887917

>>3887910
And that question I asking everyone who says the pressure on the OP question is the same in both examples.

>> No.3887919

>>3887910
Why would it matter? Twice as much water column weight, twice as much interface area, once as much pressure.

>> No.3887920

>>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?

>> No.3887921

>>3887910
*Ignoring cappilary action* that scenario gives the exact same pressure as the other two.

>> No.3887924

>>3887905

They don't have the same pressure.

>> No.3887931
File: 31 KB, 1174x648, sasasas.png [View same] [iqdb] [saucenao] [google]
3887931

>>3887892
what about this?

>> No.3887932

>>3887920
Pressure is a continuum quantity, it doesn't apply on that scale (same as density, temperature...)

>> No.3887933

>>3887920
Our assumptions only hold in completely filled, sealed containers.

These kinds of things break down at the atomic scale.

>> No.3887944

>>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.

>> No.3887946

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)

>> No.3887947

>>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

>> No.3887985
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3887985