Make a 100000 cubic km barometer so we collected energy from atmospheric pressure changing forever 🤧🙄🤔👽🙏💦💨💤💀👽😇🙏🙏🙏🙏🙏 Show calculation if we make a dam size barometer ?
Physics 3a- I really understood this problem, however, at the end I think it should be delta H =h1 -h2. If we do h2-h1, we get -13.9 which would be incorrect.
Hello Respected sir sorry i didn't understand your solution so I decided to put whole question and the 2 methods in replies of this comment , The question goes "Two communicating cylindrical tubes contain mercury . The diameter of one vessel is four times larger than the diameter of the other . A column of water of height 70 cm is poured into the narrow vessel . How much will the mercury level rise in the other vessel and how much will it sink in the narrow one ?"
Ok so method 1(Solution book method) let "x"= rise of hg in broad tube and "y"= lowering of hg in narrow tube So obviously 16x=y becz same volume liquid up and down go Ok so they measured pressure from water's base so (1)g(70)=(13.6)g(x+y) And now they solve by putting 16x= y and get answer x=0.3cm y=4.8cm
Now method which i tried: I did same thing 16x=y But I tried to equalise pressure from from the level where Hg initially was . So therefore (1)g(70-y)=(13.6)g(x). 70-y because y was decrease in narrow tube And as you can see equation different from bookish method why
You first work out the problem exactly the same as in this video. You find that the the height in the Hg column is (70 cm) (1/13.6) So you find the height difference from subtracting the 2 heights like in the video. Now the volume drop of Hg in the narrow tube equals the volume rise of Hg in the wide tube. So the height drop in the narrow tube is 16 times as much as the height rise in the wide tube. Call them H1 and H2 Then |H1| + |H2| equals the column height of the Hg you found in the first part.
Do we not consider the Atmospheric pressure here because it is applied downwards both at the left and right of the U-tube? Let's assume we closed the lid on the left side, so there would be vacuum above the water, is it then safe to say that the atmospheric downward pressure on the right must be equal the pressure the water and the mercury apply in the opposite direction?
You are correct on both accounts. We don't take into account the atmospheric pressure in this example because it is the same on both sides of the tube.
dont know if anyone will answer this question but need help on it One limb if a u-tube manometer is connected to a gas supply, the other limb being open to the atmosphere. The effect of connecting the gas supply is to depress the level of the water in that limb to 160mm below the the level in the limb open to the atmosphere. By how much does the the gas pressure differ from the atmospheric pressure and is the gas pressure greater or less than the atmospheric pressure. would appreciate it if you can make a video on this, but would understand if you won't. But someone please answer this q
One end of a U-tube that contains water is open to the atmosphere, but the other end is closed. When some air is trapped in the closed end, the difference in the height of the water, h, shown in the figure, is 14.0 cm. Assume the outside air pressure is 0.960 atm. (a) What is the absolute pressure of the trapped air? (b) What is the gauge pressure of the trapped air? How do i do this problem?
sir please answer this question. it has a diagram with it which looks like UΠU (the middle tube is smaller in height). Apparatus shown in the figure consists of two glass columns connected by horizontal sections the height of two Central columns B and C are 49 centimetre each. the two outer columns A and D are open to the atmosphere A and C are maintained at temperature of 95 degree Celsius while the coloumns B and D are maintained at 5 degree Celsius. the height of the liquid in A and D as measured from the baseline are 52.8 centimetres and 51 centimetre respectively determine the linear coefficient of thermal expansion of the liquid. I want to know at what pair of points can we equalise the pressure and why . I don't need the whole solution . (first vertical column is A then B then C then D.)
If we close the lid on one side of the U-tube, would there still be atmospheric pressure above the liquid on that side? I.e would it still be safe to ignore the atmospheric pressure and just work with gauge? Or would we have to change the way we solve the problem?
If we place a lid on one side, and then pour in the water on the other side, we would have to take into account the the pressure would increase as the gas is compressed.
Mr. Professor there is a Donnan equilibrium where water move against the force of gravity. Would You be so kind and explain the basis of this phenomenon?
