Now pull up again and now one moves amount of rope units upward. This movement moved (approximately) the whole rope upward which means that one applied a force of over distance. Then pull up on the rope with a “jerk” that moves the rope units upward. In other words, start with amount of rope hanging off of the cliff. Then think of the denoting the amount that the rope is moved and the weight of the rope that is being pulled when there is length of rope remaining to be pulled. However, this integral makes no sense with the model that uses as a segment of rope that gets lifted units.īut there is another model that actually conforms more to what we actually feel if we pull up a rope ourselves.įirst, set our variable running from top to bottom with representing the top of the cliff. Now to get to the student’s suggestion, do yet another change of variable: ![]() But the variable is dummy so we can switch back to in the second integral to obtain: Let which means so the second integral gets changed to: We can write this as a single integral if we do a change of variable in the second integral: So the total work is which, written in terms of integrals, becomes: Now let’s make our problem a bit more general: we have a rope of length and we are going to lift feet of it where, of course, Each segment gets lifted a different amount so the integrand becomes To discover why, let us remind ourselves of the model that we are using: to compute the work done by pulling up a rope, one divides the rope into small segments, each of equal weight. So the total work done is which is correct.īut a student suggested the integral which also works. My “instinct” told me to break this into two parts: This will be a diversion from the Lebesgue measure/integration posts.ĭuring today’s question and answer period, a student asked the following question: if one has a 50 foot long rope hanging off of a long cliff and one pulls up 25 feet of rope to the top, how much work does one do? Assume the rope weighs.
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