A hydraulic lift has two connected pistons with cross-sectional areas 25 cm2 and 390 cm2. It is filled with oil of density 530 kg/m3.
a) What mass must be placed on the small piston to support a car of mass 1500 kg at equal fluid levels?
kg
HELP: The pressure is constant at any height.
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?
m
c) How much did the height of the car drop when the person got in the car?
m
big piston area *p = car mass*g
little piston area * p = little mass*g
thus
car mass/big area = little mass/little area
little mass = (lttle area/big area ) car mass
little mass =(25/390)*1500
-------------------------------
enough additional mass of oil must rise in the little tube to balance 70 kg on the big piston
oil mass = (25/390)70
= 530 * 25 * h
solve for h
--------------------------
get the additional volume in the little piston
25*h
same volume down in big piston
volume = 390 * drop in big piston
a) To determine the mass that must be placed on the small piston to support the car, we can use Pascal's principle, which states that the pressure exerted by a fluid is transmitted equally in all directions.
Given:
Cross-sectional area of the small piston (A1) = 25 cm^2 = 0.0025 m^2
Cross-sectional area of the large piston (A2) = 390 cm^2 = 0.039 m^2
Density of the oil (ρ) = 530 kg/m^3
Mass of the car (m) = 1500 kg
The pressure at any height in the hydraulic lift is constant. Therefore, the pressure exerted by the car on the small piston is equal to the pressure exerted by the oil on the large piston.
Pressure1 = Pressure2
The pressure in a fluid can be calculated using the formula:
Pressure = Force / Area
Applying this to the hydraulic lift, the equation becomes:
(mass of the car x g) / A1 = (mass on the small piston x g) / A2
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the equation to solve for the mass on the small piston:
mass on the small piston = (mass of the car x A2 x A1) / (A1 x g)
Substituting the given values:
mass on the small piston = (1500 kg x 0.039 m^2 x 0.0025 m^2) / (0.0025 m^2 x 9.8 m/s^2)
mass on the small piston ≈ 6 kg
Therefore, to support the car of mass 1500 kg, a mass of approximately 6 kg must be placed on the small piston.
b) When the lift is in balance with equal fluid levels, the pressure on both pistons is equal.
Again using Pascal's principle, the pressure exerted by the oil on the small piston is equal to the pressure exerted by the person (and the car) on the large piston:
Pressure1 = Pressure2
Applying the formula Pressure = Force / Area:
(mass of the person + mass of the car) x g / A1 = (mass on the small piston + mass of the person) x g / A2
Rearranging the equation to solve for the equilibrium height difference (Δh) in the fluid levels:
Δh = (A2 x (mass of the person + mass on the small piston) - A1 x mass of the person) / (A1 x A2 x ρ x g)
Substituting the given values:
Δh = (0.039 m^2 x (70 kg + 6 kg) - 0.0025 m^2 x 70 kg) / (0.0025 m^2 x 0.039 m^2 x 530 kg/m^3 x 9.8 m/s^2)
Δh ≈ 0.249 m
Therefore, the equilibrium height difference in the fluid levels is approximately 0.249 meters.
c) To find how much the height of the car dropped when the person got in, we need to calculate the difference in fluid levels before and after.
Initially, when the car was loaded and balanced:
Pressure1 (initial) = Pressure2 (initial)
(mass of the car x g) / A1 (initial) = (mass on the small piston x g) / A2 (initial)
Now, after the person gets in and the equilibrium is reached:
Pressure1 (final) = Pressure2 (final)
((mass of the person + mass of the car) x g) / A1 (final) = (mass on the small piston + mass of the person) x g) / A2 (final)
In both cases, the pressure is the same, so we can equate the two equations:
(mass of the car x g) / A1 (initial) = ((mass of the person + mass of the car) x g) / A1 (final)
Rearranging the equation to solve for the change in height (Δh):
Δh = (A1 (initial) - A1 (final)) / (A1 (initial) x ρ x g)
Substituting the given values:
Δh = (0.0025 m^2 - 0.0025 m^2) / (0.0025 m^2 x 530 kg/m^3 x 9.8 m/s^2)
Δh = 0 m
Therefore, there is no change in the height of the car when the person gets in.