I know it's not needed in this particular problem. But it would be a good idea to mention the atmospheric presure in case somebody wants to solve a problem with one side closed.
i did a similar question, In a U-tube containing mercury, 150 mm of water was added, find the increase in height of mercury from its original position, the increase in height is 5.5 mm. using the method shown in video, the answer found is 1.10 cm, what they did was divide 1.10 cm by 2 to get the answer. this is very confusing. i can understand up until 1.10 cm. but i dont understand the logic of dividing by 2.
Draw 3 diagrams. The first with just the mercury in the tube. The second with water added on one side, but leave the mercury unchanged. The third diagram the mercury will then drop down 0.55 cm on the side that contains the water and will rise 0.55 cm on the othe side from their original positions. Then the difference in the height of the mercury is 1.1 cm, but on the one side the mercury rose 0.55 cm and on the other side it dropeed 0.55 cm.
@@MichelvanBiezen Thank you Sir, to add on if i compare it to this question in this video, 1.1 cm calculated represents this same difference mentioned in the comment as well?
A pressure gauge simply measures the gauge pressure (such as the pressure inside the tires). A manometer is a more specialized pressure gauge and is typically used for calibration of for measuring the pressure difference in a tube. For example manometers are used on airplanes to measure wind speed (speed relative to the air). A u-tube can be used for different types of pressure gauges, including manometers.
You don't want to become like me. (just ask my wife). I am just a simple person struggling through life like everyone else. But I do feel blessed that I have the opportunity to share the amazing world of science with everyone else.
Pressure Gauges used in Hydraulic Machines have got a small hole of 1 mm dia, How does the gauge shows the correct value of pressure in system, there should be a pressure drop as it is also similar to orifice?
I would need to see a schematic of the design in order to answer that question. (there are many different design of pressure gauges). Most pressure gauges measure the steady state pressure, after no additional gas enters the pressure chamber and thus the size of the hole doesn't matter.
Michel van Biezen Pressure gauge in my case is an Hydrualic pressure gauge which can measure oil pressure from 0 to 400 bar . It is similar to Pressure gauges used in hydrualic presses to measure the Main pressure and clamping pressure
I feel so dumb because I don't understand whats going on. In my science book, it gives us a diagram like yours however one end is attached with a vacuum pump and inside is a liquid. its saying that the vacuum pump is sucking out the air on the right side were the vacuum pump is, therefore the liquid is rising on the right side and on the left side it is going down. Then it proceeds to say that "Now the pressure on the outside (on the left) is compared to zero pressure (vacuum) and I cannot understand what that means.I cannot understand this with videos, would you please explain?
If both sides of the tube are exposed to the air, then you have atmospheric pressure pushing down on both sides and therefore they cancel out. If on one side (like in your example) the pressure is reduced to zero (vacuum), then you must add atmospheric pressure on one side which caused the liquid on that side to drop and on the other side to rise.
Sir why don't u take raised height after pouring water as 2h...because on the left limb when water comes down by h then in right limb water will goes up by h..hence from p¹ it will be 2h...plz reply
Good way to visualize is this: 1.) Remove the water. What do you get? Top of Hg on both sides at same level. Right. That's equilibrium. Obviously. 2.) Now pour water in one side. What happens? It goes down and other side goes up. Right? That's "still" equilibrium. Not so obvious. What's different? A stack of water, and a stack of Hg with a reference level. You can equate them and set up an equation. Enjoy...For non-engineers/scientists think of this like a teeter totter effect. I don't believe everyone viewing this video is strictly a college engr student, so taking a few steps back the basic concept of Equilibrium or Statics condition should be reviewed or pondered. This must be solidly/instinctively understood. Then the concept of liquid being able to take any shape and it's ability to support a load can be further understood. Hope this doesn't scare away the curious, but rather inspire them to search/watch more videos. Keep on.
That is why in the tube on the left, the water height above the point is 13.6 times higher than the mercury above the same point in the tube on the right. (Pressure is due to the height of the liquid above the point)
Thanks. I am not quite sure if I grasp the concept, however to do this exercise I try to imagine just the "U" formed by the mercury (excluding the water and the mercury that has gone up due to the pressure of water) after that I imagine how much of water and mercury do I have to pour in every part in order to remain balanced, as mercury is more dense than water I will need 13.6 more water.
