Calculating Mass to Support Car Weight on Hydraulic Lift

AI Thread Summary
To determine the mass required on the small piston of a hydraulic lift supporting a 1500 kg car, the relationship between the areas of the pistons is crucial. The formula used is F = mg (A1/A2), where A1 is the area of the small piston and A2 is the area of the large piston. The density of the oil is not necessary for the calculation, as the problem focuses on the area ratio. The initial attempt at the solution yielded an incorrect force of 1.261 N, indicating a need for reevaluation of the calculations. The discussion emphasizes the importance of understanding hydraulic systems as ratios rather than focusing on fluid density.
mujadeo
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Homework Statement



A hydraulic lift has two connected pistons with cross-sectional areas 5 cm^2 and 550 cm^2. It is filled with oil of density 520 kg/m3.


a) What mass must be placed on the small piston to support a car of mass 1500 kg at equal fluid levels?



Homework Equations


Is the oil density a red herring in this question?


The Attempt at a Solution



F = mg (A1/A2)

F = mg (5cm/550cm)^2

F = (1500kg)(9.81)(5cm/550cm)^2

F = 1.261N = wrong!

please help!
 
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A hydraulic system is just ratios.
In this case just the ratio of the two areas
 
> F = (1500kg)(9.81)(5cm/550cm)^2


??
Re-think...
 
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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