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Work Needed to Lift a Leaky Bucket


Date: 08/01/99 at 17:03:03
From: Allison
Subject: Calculus, physics applications

Here's the problem:

A cylindrical bucket has height of 1 ft and a radius of .5 ft, and 
weighs 4 pounds. It is attached to a rope of negligible weight. It is 
thrown down a well 100 feet deep and filled up with water to the top. 
There is a hole in the bucket, causing water to leak out at a rate 
that is proportional to the height of the water in the bucket. The 
bucket is being pulled up at a rate of 2ft/sec, and the height of the 
water in the bucket is .9 feet when the bucket reaches the top of the 
well. How much work is being done in lifting the bucket and the water 
to the top of the well? Assume the density of the water is 62 pounds 
per cubic ft.

Here is what I have done:

The bucket holds pi/4 ft^3 of water when full
rho = density of water in lbs/ft^3
density = 16/pi

It takes 50 sec to haul up the bucket, and
the rate of leakage is proportional to the height of the water

     dh/dt = -kh
     dh/h  = -kdt

Separate the variables:

      h(t) = ce^-kt

when t = 0   h(t) = 1, thus c = 1
when t = 50  h(t) = .9,  so k = .002107

x(t) = height of the bucket above the bottom of the well at time t. 
If the bucket is brought up at a rate of 2ft/sec then x(t) = 2t.

Integrate weight:

     4 + (16/pi * e^(-kx/2))dx from 0 to 100.

My final answer is 883.38.

Is this correct? Are my units right? I have done a lot of work, please 
help me, and if I am wrong tell me how and where to correct it.  

Thanks,
Allison


Date: 08/09/99 at 17:02:02
From: Doctor Fwg
Subject: Re: Calculus, physics applications

Dear Allison,

The first part of your calculation is okay, but I probably would have 
written H (as a function of t or x) like this:

     1) H = Ho Exp[-kt] = Ho Exp[-kx/v],

where H is a function of t or x (where t = x/v), v is the lifting 
velocity, and Ho is the water level when bucket is full. In your work 
above you let t = x/2 which is fine also as long as x is measured in 
feet.

For an incremental amount of work (i.e., dW) to lift the bucket 
through a distance of dx, you can write:

     2) dW = F dx = M g dx,

where F (note: F = M g) is the incremental force exerted to lift the 
pail plus water through a small distance of "dx," M is the mass of the 
pail plus water as a function of time, and g is the gravitational 
constant (about 32 ft/sec^2 or 980 cm/sec^2). But, noting that the 
mass of the pail alone is Mp (or 4.0 lb = 1818 g), and using Eq. 1, 
M can be rewritten as:

     3) M = Mp + Rho A H  = Mp + Rho A {Ho Exp[-kx/v]}
          = Mp + Rho A Ho Exp[-kx/v],

where Rho is the water's density (which is, predominantly, a function 
of the water's temperature, but here can probably be assumed to be a 
constant equal to about 62 lb/ft^3 or about 1.0 g/cm^3), A is the 
cross sectional area of the water in the bucket (note: A = 3.14159 
R^2), H is the height of the water in the bucket at time "t" (about 
1.0 ft or about 30.5 cm when t = 0.0 sec), k is a constant, and x/v 
(equal to t) is time. Note: (A H) is the volume of water in the pail 
at time t or height x, and (Rho A H) means density times volume which 
gives the mass of water in the pail at that height or time.

Now, plugging M in Eq. 3, into Eq. 2:

     4) dW = F dx = M g dx = {Mp + Rho A Ho Exp[-kx/v]} g dx.

Eq. 4 can be integrated with W varying from 0 to W and x varying from 
0 to x. Because v is constant, the result is:

     5) W = Mp g {x - 0} - (v/k)(Rho A Ho g){Exp[-kx/v] - Exp[0]}, or:

     6) W = Mp g x - (v/k)(Rho A Ho g){Exp[-kx/v] - 1.0}.

You need to be careful about the units here. It might be best to 
convert to all cgs units. Namely: grams for mass, g = 980 cm/sec^2, 
velocity in cm/sec, k in sec^(-1), Rho in gm/cm^3, and x (and Ho) in 
cm. If you use cgs units, the units of work will be in ergs, where:

     7) 1.0 erg = 1.0 g cm^2/sec^2.

The total work required will be a little less than the work needed to 
lift the mass of the bucket and the pail full of water because only a 
little water leaks out. In other words, the second term in Eq. 6 (i.e. 
 -(v/k)(Rho A Ho g){Exp[-kx/v] - 1.0} must be positive and only a 
little smaller than the term:

     8) Ww = Mw g x,

where, Ww is the work needed to lift only the water, Mw is the mass of 
only the water in the full bucket, g is the acceleration due to 
gravity, and x is height the water is lifted through.

Note: The term Exp[-kx/v] will be less than one (but not negative) for 
all positive values of x. Note also that I am using the convention in 
this problem that x is zero at the bottom of the well and increases in 
the "plus" direction as the pail is lifted.

There is at least one other way to solve this problem using t as the 
integration variable, but the way illustrated above may seem more 
straightforward to you. I hope so and I hope this illustration has 
answered your question.

With Best Wishes,
- Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Physics/Chemistry

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