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
Messages
103
Reaction score
0

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!
 
Physics news on Phys.org
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top