Pressure, force, hydraulic lift problem

AI Thread Summary
A hydraulic lift with two pistons has areas of 25 cm² and 630 cm², filled with oil of density 690 kg/m³. To support a 1600 kg car, approximately 63.5 kg must be placed on the smaller piston. When a 100 kg person enters the car, the fluid levels in the pistons become unbalanced. The discussion seeks to determine the equilibrium height difference in the fluid levels due to this added mass. Participants are encouraged to provide hints without duplicating posts.
huskydc
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A hydraulic lift has two connected pistons with cross-sectional areas 25 cm2 and 630 cm2. It is filled with oil of density 690 kg/m3.

I've found that approx. 63.5 kg of mass must be placed on the small piston to support a car of mass 1600 kg at equal fluid levels.

but then it asks...

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?

I'm stuck here...any hints??
 
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oops, I'm sorry, i couldn't find my original post, i thought it got deleted somehow...sorry about that...
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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