How Does a Hydraulic Lift Work with Different Piston Sizes and Weights?

[mrnastytime]

Homework Statement

A hydraulic lift has two connected pistons with cross-sectional areas 15 cm2 and 460 cm2. It is filled with oil of density 560 kg/m3.

Homework Equations

a)What mass must be placed on the small piston to support a car of mass 1100 kg at equal fluid levels?
A:35.86 kg
b)With the lift in balance with equal fluid levels, a person of mass 70 kg gets into the car. What is the equilibrium height difference in the fluid levels in the pistons?
A:2.72 m
c) How much did the height of the car drop when the person got in the car?

The Attempt at a Solution

Take the difference in height when its the only the car, then find difference in height of car + guy. Subtract those two. But its not working.

 

[Carid]

It seems to me that, when the guy got in the car, the car went down but the 38.5 kg weight on the other piston went up. So you have to share out that 2.72 metres between how much the car went down and how much the other piston went up.

 

[nlightner]

I would approach this problem by first understanding the principles of fluid mechanics and hydraulic systems. A hydraulic lift works based on Pascal's law, which states that the pressure applied to a confined fluid is transmitted equally in all directions. This means that the pressure applied to the small piston will be transmitted to the larger piston, resulting in a larger force being applied to the larger piston.

To calculate the mass needed to support the car, we can use the equation P1A1 = P2A2, where P1 and P2 are the pressures on the two pistons and A1 and A2 are the cross-sectional areas. We know that the pressure on the small piston is equal to the weight of the car, which is given as 1100 kg. The pressure on the larger piston will be equal to the weight of the small piston plus the weight of the added mass, which we can denote as m. So, we can set up the equation as:

1100 kg * 9.8 m/s^2 * 15 cm^2 = (1100 kg + m) * 9.8 m/s^2 * 460 cm^2

Solving for m, we get m = 35.86 kg. This is the mass that needs to be placed on the small piston to support the car.

For part b, we can use the same equation to find the equilibrium height difference between the two pistons when the person gets into the car. The only difference now is that the larger piston will have an added weight of 70 kg. So, the equation becomes:

(1100 kg + 70 kg) * 9.8 m/s^2 * 460 cm^2 = 1100 kg * 9.8 m/s^2 * 15 cm^2 + (1100 kg + 70 kg + m) * 9.8 m/s^2 * 460 cm^2

Solving for m, we get m = 35.86 kg. This is the added mass needed to balance the system with the person in the car. To find the equilibrium height difference, we can use the same equation as before, but now the added mass is known. So, we get:

1100 kg * 9.8 m/s^2 * 15 cm^2 = (1100 kg + 35.86 kg) * 9.8 m/s^2 *