Hii sir please answer my doubt I had a similar question in my book that 70 cm of water is filled in a container but i tried equalising pressure from the level where mercury initially was (before add water) and assumed that it went x distance up on other side Then I equated pressure but on left side at the level where first there was mercury came water so I equated pressure of mercury and water at level where mercury initially was but not getting answer So wanted to ask can we equate pressure of 2 different liquid at same height
If the pressure was greater at the left point, the pressure would push fluid higher on the right side, and if the pressure was greater at the right point it would push fluid higher into the left side.
+Archer June Diaz Then they have to give you other information (like the cross sectional area of the tube) or something that will allow you to calculate the height. Pressure depends on the height of the fluid column.
Sir would you know how to set this up? Even my teacher is confused. The answer we calculated us 925N, but is not correct. I tried this video since it seemed a little relevant. A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil with a density of 800 kg/m3 and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m? For purposes of this problem, neglect the weight of the pistons. Can someone explain step by step. I know Force just to keep it in place is 925.93N. I used F/A=F2/A2 Then I suppose we add this to the second calculation, but I am not sure what formula or a way to do this. I know FD=F2D2 relationship.
Green Bear The answer is 1000N Use Pascal's principle where P1 = P2 (pressure) P = F/A Thus (F1/A1) + rho*g*h = (F2/A2) (you have to take the pressure due to the liquid at a height of 1.2 m. (12,000N / (pi*0.18^2)) + (800)*(9.8)*(1.2) = (F/(pi*0.05^2))
The answer, Sir was 1.96kN. Once I found the mass of the fluid alone using the height it weight in newton force 957.62N. Then I used pascals principle but this time with the 957.62N +12000N/r of big piston = F(unknown)/r of small piston. I got 1005.55N then I added 957.62N to this and it comes out to exactly 1963.17N = 1.96kN. I am just unsure why weight was added later again. Very confusing as pascals principle is very simple and basic ration proportion to gain mechanical advantage.
Sir, I just thought about it. I think the reason we use pascals principle with the added weight force of the density of oil at 1.2m is because that force is for that oil separate and then the total force gave us the 1005N which is force needed just to keep it all in one place. So this was a two step problem. I think it kind of makes sense, but on a test there is no way I would randomly figure this out.
I can't get to the answer of this problem.... Please help: "A mercury filled U-tube manometer is used to measure the level in an open water tank. The manometer zero level is in line with the bottom of the tank. Calculate the level in the tank if the mercury is 0,5 m below the zero level" The answer is [13,1 m]
Darren Young Darren, If the mercury drops 0.5 m below the zero level on one side, it will be 0.5 m above the zero level on the other side and thus the elevation level difference will be 1.0 m Since the density of mercury is 13.6 times the density of water, this translates into an equivalent water column of 13.6m. And then the zero level of the mercury is in line with the bottom of the tank, thus when the mercury column drops 0.5 m, it will be 0.5 m below the water level of the tank and that 0.5 m of water below the tank also pushes against the mercury. Thus the water level in the tank is 13.6 m - 0.5 m = 13.1 m
If they were not equal the fluid would not be static, it would move until the pressure is the same. (If P1 was greater the fluid on the left side would push the fluid on the right side (P2) higher)
Ali,
You are correct, (I reversed h1 and h2)
Thanks
Although on the Right and Left Sides both having same liquid Hg, Why the pressures are not equal to each other where the underneath points 1 and 2 ?
CanYou Pls Make A Video explaining of How To Find Rise (Mercury) in The Other Limb of Tube From Initial. (USING SAME EXAMPLE).
Make a 100000 cubic km barometer so we collected energy from atmospheric pressure changing forever 🤧🙄🤔👽🙏💦💨💤💀👽😇🙏🙏🙏🙏🙏
Show calculation if we make a dam size barometer ?
A Titanic size barometer ?
How much energy it gives to turn the turbine
After 8 years, We are still learning from your great way of explaining!
Tank you so much.
Glad to hear that!
and he still replies to your comments too!
Thanks a lot for these great videos!
just 1 small note, Delta h = h1-h2, I believe..
Beautiful how you take things the easier way! Thank you so much
wow!! you explained this so much better than my fluid mechanics professor!! You have helped me tremendously in my engineering studies. Thank you!!
You are welcome. Glad you found our videos. 🙂
very helpful.. a great way to revise before an exam too !
ya very helpful
Thank you so much! This is one of the topics i didn't really understand by by physics lectures. Greetings from italy!
Welcome to the channel!
