• Support PF! Buy your school textbooks, materials and every day products Here!

Calculating the force resulted by pressure (integral)

  • Thread starter CivilSigma
  • Start date
  • #1
227
58

Homework Statement


" A spring of the pressure gauge shown below has a force constant of 1250 N/m , and the piston has a diameter of 1.20 cm. AS the gauge is lowered into the water in a lake, what change in depth causes the piston to move by 0.750 cm? "


Homework Equations


F = kx
dF = pressure dA
pressure = density * height * gravity

The Attempt at a Solution


My first concern is my approach at integration the correct method? Is there an easier way?

Well, I know that the force to be applied is : F = xk = 9.375 N and this force is due to the continually increasing pressure as the piston is dropped into the lake.

Thus I need to be evaluating for the upper limit of the integral

Since dF = pressure dA ,
F = ∫ pressure dA
= ∫ ρ * g * h dA
= ∫ ρ * g * (H-y) dA

I get stuck here, I know I need to integrate with respect to y. What is the relationship between dA and dy? Also why are we given the diameter of the piston.

I have attached a copy of my written work.

Any help is really appreciated, thank you for your time.

-Sakonpure6

http://imgur.com/LOU5cKR
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
Why do you think integration is the key to solving this problem?
Does the piston in the pressure gauge change in diameter with change in depth?
Pressure increases due to what variable or variables as the gauge is lowered in the water?
Are you familiar with Pascal's Law?
 
  • #3
227
58
My thought to using integration is that the force applied to the piston changes as the depth changes.

Pressure would change as the area changes correct?

Pascals Law is A1F2=A2F1 , but we are not given enough data to calculate area.

So how do you propose solving this?
 
  • #4
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
You're still confused.
You are given the diameter of the piston. You can't calculate an area from that?
Why do you think the area of the piston in the gauge changes as the gauge goes deeper in the water?

You are also misinterpreting what Pascal's Law states:

http://en.wikipedia.org/wiki/Pascal's_law
 

Related Threads on Calculating the force resulted by pressure (integral)

Replies
8
Views
1K
  • Last Post
Replies
4
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
3
Views
496
Replies
22
Views
757
Replies
4
Views
981
Replies
2
Views
963
  • Last Post
Replies
0
Views
924
Replies
3
Views
1K
Top