Hydraulic jack - Pascal's principle

Click For Summary

Discussion Overview

The discussion revolves around the application of Pascal's principle in the context of a hydraulic jack, focusing on the calculations related to forces, areas, and distances involved in lifting a mass. Participants explore the relationships between pressure, force, and area, as well as the implications of their calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents calculations for the areas of the pistons and the forces required to lift a mass, concluding that a force of 500N is insufficient to lift a 500kg mass.
  • Another participant expresses confusion about the second part of the problem, questioning whether it assumes sufficient force is available to push down a distance of 10 cm.
  • Several participants discuss the ratio of the areas of the two pistons, with one stating the ratio of diameters is 3:1, while another later calculates the area ratio to be 9:1.

Areas of Agreement / Disagreement

Participants generally agree on the calculations presented in the first part of the problem. However, there is uncertainty regarding the second part of the problem and differing views on the area ratio, indicating that multiple competing views remain unresolved.

Contextual Notes

Some assumptions about the problem's conditions are not explicitly stated, such as the sufficiency of force in the second part. The discussion also highlights dependencies on definitions and calculations that may not be fully resolved.

DevonZA
Messages
181
Reaction score
6

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.
 

Attachments

  • upload_2018-2-11_18-56-16.png
    upload_2018-2-11_18-56-16.png
    4.4 KB · Views: 1,041
Physics news on Phys.org
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
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 21 ·
Replies
21
Views
9K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K