(A long time ago, I rode my bicycle through parts of Italy and also worked as a ski instructor in La Thuille). Italy is a beautiful country.
@@MichelvanBiezen that's so cool sir!
Thank you for teaching!! Your lessons made me understand so much!!
I want your class 11 gaseous state lectures for jee mains + adv
You are a great explainer 👍👍
We do have a large number of the state of gases lectures and thermodynamics on this channel. Thank you. 🙂
ive been looking everywhere for something like this thank u
Glad you found us. Welcome to the channel! There are over 9000 videos on this channel.
Thank u very much sir
Finally your video solved my problem 😊😊
Glad the video was helpful. 🙂
Physics 3a- I really understood this problem, however, at the end I think it should be delta H =h1 -h2. If we do h2-h1, we get -13.9 which would be incorrect.
Yes, I do have them reversed. (Should have assigned y2 to the bigger value)
Simple problem, yourself can correct it
Super teaching sir do more
PHYSICS IS A MOTHER OF ALL SCIENCES
Fluid statics is a lovely subject.
I agree, but then I like all topics in physics. 🙂
best explanation
Glad you found it helpful. 🙂
Great teacher
Thank you! 😃
Much better than my teacher
Thank you!
I wish to have you as a lecture. Thanks for your help.
You are most welcome
Hello Respected sir sorry i didn't understand your solution so I decided to put whole question and the 2 methods in replies of this comment , The question goes "Two communicating cylindrical tubes contain mercury . The diameter of one vessel is four times larger than the diameter of the other . A column of water of height 70 cm is poured into the narrow vessel . How much will the mercury level rise in the other vessel and how much will it sink in the narrow one ?"
Ok so method 1(Solution book method) let "x"= rise of hg in broad tube and "y"= lowering of hg in narrow tube
So obviously 16x=y becz same volume liquid up and down go
Ok so they measured pressure from water's base so (1)g(70)=(13.6)g(x+y) And now they solve by putting 16x= y and get answer x=0.3cm y=4.8cm
Now method which i tried:
I did same thing 16x=y
But I tried to equalise pressure from from the level where Hg initially was .
So therefore (1)g(70-y)=(13.6)g(x).
70-y because y was decrease in narrow tube
And as you can see equation different from bookish method why
Thanks for your precious time if you do reply :D
You first work out the problem exactly the same as in this video. You find that the the height in the Hg column is (70 cm) (1/13.6) So you find the height difference from subtracting the 2 heights like in the video. Now the volume drop of Hg in the narrow tube equals the volume rise of Hg in the wide tube. So the height drop in the narrow tube is 16 times as much as the height rise in the wide tube. Call them H1 and H2 Then |H1| + |H2| equals the column height of the Hg you found in the first part.
Do we not consider the Atmospheric pressure here because it is applied downwards both at the left and right of the U-tube?
Let's assume we closed the lid on the left side, so there would be vacuum above the water, is it then safe to say that the atmospheric downward pressure on the right must be equal the pressure the water and the mercury apply in the opposite direction?
You are correct on both accounts. We don't take into account the atmospheric pressure in this example because it is the same on both sides of the tube.
dont know if anyone will answer this question but need help on it
One limb if a u-tube manometer is connected to a gas supply, the other limb being open to the atmosphere. The effect of connecting the gas supply is to depress the level of the water in that limb to 160mm below the the level in the limb open to the atmosphere. By how much does the the gas pressure differ from the atmospheric pressure and is the gas pressure greater or less than the atmospheric pressure.
would appreciate it if you can make a video on this, but would understand if you won't. But someone please answer this q
Excellent indeed! Best lecturer ever :)
Just a small error, i think you meant to write delta h = h1-h2 at the end.
Good catch you are correct.
Michel van Biezen he is???
Great lec, Prof !!
Greetings... A question please. Why is the pressure a standard or quantitative amount?
Not sure what you mean by "a standard or quantitative amount". Pressure by definition = F/A (force / area)
「あなたの動画はとても良いですし、メッセージがた
Thank you.
good teaching
One end of a U-tube that contains water is open to the atmosphere, but the other end
is closed. When some air is trapped in the closed end, the difference in the height of the water, h,
shown in the figure, is 14.0 cm. Assume the outside air pressure is 0.960 atm.
(a) What is the absolute pressure of the trapped air?
