Lifting an Elephant with Hydraulics: A Physics Puzzle

In summary, the conversation discusses the balancing of a 1200kg elephant on a hydraulic lift by a 70kg student and the addition of a second 70kg student. The students are on a smaller piston than the elephant, and calculations are made to determine how high the elephant will be lifted. However, it is noted that the system may not be in balance due to unequal pressure on each side. The conversation also touches on the potential impact of the weight of the pistons on the system.
  • #1
kikko
47
0

Homework Statement



The 70kg student in the figure balances a 1200kg elephant on a hydraulic lift. A second 70kg student joins the first student. How high do they lift the elephant?
diameter of the piston elephant is on = 2.0m
diameter of the piston student is on = 0.48m
The liquid inside the lift is oil, therefore rho = 900 kg/m3
Out would be elephant, In would be the students

Homework Equations


Fout/Fin = Aout/Ain
Fout=Aout/Ain x Fin - \rhoghAout

The Attempt at a Solution



(1200)(9.8) = ((pi*(2/2)^2)/(pi*(.48/2)^2))*(2)(70)(9.8) - (900)(9.8)(h)(pi)

h = .435m = 43.5 cm

This answer comes out wrong. I noticed the .435^-1 = 2.3m, and the answer is 2.3 cm, but I'm not sure if there is a relation there.

Anyways, any tips in the right direction?
 
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  • #2
Initially, the problem looks to be invalid.
Are you sure the elephant is on the larger piston?
 
  • #4
kikko said:
Yes. Here is a picture, I solved .48 for the diameter of the first piston and got it correct.

http://session.masteringphysics.com/problemAsset/1070646/3/15.P41.jpg

Yes, of course. After thinking about it a bit more that makes sense.

But, I don't see how they can be in balance.
There would have to be equal pressure on each side of the system.

On the student's side, you have:

[tex]\frac{70 kg}{(0.24)^2 \pi ~m^2} = \frac{70 kg}{(0.576) \pi ~m^2} \approx \frac{70 kg}{0.181 ~m^2} \approx 386.83 ~kg/m^2[/tex]


On the elephant's side, you have:

[tex]\frac{1200 kg}{(1)^2 \pi ~m^2} = \frac{1200 kg}{\pi ~m^2} \approx \frac{1200 kg}{3.142 ~m^2} \approx 381.97 ~kg/m^2[/tex]



So, since there is more pressure being exerted on the oil on the student's side than on the elephant's side, the elephant should be rising (at least to the extent that shortest piston's travel allows, which means either the student's piston "bottomed out," or the elephant's piston "topped out.")



Now, if the student only weight 69.12 kg, you'd have:

[tex]\frac{69.12 kg}{(0.24)^2 \pi ~m^2} = \frac{69.12 kg}{(0.576) \pi ~m^2} \approx \frac{69.12 kg}{0.181 ~m^2} \approx 381.97 ~kg/m^2[/tex]

which is the same pressure that's being exerted on the elephant's side.




... I think the weight of the pistons would affect the system though.
 
  • #5


I would first clarify the assumptions made in this puzzle. Is the hydraulic lift operating under ideal conditions, with no friction or other external forces? Is the lift able to support the weight of the elephant and two students without any structural damage? These are important factors to consider in order to accurately solve this physics puzzle.

Assuming ideal conditions, we can use the equation Fout/Fin = Aout/Ain to determine the force needed to lift the elephant. This equation states that the output force (Fout) is equal to the ratio of the output area (Aout) to the input area (Ain) multiplied by the input force (Fin). In this case, the output force is the weight of the elephant (1200kg x 9.8m/s^2 = 11,760N) and the input force is the combined weight of the two students (2 x 70kg x 9.8m/s^2 = 1,372N). The output area is the area of the elephant's piston (pi x (2m/2)^2 = pi m^2) and the input area is the combined area of the two students' pistons (pi x (0.48m/2)^2 x 2 = 0.723m^2). Plugging these values into the equation, we get:

11,760N/1,372N = (pi m^2)/(0.723m^2)

Solving for m, we get m = 2.3m. This means that the students will need to lift the elephant 2.3m off the ground in order to balance its weight.

However, this calculation does not take into account the weight of the oil in the hydraulic lift. To account for this, we can use the equation Fout = Fin - rhoghAout, where rhog is the density of the oil (900kg/m^3 x 9.8m/s^2 = 8,820N/m^3), h is the height the elephant is lifted, and Aout is the area of the elephant's piston (pi x (2m/2)^2 = pi m^2). Plugging in the values, we get:

11,760N = 1,372N - (8,820N/m^3)(9.8m/s^2)(h)(pi m^2)

Solving for
 

Related to Lifting an Elephant with Hydraulics: A Physics Puzzle

1. How is it possible to lift an elephant with hydraulics?

Hydraulics is a branch of physics that deals with the mechanical properties of fluids, such as water or oil. By utilizing the principles of fluid mechanics, specifically Pascal's law, we can use a small amount of force to create a larger force, allowing us to lift objects much heavier than the force we apply. This is how we can lift an elephant with hydraulics.

2. What is the process for lifting an elephant with hydraulics?

The process involves using a hydraulic system, which consists of a pump, a cylinder, and a fluid-filled pipe. The pump applies force to the fluid, which then travels through the pipe and into the cylinder, creating a larger force that can lift the elephant.

3. How much force is needed to lift an elephant with hydraulics?

The amount of force needed depends on the weight of the elephant. However, with the principles of hydraulics, we can create a force much greater than the force we apply. For example, with a 10:1 ratio, we can lift an elephant that weighs 10,000 pounds with just 1,000 pounds of force.

4. Are there any limitations to lifting an elephant with hydraulics?

While hydraulics can create a large amount of force, there are still limitations. The size and strength of the hydraulic system must be proportional to the weight of the object being lifted. Additionally, the surface area of the object being lifted must be able to support the weight without causing damage.

5. Are there any real-world applications for lifting objects with hydraulics?

Hydraulics are commonly used in heavy machinery, such as cranes and forklifts, to lift heavy objects. They are also used in elevators and car lifts. Additionally, hydraulics are used in construction and manufacturing industries to move and manipulate large objects with ease.

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