Hydraulic Lift

[nothing123]

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Anyone familiar with pistons and hydraulic lifts? I came about this formula, F = rho*g*(A1+A2)d2 which is used to find the additional force needed to be applied to a piston to lift a car a distance of d2 though I don't know where it's coming from. Any ideas?

Thanks.

 

[auroraboy]

Anyone familiar with pistons and hydraulic lifts? I came about this formula, F = rho*g*(A1+A2)d2 which is used to find the additional force needed to be applied to a piston to lift a car a distance of d2 though I don't know where it's coming from. Any ideas?

Thanks.

Essentially, while using hydraulic machinery, work done on both side should be the same, so

F1d1=F2d2

so F1 = (d2/d1)*F2

distance should be inversely proportional to area, so F1 = (A1/A2)*F2

 

[nothing123]

Right, but I can't seem to get the formula I posted from these relationships. If a question asked, how much force must be applied to a 0.05 m^2 piston to lift a 1000kg car resting on a 0.05 m^2 piston by 2.0 m, how would you go about the answer?

Thanks.

 

[nothing123]

Nevermind, I've figured it out. If anyone is curious, by Pascal's law, the pressure is transmitted throughout the liquid. So F/A1 = rho*g*(d1+d2). F is then just the pressure difference multiplied by A1 and substituting d2*A2/A1 for d1, we get the equation I originally posted.

 

[auroraboy]

Right, but I can't seem to get the formula I posted from these relationships. If a question asked, how much force must be applied to a 0.05 m^2 piston to lift a 1000kg car resting on a 0.05 m^2 piston by 2.0 m, how would you go about the answer?

Thanks.

sorry I dont understand this question...the car is rested on a 0.05 m^2 piston and you're pushing on 0.05 m^2 piston?
in that case A1 = A2 => F1 = F2

so force should just be m*g = 1000*10 = 10,000 N
How does the 2.0 m even come in to play? the question asked for force, not work done...even so..since A1 = A2 thus d1 = d2

I thought the whole point of hydraulics was to use difference area on both sides so total work done is same but on one side you cover more distance due to reduce area of contact.

Feel free to correct me

 

[nothing123]

Sorry, the question is a little confusing. First I meant to say that the force is on a 0.005m^2 piston (if the areas were the same, it'd kind of defeat the purpose of a hydraulic lift as you wouldn't get any force multiplication). Besides that, the question is meant to say how much additional force would you have to apply to the small piston to raise the car by 2.0 m. Everything you've said so far is correct but in figuring out the additional force, the force applied on the small piston would drive the car up until it reached an equilibrium point. At that point, the force applied would be equal to the weight of additional water moved up on the other side if you get what I'm saying...