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

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<<

<< 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) >>




no work aslso equals force x distance
 


<< How confident are you? maybe 75%, loosing 30% is not really a huge deal >>


W/o knowing the professor, I do not know.
I think that most of my calculation is ok. I don't know about the final numbers. I don't know what format. Giving every single digit makes no sense, because they are not that accurate.

I am sorry, but w/o knowing more about people normally submit huge numbers like this, I can't even guess if I would submit my answer like that.
Also, I hate submitting non SI values.

Sorry, pal, I could only help so far.
 


<< ok i got it...
since the radius is chanig what you do is use similar triangeles.
20r=22y, r = 11/10y >>


Right - could have gotten that from my post (okay, it was hidden)
 
so i get integral of 0-22 of (11y/10)^2 (dy) (62.4) (39-y)

and i get 18943016.54 ft lbs


edit: heh "W = 62.4 * 3.1415 * 32 * (125840 - 48400) = 485777031 lb ft^2 / s^2 = 20.5 10^6 J" not that this really has alot to do with the answer but joules is measured in newton meters (not in lb ft^2 / s^2)

just me being picky
 


<< edit: heh "W = 62.4 * 3.1415 * 32 * (125840 - 48400) = 485777031 lb ft^2 / s^2 = 20.5 10^6 J" not that this really has alot to do with the answer but joules is measured in newton meters (not in lb ft^2 / s^2) >>



I converted the lb ft^2/s^2 into kg m^2 / s^2 (check the math) which is equal to J
I did not know what unit he wanted it in.

Okay, web, I am pretty sure my derivation was a little sloppy. I would need to do a full differential of dW/dx...
I am gonna try that now.
Edit: Actually, that makes the thing rather nasty, and I really need to do work now. Lemme know the final answer, if you get a chance.
 


<<

<< no work aslso equals force x distance >>


True, but you have no force other than F = m * g

W = F * h = m * g * h >>



true but you dont need that for this problem, you see force is a mesaure of either in lbs or netwons or etc? aggreed?

so if we have a density in lbs/ft^3 times a volume in Ft^3, when you multiply them you get LBS (a force)!!
 


<< opps ive been integrating from 0-22 (the radius) it should be 0-20

answer i get is : 15180979.95 >>


plus you forgot to multiply with pi.
And a volume times a density is a mass, and not a force.
But at least the general math of our approaches is similar now. 🙂

Edit: okay, you figured the PI in, and you are right about mass = force in English
We agree now! Whippeee!
 
no i did mulity by pi i just forgot to type it in,

volume x density = force,

if it was a mass it would have to be in KG (correct?) and there to get that volume would have to be in KG, and density would have to be in KG, not possible
 
So, I get 15109705.4 now, too

Sorry, Alee, I needed to get what you meant. You are absolutely right with pounds = pounds-force

Hmm, maybe this is the right answer?
 
Ok, Thanks for all your help RSMemphis, I REALLY appreciated your work on this quesiton, much more than I could have asked for. I will definitly let you know the answer once I got it. I also gave you a 10 on your rating for the hard work 🙂
 


<< Ok, Thanks for all your help RSMemphis, I REALLY appreciated your work on this quesiton, much more than I could have asked for. I will definitly let you know the answer once I got it. I also gave you a 10 on your rating for the hard work 🙂 >>



Same to Alee25, because he caught my mistake... Big thanks to him - I knew there was one.
 


<< So, I get 15109705.4 now, too

Sorry, Alee, I needed to get what you meant. You are absolutely right with pounds = pounds-force

Hmm, maybe this is the right answer? >>




were are still off though....

i remember learning it this way in calc. all you need is a density x volume x distance to get work.

webnewland: you didn't put in the answer yet did you?
 


<< I tried 485777031, appeareant that was not right, So both of you got the same answer now using different ways? >>


Sorry you tried my first number. I wish you hadn't....

No, our approaches were the same.
 
could the difference be acountable for the conversion of gravity to feet (in the way you did it)

edit nevermind
 
I was thinking perhaps the problem is not as hard as we are making it to be. Cause for the second problem in this assignment, It was following.

A chain of length 6 m and of a uniform density 3 kg/m is hanging over the edge of a building. Determine the amount of work required to pull the chain up to roof level. Assume the constant of gravity is g = 9.8 m/sec2

The answer was simply integral of x*9.8*3 evaluated from 0 - 6

Perhaps there is a rather simple solution to the pump problem?
 


<< I was thinking perhaps the problem is not as hard as we are making it to be. Cause for the second problem in this assignment, It was following.

A chain of length 6 m and of a uniform density 3 kg/m is hanging over the edge of a building. Determine the amount of work required to pull the chain up to roof level. Assume the constant of gravity is g = 9.8 m/sec2

The answer was simply integral of x*9.8*3 evaluated from 0 - 6

Perhaps there is a rather simple solution to the pump problem? >>




no this problems quite different for several reasons: 1. the radius is constantly changing, 2. your are pumping the water higher from a place than what exists inside the cone,
 
I tried 15180979.95, not correct. Hmmm I just noticed that our homework system won't go past 50%, So I guess I have unlimited tries to do this now. Any other guesses, lol, i'll just shoot them in
 
really? damn, im really sorry about that. Im looking back at my old calc notes, and thats exactly the way i did it before
 
D'oh! That sucks.
I was hoping that since Alee and I got the answer idependently, it would be ok. 🙁

EDIT: I thought through an alternative way using hydrostatic pressure, but I came up with the same result.
 
how bout this answer....

2530163

or

21506388

damn if these work, that would suck.

im pretty sure our prblem is with the object pumping water UP 19 feet below the ground.
 
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