The car goes down, the piston goes up, so
h1+h2=3.86
For the conservation of volume
350 cm²*h1 = 5 cm² * h2
Solve for h1 & h2 to get
h1=0.0544 and h2=3.8056
The answer is similar to yours. I don't know if it is the number of significant figures or the sign that made your answer invalid.
Post if there are problems.
A hydraulic lift has two connected pistons with cross-sectional areas 5 cm2 and 350 cm2. It is filled with oil of density 740 kg/m3.
a) What mass must be placed on the small piston to support a car of mass 1400 kg at equal fluid levels?
20 OK
HELP: The pressure is constant at any height.
b) With the lift in balance with equal fluid levels, a person of mass 100 kg gets into the car. What is the equilibrium height difference in the fluid levels in the pistons?
HELP: Balance the pressure from the weight of the fluid and the pressure from the person's weight.
c) How much did the height of the car drop when the person got in the car?
HELP: The fluid is incompressible, so volume is conserved.
I just need help with c. I've tried
A1*h1=A2*h2
(3.5 m^2)(h1)=(.05 m^2)(3,86)
h1=.0551
and that's not correct what am I doing wrong?
1 answer
1 answer