- #1

luis20

- 50

- 0

**at rest**?

If I have a hydraulic lever and I put weight on the smaller piston, how can I calculate the acceleration and velocity at which the two pistons move to balance each other?

Thanks for any help !

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- Thread starter luis20
- Start date

- #1

luis20

- 50

- 0

If I have a hydraulic lever and I put weight on the smaller piston, how can I calculate the acceleration and velocity at which the two pistons move to balance each other?

Thanks for any help !

- #2

sankalpmittal

- 785

- 15

Is pascal law only true for fluidsat rest?

Yes Pascal's law is only true for fluids but its not at all necessary that fluids are at rest.

Pascal's law says that pressure applied at the given point in a fluid is transmitted equally in all directions and is independent of the area of the plane of the container perpendicular to the line of action of pressure, in which the fluid is filled.

Pascal has not even mentioned whether fluid is at rest or not.

If I have a hydraulic lever and I put weight on the smaller piston, how can I calculate the acceleration and velocity at which the two pistons move to balance each other?

Thanks for any help !

Take example of hydraulic press. You have a cotton bales at the plank of large area piston and you apply force on the small area piston. Let the area of large piston is A and that of smaller piston is "a". Let you apply force f on the small piston and the resultant force on larger piston comes out to be F. Then by Pascal's law , we have :

F/A = f/a

F/f = A/a

As A>a so resultant F>f. So we get "gain in force".

Also Mechanical Advantage (next time M.A. to be brief) is calculated as:

You apply f force thus imparting "P" pressure to small piston :

P=f/a

or f=Pa

Also

P=F/A

F=PA

Now M.A. = F/f = A/a

Also Work input equals work output. If change in time is held same , we have :

So Power input = Power output

FV = fv

Since you can get velocities , you can easily obtain acceleration.

- #3

luis20

- 50

- 0

Yes Pascal's law is only true for fluids but its not at all necessary that fluids are at rest.

Pascal's law says that pressure applied at the given point in a fluid is transmitted equally in all directions and is independent of the area of the plane of the container perpendicular to the line of action of pressure, in which the fluid is filled.

Pascal has not even mentioned whether fluid is at rest or not.

Take example of hydraulic press. You have a cotton bales at the plank of large area piston and you apply force on the small area piston. Let the area of large piston is A and that of smaller piston is "a". Let you apply force f on the small piston and the resultant force on larger piston comes out to be F. Then by Pascal's law , we have :

F/A = f/a

F/f = A/a

As A>a so resultant F>f. So we get "gain in force".

Also Mechanical Advantage (next time M.A. to be brief) is calculated as:

You apply f force thus imparting "P" pressure to small piston :

P=f/a

or f=Pa

Also

P=F/A

F=PA

Now M.A. = F/f = A/a

Also Work input equals work output. If change in time is held same , we have :

So Power input = Power output

FV = fv

Since you can get velocities , you can easily obtain acceleration.

But, look at this.

Imagine A is the double of "a", according to Pascal F is also the double of "f". When the smaller piston goes down "h", the larger piston goes up "h/2". So the acceleration of the larger piston has to be half the acceleration of the smaller piston. If so, how can F be the double of "f", since F=ma and this "a" is half the other one?

- #4

sankalpmittal

- 785

- 15

But, look at this.

Imagine A is the double of "a", according to Pascal F is also the double of "f". When the smaller piston goes down "h", the larger piston goes up "h/2". So the acceleration of the larger piston has to be half the acceleration of the smaller piston. If so, how can F be the double of "f", since F=ma and this "a" is half the other one?

For a moment , forget about F=ma. Now if F=2f , then according to Pascal's law , we have :

FV = fv

F/f = v/V

v=2V (Velocity ratio = M.A. =2 )

Velocity of smaller piston by which it moves is twice the velocity of which larger piston moves.

So as the two piston's start from rest and the change of time is held same , we will have :

a = v/Δt

A= V/Δt

So we can say that a=2A

Or , acceleration in smaller piston is twice of larger one.

Now , we have "F=ma"

(i) F = MA

(ii) f=ma

Since a=2A

Now on dividing the two we have :

F/f = M/2m

Now we get relationship of forces applied on pistons to the masses of pistons.

Now take the case in which F=2f

2= M/2m

4m = M

Which is of course true ! Mass of larger piston > Mass of smaller piston. M>m.

Hope this helps...

- #5

luis20

- 50

- 0

For a moment , forget about F=ma. Now if F=2f , then according to Pascal's law , we have :

FV = fv

F/f = v/V

v=2V (Velocity ratio = M.A. =2 )

Velocity of smaller piston by which it moves is twice the velocity of which larger piston moves.

So as the two piston's start from rest and the change of time is held same , we will have :

a = v/Δt

A= V/Δt

So we can say that a=2A

Or , acceleration in smaller piston is twice of larger one.

Now , we have "F=ma"

(i) F = MA

(ii) f=ma

Since a=2A

Now on dividing the two we have :

F/f = M/2m

Now we get relationship of forces applied on pistons to the masses of pistons.

Now take the case in which F=2f

2= M/2m

4m = M

Which is of course true ! Mass of larger piston > Mass of smaller piston. M>m.

Hope this helps...

Oh, I never saw FV=fv, how do you get that relationship. You came up with the conclusion that the mass of the larger piston is 4 times larger than the smaller piston, is that true ? (If area doubles, mass should double)

Thanks for the help ! :)

- #6

sankalpmittal

- 785

- 15

Oh, I never saw FV=fv, how do you get that relationship. You came up with the conclusion that the mass of the larger piston is 4 times larger than the smaller piston, is that true ? (If area doubles, mass should double)

Thanks for the help ! :)

See my post #2 again. Considering the hydraulic press to be an ideal machine (ignore friction and all...) the law of conservation of energy will hold true. So energy before and after the conversion will remain same. So Work done on the machine will be same as work done by the machine. Hence , we have :

FD=fd where D and d are distance moved respectively. On dividing both side by Δt or same change in time , we obtain :

FD/Δt = fd/Δt

FV=fv

Or Power in equals power out. Power is defined as dot product of force and velocity. Here they're in same direction so we obtain power as force times velocity.

Since we obtain that if F=2f then d=2D.

Also A=2a ( Here a is area and not acceleration !)

Then we obtain

AD = ad (Note : I used your scenario to get this relation. However its true for all scenarios : http://www.grc.nasa.gov/WWW/k-12/WindTunnel/Activities/Pascals_principle.html)

(verification :

LHS :

2ad/2 = ad = RHS)

So we can say that :

V

(the volume of fluid pushed down on the left side equals the volume of fluid that is lifted up on the right side , true for all the cases here.)

Now the area of larger piston is double of area of smaller one , but you cannot say the same thing about its mass. This is because mass is not uniform at its every point.

We can then say that

Since acceleration in smaller piston equals to twice of acceleration in larger piston ; and force applied on smaller piston is half of force generated in larger piston ,

Since acceleration in smaller piston=2 times Acceleration in larger piston , and F=M times acceleration in larger piston

and

f= m times acceleration in smaller piston

Now on dividing the two we have :

F/f = M/2m

As F=2f

So

M=4m

We get this ! You took the ratio as too much positive integer. In real machines its just difficult to get. Moreover friction etc. also play a role in calculation.

Generalizing this ,

F/f = MA/ma

Since F>f ,so MA>ma

You cannot just precisely evaluate the ratio of mass. In your scenario you took F=2f , so I said that M=4m.

Hope this helps.

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