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Can anyone help me figure out this Calculus problem involving work/force?

W

webnewland

Golden Member
Question: A conical tank, with radius 22 ft and height 20 ft, is resting on its base at ground level. Compute the amount of work done in flling this tank with water (density= 62.4 lb/ft3) assuming it is pumped from a reservoir 19 ft below ground level.

I basically using integral and the equation I got was: Integral of 9.8*62.4*(1/3pi(r^2)h) evaluated from -19 to 20.
My answer was huge, 241757105.7 and it was an incorrect answer. Can anyone help me with this problem?

Thanks
 
i coudl help with the work/force part if it was just work/force, cause i can do physics, but not calc :-\ sorry.. heres a bump though
 
Alee25: Is there anyway you can show me how you got it? is it interms of integration? Also, I am not sure of the answer, everytime i submit the answer, it is either correct or deducts 30% for each incorrect attempt (Online HomeWork).
 
ok so what i did was, you have to find the area of one slice of the cylinder, and since the area of one slice will be A=pi(r^2)(delta y).

So now you can use the integral: integral of 0-22, of Pi(22^2)(39-y)(62.4) dy <----- the 39-y is because instead of making the pump below the ground, i raised it to above the ground starting at the top of the cylinder (i think shouldnt make a difference in the answer).

so you have (area x density) = force x a distance = work



------OR
would the answer be integral of 0-22, of pi(22^2)(62.4)dy?

Edit: gimme a minute let me think about this some more, i should be able to come up with a final answer in 4 mins
 
Ok sorry i didn't see the 39 - y part, so you are basically finding the volume of the cylinder using shell method? Do you also need to include Gravity constant of 9.8?
 
no because the integral you are measuing is the series of slices which will only itegrate from 0-22

uhhh... im not using shell eqtion, gravity is not a factor in this

you want to find work, which is force x distance.


if we can find the integral of the cylinder (by using an infinte amount of disks, thus an integral), we can mulitpy it times its density to find a force (as volume (ft^3) x density (Newton/ft^3) will be force (newtons), then to find the distance, we have to find the distance at every disk height (which should be 39-y)

Edit: is said area but i meant volume
 
Okay, first of all, if you do everything in ft and pounds and such, g != 9.8 !!!!!!

g = 9.8 m/s^2 = 32 ft / s^2 (roughly, you may have to redo that)

now W = m g h = m g (39 - x) .... I do this because I want to have an integrating variable that starts from the tip

Now for the mass - this is a little tricky. Funny thing that works, you can assume the cone to be sliced into an infinite number of cylinders (works to calculate the volume of a sphere as well), which will make our life easier. The radius increases linearly, or r = (x/20 * 22), meaning when x = 0 (top) the radius is zero, when x = 20 ft, the radius r = 22 ft...

The change in mass is : dm = ro * (x/20 * 22)^2 * PI * dx (where the part after ro is the infinitesimal cylinder)

dW = ro * PI * g * (39 - x) (x/20 * 22) * dx

Integrating dW, with x = 0 ... 20 leads to:

W = integral (ro * PI * g * (47.19 x^2 - 1.21 x^3) dx)
W = ro * PI * g * (15.73 x^3 - 0.3025 x^4) with x from 0 ... 20 (zero just leads to 0, so I only need to plug in 20)
W = 62.4 * 3.1415 * 32 * (125840 - 48400) = 485777031 lb ft^2 / s^2 = 20.5 10^6 J

Okay, somewhere I think I made a mistake. Those numbers are really big. I'll give it more thought - maybe someone catches my mistake.
 
RSmemphis, your answer has the same numbers of digits as my original answer, but perhaps mine way was incorrect due to the fact that I integrated from -19 to 20. Are you able to spot any errors in your equation so far, it looks very complex and I'm having difficulty grasping it?
 
Ok im prety sure this time i got it:

You gotta find the volume of every slice of the cylinder (in ft^3), and mulitpy it times density (62.4 lb/ft3) to get a force in LBs. The you multiply this force times every height of each slice (and since its pumped from an addtional 19 feet higher (or lower) than the tank, you make it ((20+19) - y)

Since volume of every slice = (r^2)pi(delta Y) which will be equalivient in the integral to be r^2(pi)(dy) <---- thats where the dy comes from in the integral!


So the integral would be 0-22 of ((22^2)pi(dy) (in feet cubed) x 62.4 (lbs per ft cubed) x (39 - y) (in feet)

So the answer i got (and i think this should be right), is 58445772.81 ft lbs

edit: RSMemphis used alot of physics in his answer (and im 99% sure you can do it without knowing gravity because i remember doing these in calc)
 
The problem is, I don't for sure know that I made a mistake.
The way I set up the differential was slightly sloppy, I am not 100% sure if it is right.
If it were multiple choice, I'd probably know, but...

Also, in what unit are you supposed to submit your result?
 


<< ((22^2)pi(dy) (in feet cubed) x 62.4 (lbs per ft cubed) x (39 - y) (in feet) >>



No!

you used a constant radius. That is not correct.

I am actually getting slightly more confident with my answer - but I don't wanna be responsible for a 30% drop in the grade... 🙁

Edit : webnewland, at the very least you need to check my math. Half my brain is with a scientific paper right now, that I want to finish by tomorrow.
Oh and you are right, gravity is necessary, since W = m g h
 
RSmemphis, on our assignment sheets, Our teacher did not specifify the units, I copy and pasted his question here. I'm assuming interms of ft and lbs
 


<<

<< ((22^2)pi(dy) (in feet cubed) x 62.4 (lbs per ft cubed) x (39 - y) (in feet) >>



No!

you used a constant radius. That is not correct.

 >>



opps let me figure this one out again, but you still dont need gravity
 
It won't be a 30% drop in grade, just that question. In that assignment, we have 2 questions each worth 1 mark each. I lost 30% so far by attemping my answer. How confident are you? maybe 75%, loosing 30% is not really a huge deal
 


<< RSmemphis, on our assignment sheets, Our teacher did not specifify the units, I copy and pasted his question here. I'm assuming interms of ft and lbs >>


BTW, g = 32.15 ft/s^2, so if you decide to use my result, plug the numbers in again.

I wish I knew how exactly your teacher expects the answer. I would give 3 digits at most, a la 4.86 x 10 ^ 8 lb ft^2 / s^2
I don't like it when only the answer counts, but not the path to the answer.
When I grade physics, I grade 90% the path, 10% the answer
 


<< opps let me figure this one out again, but you still dont need gravity >>


Yes you do. W = m g h (work = mass * gravity * height)
 
ok i got it...

since the radius is chanig what you do is use similar triangeles.

SO since 22 radius is proportional to a 20 height, then 20/22 = r/y

20r=22y, r = 11/10y
 
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