A hydraulic lift is used to jack a 960-kg car 52 cm off the fl...

Question

A hydraulic lift is used to jack a 960-kg car 52 cm off the floor. The diameter of the output piston is 18 cm, and the input force is 380 N. What is the area of the input piston?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with Pascal's principle. This question asked, what is the area of the input piston? If a hydraulic press is used to lift a heavy crane with a mass of 970 kg to a height of 40 centimeters with an output piston diameter of 44 centimeters and an input force of 420 newtons. We're given four possible choices as our answers A is 1.6 multiplied by 10 to the negative 3 m squared. Choice B is 3.4 multiplied by 10 to the negative 3 m squared. Point C is 5.6 multiplied by 10 to the negative 3 m squared. And choice D is 6.7 multiplied by 10 to the negative 3 m squared. Now, since we have a hydraulic press that has a fluid inside of it, and we're applying a uh pressure to this fluid at the input piston and that pressure gets transmitted throughout the entire fluid all the way to the output piston. And so since we have a pressure that's being transmitted through a fluid, we want to use Pascal's principle. So recall our formula for Pascal's principle that is F one divided by 81 is equal to F two divided by A two. So here the F one and F two represent forces and a one and A two are the areas. So we'll take 0.1 to be our input piston and 0.2 to be our output piston. So we're looking for the area of our input piston. So we need to solve this for a one. And so I can do that by multiplying over by a one and dividing um back through by F two, divided by a two. So I'll have a one is equal to F one divided by F two. That fraction is multiplied by A two. Now, I do have a value for F one given to the problem. That's the 420 newtons, but I don't have values for F two and A two. So I'll need to figure out what those um are in terms of the quantities that I was given. So let's look at F two first. So for F two, that is gonna be the force that's being applied to the output piston. Well, the only force that's being acted on that piston is the weight of the heavy crane in addition to the pressure of the fluid below it. So what's pushing down against the fluid is going to be the weight of the crane we'll have F two is just equal to MG, we also need to figure out the area of the output piston. And the only thing I was given was the diameter. And so since I have a diameter, I know that the area is a area of a circle. And so I can write that as a two is equal to. And here I'd have the area circle which is pi R squared. But here I want to write that in terms of the diameter. So I'll have pi instead of radius, I'll have the D divided by two. So I've done diameter divided by two. And that quantity gets squared. So I can now plug um our F two and A two into our formula for a one. So I have a one is equal to F one divided by MG or F two. And this gets multiplied by A two. And for a two, I'll have pi multiplying the quantity of D divided by two. And that quantity is squared. Now, I have values for all the quantities on the right hand side of that equation. So I can plug them in. So I'll have a one equal two F one, we have the 420 newtons and that gets divided by the mass which is the 970 kg. That mass is multiplied by the acceleration of gravity which is 9.80 m per second squared. And then that fraction gets multiplied by pi which is multiplied by the quantity of our diameter which is 44 centimeters. But I'll need to convert that into meters by dividing by 100. That'll be 0.44 m. The diameter gets divided by two. And then that fraction is squared. Now, everything is a numerical value. On the right hand side of that equation, I can plug those into my calculator and I'll get a one is equal to 6.7 multiplied by 10 to the negative 3 m squared. So here I've just kept two significant figures. And if you look what answer choices that corresponds to this corresponds to answer D. So while this was a direct application of Pascal's principle, we did have to write our output force in this case, um RF two in terms of the weight of the crane. And we also had to write to the area of our output piston, which in this case was a two in terms of the diameter and um not the radius. So I hope that this has been useful and we'll see you in the next video.

A hydraulic lift is used to jack a 960-kg car 52 cm off the floor. The diameter of the output piston is 18 cm, and the input force is 380 N. What is the area of the input piston?

Key Concepts

Area of a Circle

Hydraulic Systems

Mechanical Advantage

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