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Work to lift the weight?

  1. Feb 19, 2007 #1
    Work to lift the weight??

    A 1000-lb weight is being lifted to a height 10 feet off the ground. It is lifted using a rope which weighs 4lb per foot and which is being pulled up by construction workers standing on a roof 30 feet off the ground. Find the work done to lift the weight?
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  3. Feb 19, 2007 #2


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    In order to get help, you should show some attempts in obtaining a solution, not to mention using the homework thread.
  4. Feb 19, 2007 #3
    I don't even know where to start this, but my idea is this:

    We have to find the weight of the 1000-lb plus the weight of the rope from 0 feet to 10 feet. Then we can use integrals to calculate it, by dividing small parts of the rope we can find the exact total work done. But then what is the purpose of this question mentioning that the worker is 30 feet from the ground?
  5. Feb 19, 2007 #4
    You need to know how high the workers are to figure out how much work is done in lifting the rope itself too (all 30 ft initially), which gets shorter as the workers pull on it. Since work is a scalar, split it into work done on pulling the weight (no integration necessary) and work done in pulling the rope (integration needed).
  6. Feb 19, 2007 #5
    so you mean here that the total work is the work to pull the weight 10 feet and the work of pulling the rope also?
  7. Feb 19, 2007 #6
    I'll give you a couple hints (in the form of questions). As you said, it's obvious that you need to use integration to solve this problem. You know that work is the integral of force with respect to distance. How can you use the linear weight density of the rope (4 lbs/foot) to find a the force required to lift the weight as a function of the distance of the weight from the ground? What will the mass of the weight-rope system be when it is on the ground, and when it's been lifted ten feet into the air? Also, how do you know what limits of integration to use?
  8. Feb 19, 2007 #7
    How can you use the linear weight density of the rope (4 lbs/foot) to find a the force required to lift the weight as a function of the distance of the weight from the ground?

    This is the question that I've been thinking of, but how?? I really have no idea
  9. Feb 20, 2007 #8
    To get the weight of the rope you multiply it's length by it's linear density.
    Add this to the weight of the object attached to the rope and you have the total weight.
  10. Feb 20, 2007 #9


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    However, the crucial point is that, as the rope is lifted, the amount of rope still be be lifted- and so the weight- varies. That's why you need an integral. You can't just multiply the weight of the rope by 10 ft.

    Since the lifters are initially 30 feet above the ground, they start having to lift 30 feet of rope. As they lift however, the length of rope reduces. After having lifted the weight x feet, the length of rope still being lifted is 30- x feet and so the weight of rope to be lifted is 4(30-x). The work done in lifting that a distance "dx" is 4(30-x)dx. Integrate that from 0 to 10 to find the work done lifting the rope.

    Of course, the work done in lifting just the weight is 1000*10= 10000 ft-lbs.
  11. Feb 20, 2007 #10
    so the total work will then be 10000 ft-lbs + the integral from 0 to 10 of 4(30-x)dx??
  12. Feb 21, 2007 #11


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    Yes, that is correct. Now do it!
  13. Mar 3, 2007 #12
    I have been working at a problem that is simmilar to this one, and I do not understand why wouldnt one integrate the weight of the cable+the weight?

    The way I was doing this problem, and I did not get the right answer I want to understand why we are doing it this way, is that I got the total weigh of the cable and added the weigh of the weight and multiply by x b/c x is a piece of my cable and then Integrate that.

    weight= 1000 lbs. cable=4 lbs/ft. height=30ft

    (4lbs/ft)(30ft)=120 lbs. (total weight of cable.)
    120 lbs.+1000 lbs.= 1120 lbs. (total weight)
    \int_{0}^{30} 1120x dx
    Why doesn't this approach give me the right answer?

    Edit: I got it. One cannot do that because the weigh of the weight is constant. the weigh of the weight does not change as it is raised.
    Last edited: Mar 3, 2007
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