A hydraulic lift has two connected pistons with cross-sectional areas 5 cm2 and 350 cm2. It is filled with oil of density 740 kg/m3.
a) What mass must be placed on the small piston to support a car of mass 1400 kg at equal fluid levels?
20 OK
HELP: The pressure is constant at any height.
b) 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?
HELP: Balance the pressure from the weight of the fluid and the pressure from the person's weight.
c) How much did the height of the car drop when the person got in the car?
HELP: The fluid is incompressible, so volume is conserved.
I just need help with c. I've tried
A1*h1=A2*h2
(3.5 m^2)(h1)=(.05 m^2)(3,86)
h1=.0551
and that's not correct what am I doing wrong?
The car goes down, the piston goes up, so
h1+h2=3.86
For the conservation of volume
350 cm²*h1 = 5 cm² * h2
Solve for h1 & h2 to get
h1=0.0544 and h2=3.8056
The answer is similar to yours. I don't know if it is the number of significant figures or the sign that made your answer invalid.
Post if there are problems.
To solve for the height difference (h1 - h2) when the person gets into the car, you can use the principle of Pascal's Law, which states that the pressure is constant in a confined fluid.
Firstly, let's establish the equation that relates the pressures between the two pistons:
P1 = P2
The pressure at the small piston (P1) is due to the weight of the car and the weight of the fluid above it:
P1 = (weight of car + weight of fluid) / A1
The pressure at the large piston (P2) is due to the weight of the fluid above it:
P2 = weight of fluid / A2
Since the pressure is constant throughout the fluid, we have:
P1 = P2
(weight of car + weight of fluid) / A1 = weight of fluid / A2
Now, let's go step by step to find the solution:
1. Solve for the weight of the fluid:
The weight of the fluid is given by the formula: weight = density * volume * g (acceleration due to gravity). The volume of the fluid can be calculated by multiplying the cross-sectional area of the large piston (A2) by the difference in heights between the fluid levels (h1 - h2):
weight of fluid = density * A2 * (h1 - h2) * g
2. Solve for the weight of the car:
The weight of the car is given as 1400 kg. You can calculate it using the formula: weight = mass * g.
weight of car = 1400 kg * g
3. Substitute the weights into the equation:
(weight of car + weight of fluid) / A1 = weight of fluid / A2
(1400 kg * g + density * A2 * (h1 - h2) * g) / A1 = (density * A2 * (h1 - h2) * g) / A2
4. Simplify the equation by canceling out the g factor and rearrange to find (h1 - h2):
(1400 kg + density * A2 * (h1 - h2)) / A1 = density * (h1 - h2)
1400 kg + density * A2 * (h1 - h2) = A1 * density * (h1 - h2)
1400 kg = A1 * density * (h1 - h2) - density * A2 * (h1 - h2)
Now, rearrange the equation to solve for (h1 - h2):
(h1 - h2) = (1400 kg) / (A1 * density - A2 * density)
Finally, substitute the given values for A1, A2, density, and the weight of the car to find the height difference (h1 - h2).