Hydraulic jack - Pascal's principle

[DevonZA]

Homework Statement

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.

 

[scottdave]

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?

 

[DevonZA]

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

 

[scottdave]

What is the ratio of the two areas?

 

[DevonZA]

3:1

 

[scottdave]

3:1

That is the ratio of diameters. What about the areas?

 

[DevonZA]

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