How Does Pascal's Principle Apply to Hydraulic Systems?

[SAT2400]

Homework Statement

Imagine a hydraulically operated dentist's chair having a mass of 200kg, and in it is stiing a 54.8kg patient. The large piston below the chair has 5cm(diameter), while the small piston, moved by a pedal on which the dentist steps, has 1cm(diameter). What's the pressure in the interconnecting fluid when the dentist operates the chair? How much force must he exert on the small piston?

2. A hydraulic lift consists of two interconnected pistons filed with a common working liquid. If the areas of the piston faces are 64cm^2 and 3200cm^2 and if 900kg car rests on the latter, how much force must be exerted to raise the vehicle very slowly?
If the car is to be raised 2m, how far must the input piston be depressed?

Homework Equations

F=mg
F/F2 =A/A2
PA= F

The Attempt at a Solution

The answers for 1 are 1.27 x10^6 Pa and 100N...
Answers for 2 are 177N and 100m.

Please help.:(! I have no idea how to get these answers :((

 

[rl.bhat]

P= F/A
Find the total weight on the larger piston. Then find the pressure using the relevant equation.

 

[kerimek]

I would approach this problem by first understanding Pascal's Principle. Pascal's Principle states that pressure applied to an enclosed fluid will be transmitted equally throughout the entire fluid. This means that the pressure exerted on the small piston by the dentist's foot will be transmitted to the larger piston and ultimately to the entire fluid in the system.

For the first problem, we can use the equation P1A1 = P2A2, where P1 and A1 are the pressure and area of the small piston, and P2 and A2 are the pressure and area of the larger piston. We know that the area of the small piston is 1cm^2 and the area of the larger piston is 5cm^2. We also know that the pressure is the same throughout the entire fluid, so we can set P1 equal to P2. We can also use the equation F = mg to calculate the force exerted by the weight of the patient on the larger piston.

Plugging in the values, we get:

P1(1cm^2) = P2(5cm^2)
P1 = 5P2

And,

F = (54.8kg)(9.8m/s^2)
F = 537.04N

Now we can use the equation PA = F to calculate the pressure in the interconnecting fluid:

P(5cm^2) = 537.04N
P = 537.04N/5cm^2
P = 1.27 x 10^6 Pa

We can also use the equation F/F2 = A/A2 to calculate the force exerted by the dentist on the small piston:

F/F2 = A/A2
F/100N = 1cm^2/5cm^2
F = (100N)(1cm^2/5cm^2)
F = 20N

For the second problem, we can use the same equation P1A1 = P2A2, but this time we know the values for P2 and A2. We can also use the equation F = mg to calculate the force needed to lift the car.

P1(64cm^2) = (1.27 x 10^6 Pa)(3200cm^2)
P1 = 2.56 x 10^4 Pa

F = (900kg)(9.8m/s^