When a second student joins the first, the piston sinks . What is...

Question

When a second student joins the first, the piston sinks . What is the second student's mass?

Accepted Answer

Hi, everyone in this practice problem. We're being asked to determine the mass of a second person that will join our uh hydraulic lift piston system. We will have an 85 kg mechanic balances a 1450 kg car on the hydraulic lift. The picture of the hydraulic lift is shown in the figure and a second person is joining the mechanic and the mechanic's piston moves downward by 38 centimeter. We're being asked to determine the mass of the second person when the incompressible fluid density is 930 kg per meter cube. The diameter of the piston um directly under the car is 1.8 m which is given in the figure. The options given for the mass of the second person are a 44 kg B 138 kg C 1000 and 265 kg. D 53 kg and E 65 kg. So first, in order for us to solve this problem, we wanna notice that the incompressible fluid density or row is 930 kg per meter cube. We take note that at first the two pistons are at the same level. So at the first condition here, the pressure of the left piston will equals to the pressure of the right piston with the left piston carrying the car and the right piston carrying the uh mechanic. When the second person joins the mechanic, the mechanic pistons sinks and the car's piston or the left piston rises. So the height difference between the two piston is going to be represented by each. And when that happens, then our left piston for the car will be on this red line here. And the right piston of the mechanic will be on the red line here. And the difference is going to be H which will equals to centimeter for the right piston. So then the uh the new condition or the second condition will then have the equation of P left plus H row G to be equals to P right, essentially the pressure on the other side of the piston on the left piston exactly at the same point as the new line or each equals 38 centimeter at this position right here. The pressure will equals to P left plus H row G. While the pressure on the right side will still remains P right. So then we wanna recall that um pressure can be calculated by taking the force divided by the area. So we wanna substitute that equation into our equation that we just got. So pressure left equals to F left divided by area left plus H row G equals to F right, divided by area, right. So we will denote all forces and areas using car and mechanic. And the car will be represented with a subscript of C and mechanic will be represented with a subscript of M and AC and AM are going to be the areas of the pistons where the objects are placed but not the area of the objects themselves. So then our equation will then equals to F C for the left divided by AC plus H row G equals to F M divided by AM. But don't forget on the second condition, we have a second person joining. So we will have to plus the right side with F P for the second person divided by A P but A P and AM are the same. So we have the same denominator here. I'm just gonna represent it with AM so that our equation will then become F C divided by AC plus H row G equals F M plus F B of that divided by AM just like. So, so we will have to solve for F B because that is what is gonna give us the mass of the second person. So rearranging this for F B will then get F C divided by AC plus H row G. Um All that in parentheses multiplied by AM minus F M or the force of the mechanic. So that will be the equation for F P or the force of the person. However, um the area for the mechanic is unknown while the area ac for the car is given or can be found by the diameter given which is 1.8 m. So in this case, we will have to use the initial situation or uh condition one where we have P left equals P right. I'm gonna write that down one more time P F equals P right. In order for us to get what the area for the mechanics piston is. So we'll use this equation and the P F as uh similarly, we can uh expand that into F C divided by AC equals B right to be F M divided by AM. And then in this case, we want to be finding an equation for AM which will then equals two uh F M multiplied by AC divided by F C. And in this case, we know everything in this uh equation right here. So we can substitute that information from the problem statement. So that AM will equals to first F M F M is just the weight of the person which is given to be 85 kg in the problem statement. So then F M will equal to 4 85 kg multiplied by 9.81 m per second squared multiply that by AC which will then be using uh just P uh pi R squared and the diameter is 1.8 m. So diameter is 1.8 m. So the R will be 0.9 m or half of that. So then ac will equals to pi multiplied by 0.9 m squared. And then of that is going to be divided by F C and F C is given to B 14 50 kg multiplied by gravitational acceleration. So 14 50 kg multiplied by 9. m per second squared. And solving for this, we will then get the area for the mechanics piston to then be 0.14917 m squared. We can use this um to then calculate and substitute this into our F B formula in order for us to get our F B. Well, so let's uh do that right now. Um Taking the formula again, we will then get F B to be equals to F C divided by AC plus H row G multiplied by AM minus F M. And then we can expand the forces into just M multiplied by G to simplify everything. So then we will then get MP multiplied by G equals two open parentheses. MC multiplied by G divided by AC plus H row G close parenthesis multiplied by AM minus M M multiplied by G. We have GS all over the equation. So we can cross this out. So I'm gonna cross out our GS just like so, and then we can then get an equation for the mass of the second person M B to be equals to open parentheses. MC divided by AC plus H row, close parentheses multiplied by AM minus M M. So we know everything in this formula so we can substitute all of our known information so that MP will equals to open parentheses. MC is the mass of the car which is 14 50 kg in the problem statement. And then AC is the area of the car's piston or the left piston which is going to be by multiplied by 0.9 m squared, which is exactly what we did previously to calculate for the area of the um ma ma piston plus H row H is uh 38 centimeters. So that will be 38 times 10 to the power of negative two m. And then row is given to be 930 kg per meter cube. In the problem statement, close parentheses, we wanna uh multiply that by AM which is uh which we just found to be 0. m squared. And then we wanna subtract that with M M or the mass of the mechanic which is going to be 85 kg. That will actually give us the mass of the second person and B to be 53 kg after doing all the calculations for this one. So the mass of the second person will be 53 kg, which will actually correspond to option D in our answer choices. So option D will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion on this one, please make sure to check out our adolescent videos on similar topics and that'll be it for this one. Thank you.

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When a second student joins the first, the piston sinks . What is the second student's mass?

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