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.