(b) What is the gauge pressure of the trapped air?
How do i do this problem?
At which end is the water level higher? At the closed end or at the open end? The difference in pressure will be (density) (g) (difference in height)
really helpful!
sir please answer this question. it has a diagram with it which looks like UΠU (the middle tube is smaller in height). Apparatus shown in the figure consists of two glass columns connected by horizontal sections the height of two Central columns B and C are 49 centimetre each. the two outer columns A and D are open to the atmosphere A and C are maintained at temperature of 95 degree Celsius while the coloumns B and D are maintained at 5 degree Celsius. the height of the liquid in A and D as measured from the baseline are 52.8 centimetres and 51 centimetre respectively determine the linear coefficient of thermal expansion of the liquid. I want to know at what pair of points can we equalise the pressure and why . I don't need the whole solution . (first vertical column is A then B then C then D.)
If we close the lid on one side of the U-tube, would there still be atmospheric pressure above the liquid on that side?
I.e would it still be safe to ignore the atmospheric pressure and just work with gauge? Or would we have to change the way we solve the problem?
If we place a lid on one side, and then pour in the water on the other side, we would have to take into account the the pressure would increase as the gas is compressed.
Mr. Professor there is a Donnan equilibrium where water move against the force of gravity. Would You be so kind and explain the basis of this phenomenon?
It is related to the pressure created by the existence of ions and a semipermeable membrane. A bit complicated to describe in a comment.
@@MichelvanBiezen Thank You.
Great explanation 👌👍,thanks ♡♡
Glad it helped.
thanks sir❤🙏
Most welcome. Glad it was helpful.
Could you please do a vedeo on determining the density of oil using hares experiment ?
That is an interesting device and experiment. We added it to the list. Thank you.
@@MichelvanBiezen Oh really ? Could you please tag the vedeo here . I'm unable to find it
Thank youuuu very. Match you really help mee
I know it's not needed in this particular problem. But it would be a good idea to mention the atmospheric presure in case somebody wants to solve a problem with one side closed.
Sir, are you assuming there's no atmospheric pressure? or that's how it actually works (vacuum at the top of the right side)?
Since there is atmospheric pressure on both sides, we can ignore it, (or just calculate the gauge pressure on both sides.
Thank you so much! It's so interesting
i did a similar question,
In a U-tube containing mercury, 150 mm of water was added, find the increase in height of mercury from its original position, the increase in height is 5.5 mm.
using the method shown in video, the answer found is 1.10 cm, what they did was divide 1.10 cm by 2 to get the answer. this is very confusing. i can understand up until 1.10 cm. but i dont understand the logic of dividing by 2.
Draw 3 diagrams. The first with just the mercury in the tube. The second with water added on one side, but leave the mercury unchanged. The third diagram the mercury will then drop down 0.55 cm on the side that contains the water and will rise 0.55 cm on the othe side from their original positions. Then the difference in the height of the mercury is 1.1 cm, but on the one side the mercury rose 0.55 cm and on the other side it dropeed 0.55 cm.
@@MichelvanBiezen Thank you Sir,
to add on if i compare it to this question in this video, 1.1 cm calculated represents this same difference mentioned in the comment as well?
Great teaching, though it can be improved by showing an experiment under the same conditions.
Also, that looks like ΔR, not Δh.
Thanks for the info!
Amazing videos!
Love you sir......
Thank you. Glad you like our videos. 🙂
What's the difference between Pressure Gauge, Manometer and U tube?
A pressure gauge simply measures the gauge pressure (such as the pressure inside the tires). A manometer is a more specialized pressure gauge and is typically used for calibration of for measuring the pressure difference in a tube. For example manometers are used on airplanes to measure wind speed (speed relative to the air). A u-tube can be used for different types of pressure gauges, including manometers.
Excellent lesson, thank you!
are we neglecting ATM pressure?
Since there is atmospheric pressure on both sides, it cancels out.
@@MichelvanBiezen but its acting on different heights so its not the same on both sides?
I love you???????? So MUCH??????? you make everything so simple!!!!!!!!!!!!! how can I become like you
You don't want to become like me. (just ask my wife). I am just a simple person struggling through life like everyone else. But I do feel blessed that I have the opportunity to share the amazing world of science with everyone else.
thank you so much sir..
