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MathHelp's question at Yahoo! Answers regarding the computation of work done hauling ropes

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MarkFL

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Feb 24, 2012
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Here is the question:

Calculus: Work problems help needed?

Need help on this work problem

You are on top of the roof of 120 meter tall building with a 120 meter rope weighing 5kg. Another person on the ground at the base with another 120 meter metal rope weighing 60kg and a bucket with weight of 4kg holding 40kg of sand. Acceleration of gravity is 9.81 m/sec^2.

a) You drop one end of the rope off the building and the other person attaches the end of his metal rope to your rope. How much work do you do pulling the rope back up the roof so that the metal rope is dangling from the edge of the roof down to the ground.

b)Your friend then attaches the end of the metal rope to the bucket. How much work do you do pulling bucket and the metal rope up to the roof?
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Re: MathHelp's question at Yahoo! Answers regading the computation of work hauling ropes

Hello MathHelp,

Let's let:

$h$ = the height of the building.

$\rho_1$ = the linear mass density of the less dense rope.

$m_1$ = the total mass of the less dense rope.

$\rho_2$ = the linear mass density of the more dense rope.

$m_2$ = the total mass of the more dense rope.

$\ell$ = the mass of the load hauled at the end of the second rope in part b).

a) Once the two ropes are fixed together, the person at the top of the building begins hauling up the two rope system, with the lighter rope essentially being replaced with the heavier rope. Let's orient a vertical axis $y$ with the origin at the ground. Initially, the point where the two ropes are attached is at the origin.

At some arbitrary point in time during the process of hauling the rope up, this point of attachment is at the point $y$, where $y$ is the length of heavier rope being suspended, and $h-y$ is the length of lighter rope being suspended. The incremental amount of work to move this point up a height $dy$ is then given by:

\(\displaystyle dW=g\left(\rho_1(h-y)+\rho_2y \right)\,dy=g\left(\rho_1h+\left(\rho_2-\rho_1 \right)y \right)\,dy\)

The summation of these increments of work by integration reveals:

\(\displaystyle W=g\int_0^h \rho_1h+\left(\rho_2-\rho_1 \right)y\,dy=g\left[\rho_1hy+\frac{1}{2}\left(\rho_2-\rho_1 \right)y^2 \right]_0^h=g\left(\rho_1h^2+\frac{1}{2}\left(\rho_2-\rho_1 \right)h^2 \right)=\frac{gh^2\left(\rho_1+\rho_2 \right)}{2}\)

Now, since we have:

\(\displaystyle \rho_1=\frac{m_1}{h},\,\rho_2=\frac{m_2}{h}\)

we may write:

\(\displaystyle W=\frac{gh\left(m_1+m_2 \right)}{2}\)

Plugging in the given data:

\(\displaystyle g=9.81\frac{\text{m}}{\text{s}^2},\,m_1=5\text{ kg},\,m_2=\text{ 60 kg},\,h=120\text{ m}\)

we find:

\(\displaystyle W=\frac{\left(9.81\dfrac{\text{m}}{\text{s}^2} \right)\left(120\text{ m} \right)\left((5+60)\text{ kg} \right)}{2}=38.259\text{ kJ}\)

b) This time, the load is decreasing as the rope hauled up is not replaced. So we may state the increment of work with the bucket at $y$ is:

\(\displaystyle dW=g\left(\ell+\rho_2(h-y) \right)\,dy=g\left(\left(\ell+\rho_2h \right)-\rho_2y) \right)\,dy\)

The summation of these increments of work by integration reveals:

\(\displaystyle W=g\int_0^h \left(\ell+\rho_2h \right)-\rho_2y\,dy=g\left[\left(\ell+\rho_2h \right)y-\frac{1}{2}\rho_2y^2 \right]_0^h=gh\left(\left(\ell+\rho_2h \right)-\frac{1}{2}\rho_2h \right)=\frac{gh\left(2\ell+m_2 \right)}{2}\)

Plugging in the given data:

\(\displaystyle g=9.81\frac{\text{m}}{\text{s}^2},\,h=120\text{ m},\ell=44\text{ kg},\,m_2=\text{ 60 kg}\)

we find:

\(\displaystyle W=\frac{\left(9.81\dfrac{\text{m}}{\text{s}^2} \right)\left(120\text{ m} \right)\left(\left(2\cdot44+60 \right)\text{ kg} \right)}{2}=87.1128\text{ kJ}\)