Pascal’s Law

Pascal’s Law

The picture that you see in front of you is a
picture of a very special instrument. This instrument is known as a hydraulic jack. Now what is the job of a hydraulic jack?
TA hydraulic jack is used to lift up heavy objects with the application of a very
little amount of force. This hydraulic jack is often used by car mechanics in
order to lift up cars and buses. Now you must be knowing that a car or even a
bus will be very very heavy, but you will find that these mechanics by applying a
minimum amount of force can lift up these cars and buses quite easily. So how do you think
this happens. How do you think that with the help of
these instruments which are known as hydraulic jacks a car or even a bus can
be lifted up very easily? Well, let us find out. We will now discuss that the working
principle behind hydraulic jacks is what enables the mechanics to lift cars or
buses in a very easy manner. So in order to do that first we are going
to learn about a very interesting concept. That interesting concept leads
us to another important conclusion. Now if you notice this video over here, you
will find that as more and more air is being pumped into the balloon, the
balloon is increasing or expanding in size and if you observe closely you will
find that the expansion in size of the balloon is taking place in every
direction because when gas molecules are coming in, they are exerting pressure on
the inside walls of the balloon and due to this pressure exerted by the gas
molecules, it is expanding equally in all directions. So from this we arrive at a very
important conclusion. But before we do that, let us consider another animation. Over here you will find that football is
being pumped with water, that is, water is being placed inside the football. Now once
water is filled, we are puncturing the football in six different places and from these six
different places we will find that water is trickling out or leaking with
the same amount of pressure if we put our fingers there. So now if we consider an instance of the
football where water was trickling out, if you place your finger at this point
or even at this point or let’s say at this point, you will find that the
pressure being applied on your finger by the water trickling out is the same. So
what conclusion does this lead to? Just like in the case of the gas
molecules exerting pressure on the inside of the balloon as well as the
water exerting pressure in order to come out from the football, the pressure that
was exerted by these fluids is the same and this leads us to a very important
law known as Pascal’s law. This Pascal’s law was first given by scientist Blaise
Pascal and this law has been named after him. What does it state? It
states that whenever any pressure is applied to any part of the boundary of a
confined fluid, it is transmitted equally in all
directions, irrespective on the area on which it is
acting and it always acts at right angles to the surface of the containing vessel.
So simplifying this I’ll tell you that, for example, the balloon when air was being
pumped inside the balloon. When air was being pumped inside the balloon
more and more gas molecules were entering the balloon and colliding with the inner
walls. These collisions were taking place at right angles. So we can say the force
being exerted by the gas molecules on the inside walls of the balloon was at
right angles. So as a result what happened? The balloon increased in size
and since this increase was uniform in all directions, we can say that the force
per unit area or the pressure exerted by the gas molecules was the same in every
direction and as you can clearly see this balloon is a confined container. So
thus Pascal’s law holds in that case. Even in the case of the football, a
football is an enclosed container in which the fluid, that is, water was
confined. So pressure applied by the fluid was transmitted
equally in every direction or in other words if we apply pressure to any part
in the fluid, it will be transmitted equally in all directions irrespective of the area on which it is
acting. So with the help of Pascal’s law let us find out how we can find out the
working principle of the hydraulic jack. Now first let us consider a scenario
where a water pumping station is pumping water to four houses A, B, C and D at a pressure
of 20 Pascal’s. Now the question is, this water pumping
station that is pumping water at 20 Pascal’s, this water is reaching each of
the homes, which home do you think receives water with the maximum amount
of water pressure? You will be surprised to know that the
answer is all these homes receive water at equal pressure, that is, the same
pressure with which the water pumping station is pumping the water, twenty
Pascal’s. Why, because according to Pascal’s law pressure applied to any
part of an enclosed fluid is transmitted equally in all directions. In the case of
the water pumping station the fluid was enclosed in the pipes that were
connecting the houses to the water pumping station. So since it was an
enclosed fluid, the pressure was transmitted equally in
every direction and this is why water reaches with equal pressure at all the
homes. So now we look at the working principle
of the hydraulic jack and clear your doubts on how a small force can lift up a heavy
object. So we consider a heavy object, which is a
typical small car. Now the weight of a car, it weighs around nine hundred Kgs, a typical
small car. Obviously if you consider a larger car, it will have a greater mass. Now
the weight of the car can be easily find out if you multiply mass with the
acceleration due to gravity. For ease of calculation, we consider
acceleration due to gravity 10 m/s square and the weight of the car is
thus 9000 Newton. Now it is common knowledge that gravity acts in the
downward direction. So the weight of the car will be a force which will be acting
in the downward direction. Now in order to lift up this particular car which has
a weight of 9,000 Newton acting in the downward direction, we must apply a force
in the upward direction and this force will have to be greater than nine thousand Newton. Now nine thousand Newton, as you can
see is a considerable amount of force. Now you might be thinking that do we have to
apply that amount of force in order to lift this up. The answer is no. So let us
find out how much force will be able to lift up a car weighing 900 kgs. So consider the diagram over here. This schematic
represents a hydraulic press. Now in a hydraulic press we have three parts. We
have two cylindrical tubes, tube P and tube Q and they have pistons
attached to them, piston one and piston two. These two tubes are
connected by a horizontal tube R and inside this tube there is a liquid. Now as you
can see, piston one has a smaller cross-sectional area and piston two has a
larger cross-sectional area. The reason behind which will be soon clear to you. So the working of the hydraulic press is
such that the force to be applied by us is applied on piston one and the object
