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:
$$dW=g\left(\rho_1(h-y)+\rho_2y \right)\,dy=g\left(\rho_1h+\left(\rho_2-\rho_1 \right)y \right)\,dy$$
The summation of these increments of work by integration reveals:
$$W=g\int_0^h \rho_1h+\left(\rho_2-\rho_1 \right)y\,dy=g\left[\rho_1hy+\frac{1}{2}\left(\rho_2-\rho_1 \right)y^2 \right]_0^h=g\left(\rho_1h^2+\frac{1}{2}\left(\rho_2-\rho_1 \right)h^2 \right)=\frac{gh^2\left(\rho_1+\rho_2 \right)}{2}$$
Now, since we have:
$$\rho_1=\frac{m_1}{h},\,\rho_2=\frac{m_2}{h}$$
we may write:
$$W=\frac{gh\left(m_1+m_2 \right)}{2}$$
Plugging in the given data:
$$g=9.81\frac{\text{m}}{\text{s}^2},\,m_1=5\text{ kg},\,m_2=\text{ 60 kg},\,h=120\text{ m}$$
we find:
$$W=\frac{\left(9.81\dfrac{\text{m}}{\text{s}^2} \right)\left(120\text{ m} \right)\left((5+60)\text{ kg} \right)}{2}=38.259\text{ kJ}$$
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:
$$dW=g\left(\ell+\rho_2(h-y) \right)\,dy=g\left(\left(\ell+\rho_2h \right)-\rho_2y) \right)\,dy$$
The summation of these increments of work by integration reveals:
$$W=g\int_0^h \left(\ell+\rho_2h \right)-\rho_2y\,dy=g\left[\left(\ell+\rho_2h \right)y-\frac{1}{2}\rho_2y^2 \right]_0^h=gh\left(\left(\ell+\rho_2h \right)-\frac{1}{2}\rho_2h \right)=\frac{gh\left(2\ell+m_2 \right)}{2}$$
Plugging in the given data:
$$g=9.81\frac{\text{m}}{\text{s}^2},\,h=120\text{ m},\ell=44\text{ kg},\,m_2=\text{ 60 kg}$$
we find:
$$W=\frac{\left(9.81\dfrac{\text{m}}{\text{s}^2} \right)\left(120\text{ m} \right)\left(\left(2\cdot44+60 \right)\text{ kg} \right)}{2}=87.1128\text{ kJ}$$