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Math Help - Application of integration : Work problem

  1. #1
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    Application of integration : Work problem

    A chain of mass 20 kg and length 10 meters initially lies on the ground. Let P be the fixed point on the chain which is 7 meters from one end of the chain.Assuming frictional force is negligible use integration in order to determine the minimum amount of work done in lifting chain at point P so that the final height is a height of 15 meters above ground level.


    My attempt :


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  2. #2
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    Quote Originally Posted by nyasha View Post
    A chain of mass 20 kg and length 10 meters initially lies on the ground. Let P be the fixed point on the chain which is 7 meters from one end of the chain.Assuming frictional force is negligible use integration in order to determine the minimum amount of work done in lifting chain at point P so that the final height is a height of 15 meters above ground level.
    I don't know anything about physics; however, I know the math. If the chain is 10 meters long and weighs 20kgs, and we assume the mass to be uniformly distributed, then the mass per linear displacement is 2kgs/meter. Now, let's analyze this problem with Calculus. Essentially, calculus works through taking dynamic relationships down to a level where they are static. At this level, we can construe things as being constant. We evaluate these little pieces and put them together to arrive at an answer for the event as a whole. This, in essence, is the concept of integration.

    So, work is force over a distance. Well, the mass of every infinitesimal piece of chain is simply 2dx kgs. What about the distance? If I read your problem correctly, you want to take the point 7 meters from the end and lift it so that that point is 15 meters above the ground which means that the bottom of the chain itself is 8 meters above the ground. Each infinitesimal unit of chain is therefore lifted 18-x units in the air. Integrate from 10 to 3 because the first 3 meters of chain have a different distance function. The answer is 161 I didn't take gravity into account. Again, I don't know the physics. Now, the first 3 meters of chain need to lifted as well. They have a different distance function. Theirs is 12+x. Integrate that function with 2 as a constant mass factor from 3 to 0 and get 81 for an answer. Add the two 242 is your answer without gravity as a factor.
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  3. #3
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    Quote Originally Posted by Bilbo Baggins View Post
    I don't know anything about physics; however, I know the math. If the chain is 10 meters long and weighs 20kgs, and we assume the mass to be uniformly distributed, then the mass per linear displacement is 2kgs/meter. Now, let's analyze this problem with Calculus. Essentially, calculus works through taking dynamic relationships down to a level where they are static. At this level, we can construe things as being constant. We evaluate these little pieces and put them together to arrive at an answer for the event as a whole. This, in essence, is the concept of integration.

    So, work is force over a distance. Well, the mass of every infinitesimal piece of chain is simply 2dx kgs. What about the distance? If I read your problem correctly, you want to take the point 7 meters from end and lift it so that that point is 15 meters above the ground which means that the bottom of the chain itself is 8 meters above the ground. Each infinitessinal unit of chain is therefore lifted 18-x units in the air. That's the integral. The answer is 161 I didn't take gravity into account. Again, I don't know the physics.

    Point P is at 3 meters and we are lifting it to a height of 15 meters from ground level. The answer in the book is 2300 joules
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  4. #4
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    Quote Originally Posted by nyasha View Post
    Point P is at 3 meters and we are lifting it to a height of 15 meters from ground level. The answer in the book is 2300 joules
    Please, read my amended post above. I only integrated the larger portion of the chain and not the entire chain. The answer in your quote is incomplete. I computed an answer of 242 without taking into account gravity as I am unsure about the physics. OK I added gravity at 9.8 as a factor to the two integrals and arrived at an answer of 2371.6. Your book must have rounded.
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