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Calculating the force resulted by pressure (integral)
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[QUOTE="CivilSigma, post: 4983941, member: 463654"] [h2]Homework Statement [/h2] " 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? "[h2]Homework Equations[/h2] F = kx dF = pressure dA pressure = density * height * gravity [h2]The Attempt at a Solution[/h2] 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 [URL]http://imgur.com/LOU5cKR[/URL] [/QUOTE]
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Calculating the force resulted by pressure (integral)
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