Most welcome
Pressure Gauges used in Hydraulic Machines have got a small hole of 1 mm dia, How does the gauge shows the correct value of pressure in system, there should be a pressure drop as it is also similar to orifice?
I would need to see a schematic of the design in order to answer that question. (there are many different design of pressure gauges). Most pressure gauges measure the steady state pressure, after no additional gas enters the pressure chamber and thus the size of the hole doesn't matter.
Michel van Biezen Pressure gauge in my case is an Hydrualic pressure gauge which can measure oil pressure from 0 to 400 bar . It is similar to Pressure gauges used in hydrualic presses to measure the Main pressure and clamping pressure
Thanks
Thax
You are welcome.
such great videos sir,thank you!..... actually i watched all of them on fluid static and dynamics..
Wow! That will keep you busy for a while.
Me too. It helps me with my FE exam preparation
I feel so dumb because I don't understand whats going on. In my science book, it gives us a diagram like yours however one end is attached with a vacuum pump and inside is a liquid. its saying that the vacuum pump is sucking out the air on the right side were the vacuum pump is, therefore the liquid is rising on the right side and on the left side it is going down. Then it proceeds to say that "Now the pressure on the outside (on the left) is compared to zero pressure (vacuum) and I cannot understand what that means.I cannot understand this with videos, would you please explain?
If both sides of the tube are exposed to the air, then you have atmospheric pressure pushing down on both sides and therefore they cancel out. If on one side (like in your example) the pressure is reduced to zero (vacuum), then you must add atmospheric pressure on one side which caused the liquid on that side to drop and on the other side to rise.
@@MichelvanBiezen thank you! this really helped for my exam yesterday!
You didn't change the unit?
cm to m
We didn't have to since the units cancelled out.
What if one tube's radius was the smaller than the other one?
Since pressure only depends on density and height, it would not make any difference.
Thank you!!
Why is it that the atmospheric pressure is not included in this ?
because it is the same on both sides, so it cancels out.
Thank you so much 🙏🏾
great sir
You're Welcome!
Sir why don't u take raised height after pouring water as 2h...because on the left limb when water comes down by h then in right limb water will goes up by h..hence from p¹ it will be 2h...plz reply
That is not the case if the density of the 2 liquids is not the same.
Thank yOu
Thanks a lot
Happy to help
please help what if the two sectional areas of the tube are not the same size??
+Ahmed Said The cross sectional areas of the tube don't matter. P = density x height x g (only the height matters)
Sir, this came in jee advance this year
Can you please explain the reason foe p2 being =p1.
By Pascal's principle. The pressure inside a fluid must be the same everywhere if at the same level.
Good way to visualize is this: 1.) Remove the water. What do you get? Top of Hg on both sides at same level. Right. That's equilibrium. Obviously. 2.) Now pour water in one side. What happens? It goes down and other side goes up. Right? That's "still" equilibrium. Not so obvious. What's different? A stack of water, and a stack of Hg with a reference level. You can equate them and set up an equation. Enjoy...For non-engineers/scientists think of this like a teeter totter effect. I don't believe everyone viewing this video is strictly a college engr student, so taking a few steps back the basic concept of Equilibrium or Statics condition should be reviewed or pondered. This must be solidly/instinctively understood. Then the concept of liquid being able to take any shape and it's ability to support a load can be further understood. Hope this doesn't scare away the curious, but rather inspire them to search/watch more videos. Keep on.
Thank
you
sir . :)
Thank you
Doctor , why we didn't measure atmosphere pressure
There is difference in the height , so atmosphere pressure are not equal
The difference in the atmospheric pressure would be negligible.
Michel van Biezen thank you for your help
Why is it 1/13.6?
That is the ratio of the density of water to the density of mercury.
I cant understand why at two points at the same height the pressure is the same...Can anyone explain?
If you have a single bottle of wáter, 1 meter hight and a bottle of mercury of 1 meter high, at 0,5 m the wont weight the same...
That is why in the tube on the left, the water height above the point is 13.6 times higher than the mercury above the same point in the tube on the right. (Pressure is due to the height of the liquid above the point)
Thanks. I am not quite sure if I grasp the concept, however to do this exercise I try to imagine just the "U" formed by the mercury (excluding the water and the mercury that has gone up due to the pressure of water) after that I imagine how much of water and mercury do I have to pour in every part in order to remain balanced, as mercury is more dense than water I will need 13.6 more water.
