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Pad 8:20 PM * 96% - a flipitphysics.com So iPhone X -... Ticket Det..code to ge Melonecha... ColorWare Homew..Ilinois.gov... Sunday, July 15 8:20 PM saloumd@dupage.edu account log FlipltPhsics 1201-014 Summer 2018 macmillan learning Unit 13: Homework Problems Homework: Fluids Print Assignment View Deadline: 100% until Sunday, July 15 at 11:59 PM Interactive Example Floating Cylinders Hydraulic Lift 1 2 3 A hydraulic lift has two connected pistons with cross-sectional areas 5 cm2 and 450 cm2. It is filled with oil of density Interactive Example Bobber 540 kg/m Standard Exercise Multi Tube 1) What mass must be placed on the small piston to support a car of mass 1500 kg at equal fluid levels? Standard Exercise Wet Penguins Standard Exercise Aspiratonr 2) With the lift in balance with equal fluid levels, a person of mass 90 kg gets into the car. What is the equilibrium height difference in the fluid levels in the pistons? Standard Exercise Faucet m Submit Help Standard Exercise Dam and Plug 3) How much did the height of the car drop when the person got in the car? Standard Exercise Water Stream from Can m SubmitHelp Standard Exercise Apparent Weight in Water Standard Exercise Hydraulic Lift Copyright o 2018 Freeman Worth Publishers - a division of Macmillan Learning. About I Tech Support Find Your Local Sales Rep Terms Of Use | Privacy Policy mac learn

Explanation / Answer

(1) Given in the problem -

Cross-sectional of the small piston: A1 = 5 cm^2
Cross-sectional of the large piston: A2 = 450 cm^2
Mass on the small piston: m1 = ?
Mass on the large piston: m2 = 1500 kg

Apply Pascal's Principle:
F1/A1 = F2/A2
m1*g/A1 = m2*g/A2
m1 = m2*(A1/A2)

Put the values given above -

m1 = 1500 kg (5 cm^2 / 450 cm^2) = 16.67 kg

(2) Now, when a person enter into the car, the person on the small piston will be elevated a heigh h1 over initial level and the car is going down a lenght h2, and height difference in the fluid levels in the pistons is H = h1 + h2.
Mass on the small piston: m1 = 16.67 kg
Mass on the large piston: m2 = 1500 + 90 = 1590 kg

So, we have -

Pressure by 16.67 kg person + hydrostatic pressure by column of fluid = pressure by (90 kg person and 1500 kg car)
m1*g*A1 + d*g*H = m2*g*A2
H = (m2*A2 - m1*A1)/d

Put the values -

H = {(1590 kg * 0.045 m^2) - (16.67 kg * 0.0005 m^2)}/(540 kg/m³)
H = ( 71.55 - 0.008335) / 540

= 0.1325 m

(3) Volume displaces is the same in both pistons:
A1*h1 = A2*h2
A1*(H - h2) = A2*h2
A1*H = h2*(A1 + A2)
h2 = H*{A1/(A1+A2)}

Put the values -

h2 = 13.25 cm (5 cm^2 / 455 cm^2)
h2 = 0.1456 cm = 0.001456 m

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