Different pressures.

>>3886872
are you a troll?

its the same pressure.

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

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

There's more pressure on the right right.

>>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.

They have the same pressure even at the sides.

>>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!

>>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.

>>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.

depends

>>3886948
Here is a hint for you.

Hydrostatic Paradoxon in da house, bitches.

>>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.

>>3886948
so you think this would work?

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

A fucking housewife could get this right you ape.

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.

>>3886992
*left

>>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?

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.

sup?

>>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.

>>3887025
Aw shit nigga,

>>3887006
>2 different objects
>same area?

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

>>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.

>>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)

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

p = rho * g * h

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

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

>>3887025

DUN DUN DUN

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

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

>>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.

>>3887061
You are confusing pressure with force.

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

>>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.

>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?

the this will happen!

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

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

>>3887088
add a turbine and free energy!

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

>>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.

>>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.

>>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

the /g/ thread is much more productive.

>>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.

P = gamma(h)

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

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

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

BRB PATENT OFFICE

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

>>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

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?

>>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

>>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.

>>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?

>>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.

>>3887061
That is what happens, though exaggerated.

Two words:
Capillary action

>>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.

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.

>>3887194
[x] Hydrotold!

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

>>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.

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

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.

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.

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

>>3887245
yes it will.

>>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.

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

>>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

>>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

>>3887201
Someone answer me.

>>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?

>>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.

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

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

>>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.