Hii sir please answer my doubt
I had a similar question in my book that 70 cm of water is filled in a container but i tried equalising pressure from the level where mercury initially was (before add water) and assumed that it went x distance up on other side
Then I equated pressure but on left side at the level where first there was mercury came water so I equated pressure of mercury and water at level where mercury initially was but not getting answer
So wanted to ask can we equate pressure of 2 different liquid at same height
If they are in same "U-tube"
SInce pressure is equated as follows: P = density x g x h, if the density is not the same then the pressure cannot be the same for the same height.
I still don't understand why the pressure is equal at both points.
If the pressure was greater at the left point, the pressure would push fluid higher on the right side, and if the pressure was greater at the right point it would push fluid higher into the left side.
it is very simple problem
best the other
Thank you
how is it that Pressure at those 2 points is the same, when 1 has water on top & 2 has mercury..?
The column of water is 13.6 times as high since the density of mercury is 13.6 time as high as water
Where's the problem?
?
please help. What if we are given by mass not length. For example 100grams instead of 15 cm
+Archer June Diaz
Then they have to give you other information (like the cross sectional area of the tube) or something that will allow you to calculate the height.
Pressure depends on the height of the fluid column.
Sir would you know how to set this up? Even my teacher is confused. The answer we calculated us 925N, but is not correct. I tried this video since it seemed a little relevant.
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil with a density of 800 kg/m3 and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m? For purposes of this problem, neglect the weight of the pistons.
Can someone explain step by step.
I know Force just to keep it in place is 925.93N. I used F/A=F2/A2
Then I suppose we add this to the second calculation, but I am not sure what formula or a way to do this. I know FD=F2D2 relationship.
Green Bear
The answer is 1000N
Use Pascal's principle where P1 = P2 (pressure)
P = F/A
Thus
(F1/A1) + rho*g*h = (F2/A2) (you have to take the pressure due to the liquid at a height of 1.2 m.
(12,000N / (pi*0.18^2)) + (800)*(9.8)*(1.2) = (F/(pi*0.05^2))
The answer, Sir was 1.96kN. Once I found the mass of the fluid alone using the height it weight in newton force 957.62N. Then I used pascals principle but this time with the 957.62N +12000N/r of big piston = F(unknown)/r of small piston. I got 1005.55N then I added 957.62N to this and it comes out to exactly 1963.17N = 1.96kN. I am just unsure why weight was added later again. Very confusing as pascals principle is very simple and basic ration proportion to gain mechanical advantage.
Sir, I just thought about it. I think the reason we use pascals principle with the added weight force of the density of oil at 1.2m is because that force is for that oil separate and then the total force gave us the 1005N which is force needed just to keep it all in one place. So this was a two step problem. I think it kind of makes sense, but on a test there is no way I would randomly figure this out.
I can't get to the answer of this problem.... Please help:
"A mercury filled U-tube manometer is used to measure the level in an open water tank. The manometer zero level is in line with the bottom of the tank. Calculate the level in the tank if the mercury is 0,5 m below the zero level" The answer is [13,1 m]
Darren Young
Darren,
If the mercury drops 0.5 m below the zero level on one side, it will be 0.5 m above the zero level on the other side and thus the elevation level difference will be 1.0 m
Since the density of mercury is 13.6 times the density of water, this translates into an equivalent water column of 13.6m.
And then the zero level of the mercury is in line with the bottom of the tank, thus when the mercury column drops 0.5 m, it will be 0.5 m below the water level of the tank and that 0.5 m of water below the tank also pushes against the mercury.
Thus the water level in the tank is 13.6 m - 0.5 m = 13.1 m
Michel van Biezen Thank you very much sir for helping i fully understand now....
Thank You again!
nice
Thank you so much. Nice bowtie
thanks but h1=15 not h2 you see that
Yes, I should have placed absolute value signs around it. Thanks.
What is the gauge pressure?
Gauge pressure = total pressure - atmospheric pressure
American Teacher using the Metric System? What happened :D good video!
That's why we teach both.
Can someone please tell me why p1 equals to p2?? Pleaseee
If they were not equal the fluid would not be static, it would move until the pressure is the same. (If P1 was greater the fluid on the left side would push the fluid on the right side (P2) higher)
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