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Pressure and transmission of pressure in liquids

by sgstudent
Tags: liquids, pressure, transmission
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sgstudent
#1
Apr13-12, 12:21 PM
P: 645
I understand that pressure is F/A. But whenever they ask me to find the pressure of a book acting on the table, since the object in question is actually the table so is the force the reaction force on the table by the book and not actually weight since weight is an internal force?

Again, with the hydraulic system information in my textbook a small force acting on a small piston gives rise to a large force in the large piston as the pressure is the same. But I'm quite confused as they say that a forfeit is exerted on the piston. I thought that the object that we have to take into perspective is the liquid? Also since my force is acting the piston that force's area has to be the same as the piston's or else the pressure won't be the same since P=F/A. I'm not very sure about this since all they say is the force on the piston so I don't quite understand it in detail.

Lastly, pressure in a liquid acts at all directions. What does this mean? Does it mean that when I use P=hpg to get the pressure in the liquid (taking that there is no atmospheric pressure above) that pressure acts at all directions? But won't that mean that the force is infinite since the pressure acts at all direction so P=F/A is infinite as at one point upwards pressure is same as downwards.

Thanks guys for the help! Hope to hear (or read) from you guys soon!
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tiny-tim
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Apr13-12, 02:02 PM
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hi sgstudent!
Quote Quote by sgstudent View Post
I understand that pressure is F/A. But whenever they ask me to find the pressure of a book acting on the table, since the object in question is actually the table so is the force the reaction force on the table by the book and not actually weight since weight is an internal force?
the pressure of one body on another is always reaction force / area

(btw, weight is an external force it comes from the Earth)
Again, with the hydraulic system information in my textbook a small force acting on a small piston gives rise to a large force in the large piston as the pressure is the same. But I'm quite confused as they say that a forfeit is exerted on the piston. I thought that the object that we have to take into perspective is the liquid? Also since my force is acting the piston that force's area has to be the same as the piston's or else the pressure won't be the same since P=F/A. I'm not very sure about this since all they say is the force on the piston so I don't quite understand it in detail.
i don't understand your question

the smaller piston moves a longer distance, so it can do the same work with less force

so it's like a lever, where again the forces are inversely proportional to the distance moved
But won't that mean that the force is infinite since the pressure acts at all direction so P=F/A is infinite as at one point upwards pressure is same as downwards.
sorry, i don't understand
256bits
#3
Apr13-12, 02:38 PM
P: 1,425
Lastly, pressure in a liquid acts at all directions. What does this mean? Does it mean that when I use P=hpg to get the pressure in the liquid (taking that there is no atmospheric pressure above) that pressure acts at all directions? But won't that mean that the force is infinite since the pressure acts at all direction so P=F/A is infinite as at one point upwards pressure is same as downwards.
If you calculate or measure the pressure at a depth h, yes the pressure will be the same in all directions. If you can imagine lowering a very small cube down to the depth h then the pressure on all faces of this small cube is the same ie front, back, left, right, top and bottom. Since the force on all faces of the cube would be also the same. No matter which way you orient the small cube the pressure will be the same on all of its faces - there is no preferred direction as far as pressure is concerned.

I am puzzled how you managed introduce infinity.

sgstudent
#4
Apr13-12, 10:11 PM
P: 645
Pressure and transmission of pressure in liquids

Quote Quote by tiny-tim View Post
hi sgstudent!


the pressure of one body on another is always reaction force / area

(btw, weight is an external force it comes from the Earth)


i don't understand your question

the smaller piston moves a longer distance, so it can do the same work with less force

so it's like a lever, where again the forces are inversely proportional to the distance moved


sorry, i don't understand
Hi tiny Tim the part I'm confused about is that since we have to use reaction force/area then what does it mean by we apply a force onto a piston? Since the force that we exert is smaller than the surface area of the piston so I don't understand how can we calculate the pressure acting on the piston. Also, I thought the object that we should focus on is the water so why isn't it the force acting on the water?

Also for the other part, I meant that in a single point underwater pressure is outwards in all direction at that point right? So at the centre of it, from the left right up down the pressure is the same right? So won't the force be infinite?

Thanks for the help guys!
haruspex
#5
Apr14-12, 12:57 AM
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Quote Quote by sgstudent View Post
Since the force that we exert is smaller than the surface area of the piston
Thanks for the help guys!
How can a force be less than an area?
Perhaps you mean the force you exert on the first piston is exerted over a smaller area than the area of the piston against the fluid? Irrelevant.
You somehow exert a force F1 on the first piston.
That piston spreads the force over an area A of fluid, generating pressure P = F1/A.
The pressure is transferred throughout the fluid.
At the second piston the pressure acts over area B, generating force F2 = P*B = F1*B/A.

Regarding pressure at depth, consider a small cube, area C on each face.
Force on each face F = C*P.
As C tends to zero, F tends to 0, not infinity.
tiny-tim
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Apr14-12, 03:33 AM
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hi sgstudent!
Quote Quote by sgstudent View Post
Hi tiny Tim the part I'm confused about is that since we have to use reaction force/area then what does it mean by we apply a force onto a piston? Since the force that we exert is smaller than the surface area of the piston
as haruspex says, how can a force be less than an area?

