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Math Help - Work Problem

  1. #1
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    Work Problem

    A heavy rope, 50 ft. long, weighs .5 lb/ft and hangs over the edge of a building 120 ft. high.

    The first part of the question asks you to approximate the work by a Riemann sum and then an integral for how much work is required to pull the rope to the top of the building. I understand and got the correct solution for that part. I'm having trouble with part b) which asks:

    How much work is done by pulling half of the rope to the top of the building? Intuitively, I thought it would just be the definite integral from 25 to 50 of the expression (.5x dx) where x is how far the rope is moved. This yields the correct answer of 468.75 ft-lb. However, the solution manual I have solves it in a different way and I have no idea how they did it and would like some help understanding it. Also, because of how the problem was setup, I didn't even know if my answer was an actual way to solve the problem or just a lucky fluke that it came out to be the same value.

    Their solution breaks up the work into W1 and W2, the work done by lifting the top half of the rope and the bottom half of the rope, respectively. The integral for the top is W1 = definite integral from 0 to 25 of (.5xdx), the same expression that I used in my answer, but using different starting points and endpoints. Their integral for W2 = the definite integral from 25 to 50 of (25/2 dx). They then add these two integrals together to get the same answer I did, 468.75.

    I have looked at this for like 20 minutes and do not understand a) why the problem needs to be broken up into two integrals, b) what their axes system is since the top half goes from 0 to 25 and the bottom half goes from 25 to 50, and c) if my solution is actually correct or just correct by accident.

    Any help would be appreciated! Sorry that I don't know how to use the math tags. If anyone could direct me to where that is explained, I would also be obliged.
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  2. #2
    Senior Member
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    The rope hangs.
    x is any point on the rope.
    q mass density of the rope to unit length.
    On the top of the building x=0, whole lenght of the rope is L.
    So 0 <= x <=L.

    The work done to lift dx of the rope at position x is
    dW = dm g x
    where
    dm = q dx.
    So
    dW = g q x dx.

    The work done to lift all the rope is
    W(all)=\int_0^L {dW}=\int_0^L {g q x dx}=gqL^2/2
    The work done to lift upper half is
    W(up)=\int_0^{L/2} {dW}
    The work done to lift buttom half is
    W(bot)=\int_{L/2}^{L} {dW}
    It is clear that
    W(all)=W(up)+W(bot)
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  3. #3
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    That makes sense to me, but maybe I am missing something here.

    The solution that I posted indicates that a different integral is to be used for the bottom half of the rope than the top half of the rope. In the solution you posted, both the integrals for top and bottom have the same function (dW), so there would be no need to separate them since you could just combine them to form the definite integral from 0 to L of dW. In the solution I posted, they use a different integral for the top half than the bottom half. What you refer to as dW is only used for the top half.
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  4. #4
    Senior Member
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    Please draw a graph of f(x).
    Integral is the area under f(x).
    Their solution means the division of this area into two parts.

    W=\int_{L/2}^L {f(x)dx}=triangle+rectangle
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