A 55 kg cheerleader uses an oil-filled hydraulic lift to hold four 110 kg football players at a height of 1.0 m. If her piston is 16 cm in diameter, what is the diameter of the football players' piston?

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Hey, everyone. Let's go through this practice problem. A mechanic standing on a hydraulic lift supports 3 kg motorcycles at a height of 1.2 m above the mechanic's piston. The hydraulic lift is filled with oil and the mechanic's piston has a radius of six centimeters. Calculate the radius of the piston lifting the motorcycles. If the mechanic's mass is 72 kg, assume that oil has a density of 900 kg per cubic meter were given four, multiple choice options to choose from. Option A 0.601 m. Option B 0.548 m, option C 0.147 m and option D 0.134 m. So the first thing I'm going to do is draw a sketch of the hydraulic lift. So there's going to be a thinner piston on one side and a thicker piston on the other. And I'm going to label the small piston as A and the large piston as piston B. We're told that this smaller piston has a radius of six centimeters. So that has a radius of R sub A and the larger piston has the unknown radius. R sub B that we're trying to find. Lastly, we're told that when the mechanic stands on piston, A piston B rises by some height that I'm going to call H and so there's going to be a pressure difference due to the height and that H is given to us in the problem as 1. m. Now, the key to this problem is to know that at any point where the heights are the same, the pressures are also the same. So for convenience, I'm going to focus on the pressures right at the height of piston A. So that way we can separate the pressure difference due to the additional height of piston B separately. But piston, but the pressure at piston A P sub A is equal in magnitude to P sub B or the pressure on the piston B side. That is at the same height as piston A. Since we're being asked to find the radius of the large piston, we can relate that to the pressure using the basic pressure formula and recall that it states that the pressure on the surface is equal to the force divided by the surface area over which that force is distributed. For P sub A, the pressure on piston A is going to be equal to the force acting on piston A divided by the surface area of piston, A for piece, sub B, it's almost the same idea. It's the force acting down on piston. B divided by the surface area of piston B. But since there is a height difference, we also have to account for the pressure added due to that height difference in the fluid. And recalled the pressure difference due to a change in height within a fluid is equal to the density of the fluid multiplied by the gravitational acceleration. G multiplied by that height difference. That is a pressure difference due to a height. So we'll have to add that to P sub B since that's contributing to what's causing the pressure at the, the line that's at the height of piston A. So, so P sub B, the pressure piston D is F sub B divided by A sub B plus the density of the oil multiplied by the gravitational acceleration multiplied by the H that was given to us earlier. Since we've already established that P sub A and P sub B should be equal, we can set them equal to each other and then expand out with their formulas. So P sub A can be rewritten as F sub A divided by a sub A and it's equal to F sub B divided by a sub B plus row GH. Now, in order to solve this for radius, in terms of the variables we've been given. There are two things we need to recall. First recall that the surface area of a circle is equal to pi multiplied by the square of the radius of the circle. So let's incorporate this into our force equation. The other thing we need to be aware of is what the force acting on the pistons are. Since the forces are due to the weights of the mechanic and the motorcycles, we'll multiply the mass of each of those things acting on each piston multiplied by the gravitational acceleration. G. So instead of F sub A, I'm going to write M sub a multiplied by G. And we're given M sub A in the problem since we're told what the mass of the mechanic is, and that's gonna be divided by the surface area of piston A which is pi multiplied by the radius of piston A squared. And this is equal to the same thing before the piston side for, for the piston B. So it's going to be, and the problem tells us the mass of one motorcycle, but we are told that there are three motorcycles resting on piston M. So I'm going to say that the mass acting down on piston M is three multiplied by M sub B, the mass of one motorcycle. This is multiplied by G and then divided by pi multiplied by R sub B squared. Then this is plus the density of the oil multiplied by the gravitational acceleration multiplied by the depth. Now one thing we can do to simplify this right off the bat is cancel out the gravitational acceleration since it appears in every single term. Since everything in our equation is now known, except for our sub B, the radius, we're trying to find. The last step is to algebraically solve this equation for our sub B. And with the way this equation is set up that can be a little tricky. So the first thing I'm going to do is get our sub B into the numerator by multiplying all three, both sides of the equation, all all three terms by pi R sub A squared. So the leftmost term just becomes a simple M sub A and this is equal to three MC B R sub A squared. The pies have now canceled out that's divided by R sub B squared, all plus pi R sub A squared multiplied by the density multiplied by the height difference. For the next step, I'm going to get the term that contains R sub B on its own by subtracting from both sides of the equation. The pi R sub A squared row H term. So we find that three M sub BR sub A squared divided by R sub B squared is equal to M sub A minus pi R sub A squared row H and all that's left for us to do to solve for R sub B squared is multiply both sides of this equation by R sub B squared. And then divide both sides of the equation by M sub A minus pi R sub A squared row H. So the next thing we find is that rab squared is equal to three M sub B R sub A squared divided by M sub A minus pi R sub A squared row H. And then to solve for just R sub B on its own, we'll take the square root of both sides. So R sub B is equal to the square root of three M sub BR sub A squared divided by M sub A minus pi R sub A squared row H. And this is the final equation that we'll use to solve this problem. Notice that everything in the equation, all the variables have been given to us where M sub B is given to us in the problem as 100 kg, three multiplied by 100 and 20 kg multiplied by the square of radius A. So that's six centimeters or 0.06 m. That's squared all divided by M sub A or 72 kg minus pi R sub A squared multiplied by the density of oil which is given to us in the problem as 900 kg per cubic meter multiplied by the height difference which is given to us as 1.2 m. So now if we put this into a calculator, then we find a radius of about 0.147 m. And so that is the answer to this problem. And if you look back back up at our multiple choice options, we can see that option. C says exactly that 0.147 m. So that is the answer to this problem. I hope this video helped you out if you'd like more experience, please check out some of our other tutoring videos which will give you more practice with these types of problems. But that's all for now. I hope you all have a lovely day. Bye bye.

A 55 kg cheerleader uses an oil-filled hydraulic lift to hold four 110 kg football players at a height of 1.0 m. If her piston is 16 cm in diameter, what is the diameter of the football players' piston?

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Key Concepts

Hydraulic Systems

Pressure

Area of a Circle