Hydraulic jack - Pascal's principle

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SUMMARY

The discussion centers on the application of Pascal's principle in hydraulic jacks, specifically analyzing the force and area relationships involved. The calculations demonstrate that a force of 500N is insufficient to lift a 500kg mass, as a force of 543N is required. The area calculations for the pistons reveal a ratio of 9:1 between the larger and smaller pistons, confirming the effectiveness of the hydraulic system. The participants clarify the implications of these calculations regarding the distance moved on the mass side when a specific distance is pushed on the small piston.

PREREQUISITES
  • Understanding of Pascal's principle in fluid mechanics
  • Familiarity with basic physics equations related to force and area
  • Knowledge of area calculations for circular pistons
  • Ability to perform unit conversions and calculations involving weight and distance
NEXT STEPS
  • Study the applications of Pascal's principle in various hydraulic systems
  • Learn about the design and efficiency of hydraulic jacks
  • Explore advanced calculations involving hydraulic pressure and force distribution
  • Investigate real-world examples of hydraulic systems in engineering
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Students studying physics, engineers working with hydraulic systems, and anyone interested in the practical applications of fluid mechanics.

DevonZA
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Homework Statement


upload_2018-2-11_18-56-16.png


Homework Equations


P1=P2

P=F/A

F=PA

F1/A1=F2/A2

F2 = F1(A2/A1)

F2 = W(A2/A1)

W=mg

W=F*d

The Attempt at a Solution



A1= pi/4 * d2 = pi/4 * (0.15)2 = 0.0177m2

A2 = pi/4 * d2 = pi/4 * (0.05)2 = 0.00196m2

P1=P2

P=F/A

F=PA

F1/A1=F2/A2

F2 = F1(A2/A1)

F2 = W(A2/A1)

W=mg

= 500 * 9.81

= 4905N

F2 = W(A2/A1)

= 4905 (0.00196/0.0177)

= 543N

1.1) Therefore the force of 500N cannot lift the 500kg mass because 543N is required to do so.

1.2) W=F*d

= 543 x 0.01

= 5.43J

W= F*d

5.43 = 4905 x d

d = 5.43/4905

d = 0.001m

= 1mm is the distance depressed.
 

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I am confused. What is your question for us?

Your part 1 is correct. I am not sure what they want in 2nd part. Are they assuming that now you do have sufficient force and you push down a distance of 10 cm?
 
scottdave said:
I am confused. What is your question for us?

Your part 1 is correct. I am not sure what they want in 2nd part. Are they assuming that now you do have sufficient force and you push down a distance of 10 cm?

I only want to confirm that my answers are correct.

In 1.2 I believe they want to know the distance moved on the side of the mass when 10mm pushed down on the small piston side
 
What is the ratio of the two areas?
 
3:1
 
DevonZA said:
3:1
That is the ratio of diameters. What about the areas?
 
A1= pi/4 * d2 = pi/4 * (0.15)2 = 0.0177m2

A2 = pi/4 * d2 = pi/4 * (0.05)2 = 0.00196m2

Ratio = 0.0177/0.00196
= 9

Therefore ratio is 9:1
 

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