the force on "our" end of the piston (where we're pushing it) is equal and opposite to the force on the water end of the piston

the pressure is that force divided by the area of the piston
so I don't understand how can we calculate the pressure acting on the piston. Also, I thought the object that we should focus on is the water so why isn't it the force acting on the water?
the force on "our" end of the piston is the same as the force on the water
Also for the other part, I meant that in a single point underwater pressure is outwards in all direction at that point right? So at the centre of it, from the left right up down the pressure is the same right? So won't the force be infinite?
if you have a sphere, the force will be the same all over, and will be proportional to the surface area of the sphere

as you reduce the sphere to a point, the force reduces to zero (because the area reduces to zero, and the force is proportional to the area)

btw, it you consider the force of the water on just one half of the sphere, that's calculated in exactly the same way as the force of your body on the spherical top of your femur

(http://www.arthursclipart.org/skelet...ip%20joint.gif)
sgstudent
#7
Apr14-12, 07:43 AM
P: 645
Oh, so the force acting on the piston is equal to the force acting on the water? Why is this so, since if I have a 10N force on the water there will be a reaction force of 10N on the piston so won't I have to apply 20N of force rather than 10N so I'm confused about this part.

Thanks for the help guys!

BTW tiny Tim, you seem to be able to pick out the bits that I'm confused about really quickly, that's amazing! Thanks for the help!
tiny-tim
#8
Apr14-12, 07:57 AM
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Quote Quote by sgstudent View Post
Oh, so the force acting on the piston is equal to the force acting on the water? Why is this so, since if I have a 10N force on the water there will be a reaction force of 10N on the piston so won't I have to apply 20N of force rather than 10N
are we talking about a particular device?

usually there's two pistons, we push one, and the other pushes a wheel etc

if we push the first piston with 10 N, the water pushes back on the first piston with 10 N

at the other end, the same water pushes on the second piston with 20 N, and the water pushes back on the second piston with 20 N

(so it's geared: we push the first piston, and the second piston moves half as far with twice the force)
sgstudent
#9
Apr14-12, 08:13 AM
P: 645
Huh, I don't understand the 'I push with 10N so the water push back with 10N part.' Why would the water push back with an equal but opposite force? Thanks for the help!
tiny-tim
#10
Apr14-12, 08:32 AM
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Quote Quote by sgstudent View Post
Why would the water push back with an equal but opposite force?
newton's third law?
sgstudent
#11
Apr14-12, 09:07 AM
P: 645
But if I push the piston down won't the reaction force be on my hand? I think I'm quite confused here the image I'm picturing looks sorta like this: http://science.howstuffworks.com/tra...hydraulic1.htm so I don't quite get the action reaction pair here..
tiny-tim
#12
Apr14-12, 09:37 AM
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Quote Quote by sgstudent View Post
But if I push the piston down won't the reaction force be on my hand?
to make things easy, let's assume that you put your whole weight on the piston

so the force from you on the top of the piston is mg

and the force from the piston on you is also mg

the force from the bottom of the piston on the water is mg

and the force from the water on the piston is mg
everything is mg (until you get to the other side of the water, where the diameter is different)
sgstudent
#13
Apr14-12, 10:56 AM
P: 645
Oh so the piston has no net force? Actually are you able to deduce these stuff just like this because previously in most dynamics question we treat two bodies differently like when I push a box across a table my hand is one object and the book as well so if I push 10N on the book there will be a reaction force of 10N so just from this information I don't really know about the other forces acting on my hand to balance the reaction force which is opposing me. So I'm pretty confused about this part here. Thanks for the help tiny tim!
tiny-tim
#14
Apr14-12, 11:04 AM
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Quote Quote by sgstudent View Post
Oh so the piston has no net force?
if the piston is moving steadily, its acceleration is zero, so the net force must be zero
Actually are you able to deduce these stuff just like this because previously in most dynamics question we treat two bodies differently like when I push a box across a table my hand is one object and the book as well so if I push 10N on the book there will be a reaction force of 10N so just from this information I don't really know about the other forces acting on my hand to balance the reaction force which is opposing me. So I'm pretty confused about this part here.
i'm not following you

there's a 10 N force all the way along, finishing where you, or something, is pushing against the Earth
sgstudent
#15
Apr14-12, 12:20 PM
P: 645
Oh sorry for being vague. I mean like when I push a box with 10N, my hand experiences a 10N force too which is the N3L reaction force. But I'm not sure if the force I apply to my hand will be 10N or greater than 10N. So that's why I'm confused about why I can assume all the forces in the piston and the water like the weight example you gave just now. Thanks for the help!
tiny-tim
#16
Apr14-12, 12:49 PM
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Quote Quote by sgstudent View Post
I'm not sure if the force I apply to my hand will be 10N or greater than 10N.
the horizontal component of the force you apply to your hand will be 10 N

(this is because your hand is not accelerating, so if there's 10N on one side, there must be 10 N on the other side, net horizontal force 0)
sgstudent
#17
Apr14-12, 01:16 PM
P: 645
But how do I know if my hand is accelerating since I might apply even more force to counteract the reaction force such that my hand will accelerate at the same rate as the object in question. So I'm guessing that there is no one answer for this and it depends on the person? But then if this is so then how can the pressure in the piston like the wright example you gave be justified? Thanks for the help tiny tim!
tiny-tim
#18
Apr14-12, 02:08 PM
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Quote Quote by sgstudent View Post
But how do I know if my hand is accelerating since I might apply even more force to counteract the reaction force such that my hand will accelerate at the same rate as the object in question.
your hand will always have the same acceleration as the object it is in contact with

(as i said, the forces on either side of your hand will only be equal if there is no accceleration)
So I'm guessing that there is no one answer for this and it depends on the person? But then if this is so then how can the pressure in the piston like the wright example you gave be justified?
the piston will on average have zero acceleration

(and in a water system like this, the acceleration would be very small anyway)


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