Can the output forces for F3 and F4 be the same as F2 in hydraulic lifting?

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Discussion Overview

The discussion revolves around the calculation of output forces F3 and F4 in a hydraulic lifting system, specifically in relation to an input force F1 and an area A1, with a focus on whether these output forces can be equal to an output force F2, as determined by Pascal's principle. The scope includes theoretical considerations and practical calculations in the context of hydraulics.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that the output force F2 can be calculated using the formula F2 = A2/A1 x F1, based on Pascal's principle.
  • Another participant mentions that hydraulic pressure acts on all inner surfaces of area A2, noting that some forces may cancel due to opposite directions, leading to a net result.
  • A different participant emphasizes that only the "moving areas" should be considered, arguing that forces on the top surfaces of A2 in cases 2 and 3 do not contribute to mechanical work as they do not produce movement.
  • There is a repeated assertion regarding the irrelevance of forces on the top surfaces for mechanical work, questioning why a piston would not move if it is free to do so.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of forces acting on the top surfaces of A2 and their contribution to the movement of the piston. There is no consensus on how to calculate the output forces F3 and F4 or whether they can be equal to F2.

Contextual Notes

Some assumptions regarding the conditions under which the forces act and the definitions of the areas involved may be missing. The discussion does not resolve the mathematical steps necessary for calculating F3 and F4.

ucody0911
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User has been reminded to post schoolwork problems in the Homework Help forums and to always show their Attempt at the Solution
Hello All,
pls see picture , input force F1 and area A1 and second area A2 are all same for 3 cases
for the 1st case , output force F2 should be F2=A2/A1 x F1 , by pascal's principle ,
what about output forces of F3 and F4 ? can be same as F2 ? how i calculate them ?

thanks
 

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Hydraulic pressure applies on all the inner surfaces, upper, base and the sides of A2. Some forces by pressure cancel because they have opposite directions and you would get the net results.
 
Last edited:
The "moving areas" are the only ones to be considered.
The force on the top surfaces A2 of cases 2 and 3 produces no movement; therefore, no mechanical work.
 
Lnewqban said:
The "moving areas" are the only ones to be considered.
The force on the top surfaces A2 of cases 2 and 3 produces no movement; therefore, no mechanical work.
Presumably the red open outline represents a free-to-move piston. Why wouldn't the piston move?
 
Last edited:
ucody0911 said:
Hello All,
pls see picture , input force F1 and area A1 and second area A2 are all same for 3 cases
for the 1st case , output force F2 should be F2=A2/A1 x F1 , by pascal's principle ,
what about output forces of F3 and F4 ? can be same as F2 ? how i calculate them ?

thanks

Please re-post this in the Homework Help, Introductory Physics forum and use the hints you have been given here to show your work toward a solution. This misplaced schoolwork thread is now locked.
 

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