Pascal's principle hydraulic lift with torque involved

In summary, the conversation discusses using a hydraulic lift with torque involved to support a mass using two pistons with different masses and areas. The goal is to find the magnitude of the force needed to be applied to one of the pistons to support the mass. The conversation includes equations for pressure and torque and walks through the steps to find the correct answer. It is important to pay attention to units when solving this type of problem.
  • #1
fordy314
6
0
Pascal's principle hydraulic lift with torque involved(need help with lever)

Homework Statement

A hydraulic jack is used to support a mass which has a mass of 30.7 kg.
13YOy.jpg

The mass is supported by piston 2 which has a mass of 0.250 kg and an area of 40.2 cm2 . A force is applied by the lever on piston 1 which has a mass of 0.085 kg and an area of 14.2 cm2.

The distance between the hinge and piston 1 is 4.47 cm. A force is applied a further 16.4 cm from piston 1.

What is the magnitude of the force F which must be applied to support the mass? You may neglect the mass of the fluid. You may also wish to TORQUE to someone about this problem.

Homework Equations


ρ=m/v
P=F/A
P=Po+ρgh
Any relevant torque equations (not that far yet)

The Attempt at a Solution


I'm pretty lost here, but I know that I need to find the force exerted by piston1.
The force required to lift the mass and piston2 is:

30.95kg*9.81m/s^2 = 303.6N
P=F/A=303.6N/(pi*.402^2)=598(k?)Pa
This is where I get lost. normally I would take P=Po+ρgh to find the pressure on piston 1 to find the force needed to push against it, but I don't have the height. Am I missing something, or do you think that I'm supposed to assume that they're at the same height?

Thanks.
 
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  • #2
It looks like you're supposed to assume they're at the same height. The pressure from piston 1 needs to equal the pressure from piston 2. Then you can find the force that the lever needs to apply to the piston to create that pressure. Using the law of the lever, you can then find the force that needs to be applied to the end of the lever.
 
  • #3
Ok, then FPiston1=PA1=37.9N

37.9N-(.085kg*9.81m/s^2)=37.1

Now do I want to use the equation F1*d1=F2*d2?

That would be (37.1*.0447m)/.164m=10.1N, but my answer comes up wrong. Any suggestions?
 
  • #4
fordy314 said:
Ok, then FPiston1=PA1=37.9N

37.9N-(.085kg*9.81m/s^2)=37.1

Now do I want to use the equation F1*d1=F2*d2?

That would be (37.1*.0447m)/.164m=10.1N, but my answer comes up wrong. Any suggestions?

For the area of piston 2, you used:
A=pi*.402^2
When the problem states that the area is .402 meters (not the radius as you undoubtedly thought).
and yes, you've got the right lever equation.
Also if you're still wondering, the pressures are pascals (not kPA).
 
  • #5
Nessdude14 said:
For the area of piston 2, you used:
A=pi*.402^2
When the problem states that the area is .402 meters (not the radius as you undoubtedly thought).
and yes, you've got the right lever equation.
Also if you're still wondering, the pressures are pascals (not kPA).

Thanks so much. Of course its always that units that'll get me.
 

1. What is Pascal's principle?

Pascal's principle states that pressure applied to a confined fluid will be transmitted equally in all directions throughout the fluid.

2. How does a hydraulic lift use Pascal's principle?

A hydraulic lift uses Pascal's principle by applying a small amount of force to a small piston, which then transfers the pressure to a larger piston, resulting in a larger output force.

3. How is torque involved in a hydraulic lift?

Torque is involved in a hydraulic lift because the fluid pressure created by the smaller piston is used to rotate a larger piston, creating a rotational force or torque.

4. What are the advantages of using a hydraulic lift with torque involved?

The use of a hydraulic lift with torque involved allows for the lifting of heavy objects with minimal effort, as the pressure of the fluid can be multiplied to create a larger output force.

5. Are there any limitations to using a hydraulic lift with torque involved?

One limitation of using a hydraulic lift with torque involved is that the system can only generate as much force as the input pressure allows. Additionally, the system can be prone to leaks and may require regular maintenance.

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