to be lifted, that is, the heavy object is kept on piston two. We have seen that
piston one has a small cross-sectional and piston two has a large
cross-sectional area. So let us find out the reason for applying the force on
piston one and keeping the object on piston two. The assumption that we make is
both pistons have a square shape or in other words if we consider piston one,
then the length and breadth of piston one will be the same because it’s a
square and if we consider piston two then the length and breadth of piston two
will also be the same. However note that the length and breadth of piston one
will not be similar to the length and breadth of piston two. We have only
considered that each of these pistons is a square and the length
and breadth of the respective pistons are same. So let us consider that the length of
the side of piston one is one centimetres or 0.01 metres and length of side
piston two is equal to 10 centimetres or 0.1 metres. So how can we find out
the area of these pistons? Simply by squaring these quantities. So I square this quantity and I get
the area of cross-section or the area, surface area of piston one, that is, 0.0001 metre square. And if I square this quantity, I will get the area or the
surface area of piston two, which is nothing but 0.01 metre square. So as you can clearly see,
piston two has greater area as compared to piston one. So now I write the respective areas
A1 0.0001 metre square and A2 0.01 metre square beside the respective pistons or quick
reference. So now we’re going to a apply force on
piston one, as I told you earlier. Now we are applying a force of around hundred
Newton’s. You might be wondering how much of a force is hundred Newton’s or
how big is a force of hundred Newton. Will we be able to apply a force
of hundred Newton’s? Well, let us find out. Hundred Newton force is a force which is
roughly equal to the weight of ten thick books. So when we are applying a force of
hundred Newton’s on piston one, it is equivalent to placing ten thick books on
piston 1. So let us find out when we are applying a force of hundred Newton’s
what force we are getting on piston two. So the force on piston 1 is 100 Newton’s
and the area of cross-section of piston one is .0001 metre square? Now we have
studied that pressure is equal to force per unit area or
mathematically force divided by area. Thus I put the value of hundred Newton in place
of force and 0.0001 metre square in place of area and I get 10 lakh Newton per meter square 10 lakh Pascal. 10 lakh Pascal is the
pressure which is being applied on the liquid column just below piston one. Now according to Pascal’s law this
pressure which is applied to the fluid below piston one will be equally
transmitted in every direction. Why, because as you can clearly see from the
schematic of the hydraulic press, this fluid has been kept enclosed. It is a
confined fluid. So when I’m applying the pressure of 10 lakh Pascal below piston
one. This pressure will also be transmitted to piston two. Now the area of piston
two is 0.01metre square. So since pressure applied to any part of an enclosed fluid will be
equally transmitted this pressured be will be the same as this particular
pressure, that is, 10 lakh Pascal’s. So now we have that 10 lakh Pascal’s pressure is acting on
piston two which has an area 0.01 metre square .Now if I rearrange equation
for pressure. As pressure into area Equals force. Because force by area was pressure so I simply cross
multiplied area. So pressure into area will give us force we find that the pressure acting is 10 lakh
Pascal and the area of the piston is the 0.01 metre square so if I multiplied these two
quantities, that is, If I rearrange this equation and multiplied,
I will find that I’m getting a force of 10,000 Newton’s. So by applying a
force of hundred Newton’s on piston 1, I’m able to get an upward force of
10,000 Newton’s on piston two. Now, this force that I’m applying is in the
downward direction and the force on Piston two that I’m getting is in the
upward direction. So as a result what happens, the force on
piston two 10,000 newton’s is in the upper direction. Now, if we place the car that we had
considered which had a weight of 9,000 Newton’s. What do you think will happen?
Over here, the downward forces 9000 Newton’s and the upward force is 10,000 Newton’s.
Which one is greater? Obviously, the upward force 10,000 Newton’s and as a result, the resultant
force or the net force will be in the upward direction. Due to this, Since the net force is in the
upward direction with minimum effort, that is, it’s just a force of hundred Newton’s
we are able to lift the car which is 9000 Newton’s and this is how using a
Hydraulic press or hydraulic jack comes in very handy. So, we see that hundred Newton’s was
sufficient to lift a car which made nine hundred Kgs of very heavy object. So in
other words we can see that you place Ten thick books on piston 1 it will be
enough to lift the car of 900 kgs. No matter how mind-boggling and funny it
might seem this is actually fact and this fact is what Pascal’s law tells us So, what do you think will happen if we
increase the area on piston two? If we increase the area Piston two, keeping this area the same we
are going to get an even greater upward force. So if you place even a heavier
car that has a mass greater than 900 Kg or even if we keep a bus on
Piston two we will be able to lift it with just hundred Newton’s of force on
piston one and so by applying the same amount of pressure on a larger area of piston two,
we will get a much larger upward force and we will be able to lift an even
heavier object than the 900kg car that we have considered. So this is how a
mechanic is able to lift up heavy objects with minimum effort applied to
the hydraulic press or the hydraulic jack.

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89 thoughts on “Pascal’s Law

  1. Nice video sir.. I have a doubt that in case of football how come the pressure is same at all points..? As studied earlier pressure of liquid at bottom is high than at the top.. Please explain

  2. I'm sorry but you're a book worm. Try to let step in your imagination and give the answers that a book can't provide.

  3. Thank you so much sir. You are doing a very good job . Thank you sir. I'm very very satisfied.๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘I also agree with the question that pressure is always greater at the bottom than at the top.Since pressure of the fluid depends upon height.

  4. As per previous remarks of yours , pressure in fluids dependent on height…but in foot ball it doesn't happen…can you reply me with and..thx again n advance .

  5. some of your explanations are nice and some of the explanations are not
    but your biology videos are nice and good too

  6. Thank you so much sir…you cleared all my doubts regarding this topic..

  7. Imagine if khan academy aumsum time and only this person from delta step mixes and teach us something……… I swear I would be more smarter than Einstein ๐Ÿ˜‚

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