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Finding the work done on a cutting tool

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  1. Feb 17, 2015 #1
    1. The problem statement, all variables and given/known data
    A cutting tool under microprocessor control has several forces acting on it. One force is F⃗ =−αxy2j^, a force in the negative y-direction whose magnitude depends on the position of the tool. The constant is α = 2.70N/m3 . Consider the displacement of the tool from the origin to the point x= 2.95m , y= 2.95m .

    Part A
    Calculate the work done on the tool by F⃗ if this displacement is along the straight line y=x that connects these two points.

    Part B
    Calculate the work done on the tool by F⃗ if the tool is first moved out along the x-axis to the point x= 2.95m , y=0 and then moved parallel to the y-axis to x= 2.95m , y= 2.95m .


    3. The attempt at a solution
    So I was simply trying to use the W=force*distance, and I tried to use the vector equation for force by plugging in the coordinate that it would end up at and then for distance I used the distance from the origin to the point (2.95, 2.95) for the first part, and then I broke Part B into two parts, calculating work from origin to (2.95, 0) and then work from there to (2.95, 2.95).

    It didn't work, and it is due by 11:59 PM. I am really confused. (Please, if you use calculus in your replies, please explain in detail. My calculus is very weak right now...)
     
  2. jcsd
  3. Feb 17, 2015 #2

    haruspex

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    Please post your working. It is hard to say where you went wrong from merely an outline of what you did.
    Also, this is very much an international forum. I don't know what time zone you are in.
     
  4. Feb 17, 2015 #3
    Hello

    I am in the Eastern Time Zone by the way, so about 51 minutes left (!)

    For the first one, I took W=F*D and took F⃗ =−αxy^2 j-hat, plugged in α=2.70, x=2.95, y=2.95, and then multiplied that by the sqrt(2*2.95^2). I got -289.

    For the second one, I did the same thing but in two parts. The first part I plugged in α=2.70, x=2.95, y=0, and then multiplied that by 2.95. Then I added the product of the next part, which I plugged in α=2.70, x=2.95, y=2.95, and multiplied that by 2.95.

    The homework system tells me I am wrong. Please correct me.
     
  5. Feb 17, 2015 #4

    haruspex

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    Substituting those coordinates in the expression for F gives you the force at those coordinates. You need to integrate the force along the path to those coordinates.
     
  6. Feb 17, 2015 #5
    I do not know how to integrate the force along the path...Could you explain in detail?
     
  7. Feb 17, 2015 #6

    haruspex

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    Suppose you move a distance dx in the x direction. To stay on that path you must also move dx in the y direction. Express that as a vector. The work done by the force is the dot product of F with that vector. Compute that dot product and integrate over the range of x.
     
  8. Feb 17, 2015 #7
    Please look at my math

    Could I write it as the integral from 0 to 2.95 of (-2.7*x*y^2) dx?
     
  9. Feb 17, 2015 #8
    Or would it be the integral of -2.7*y^2*dx from 0 to 2.95?
     
  10. Feb 17, 2015 #9

    haruspex

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    In the OP you wrote ##\alpha xy2\hat j##. I gues that meant y2. If so, yes, but remember that y is a known function of x along this path.
     
  11. Feb 17, 2015 #10
    I do not know what you mean by y being a known function of x. Was my first integral correct, and then if I plug in the value of that integral into the W=F*D equation, then it works?
     
  12. Feb 17, 2015 #11

    haruspex

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    What does it tell you about the path in part A?
     
  13. Feb 17, 2015 #12
    OH

    so y=x. Then for my final integral, it is the integral of x^3 dx from 0 to 2.95? (I may have that completely wrong.)
     
  14. Feb 17, 2015 #13

    haruspex

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    Yes, not forgetting alpha.
     
  15. Feb 17, 2015 #14
    So it would be the integral of -2.7x^3 dx from 0 to 2.95

    Then for part B,

    is the first integral -2.7*0*dx from 0-2.95? and then the second one...since it is dy, then I could do -2.7*x*2y*dy from 0 to 2.95?
     
  16. Feb 17, 2015 #15

    haruspex

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    I don't understand the 2y. What happened to y2?
     
  17. Feb 17, 2015 #16
    I thought since it was dy, I would not so anything to the x, but the change would happen to the y part, unless I was supposed to change it to (1/3) y^3? My calculus is really rusty
     
  18. Feb 17, 2015 #17
    I calculated the integral for the first part and got -51.12, and then multiplied that by the sqrt(2*(2.95^2)) and got -213.3. It is still wrong. I do not know what to do.
     
  19. Feb 17, 2015 #18

    haruspex

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    For this part of the question, it's the same algebraic form of F, but as you move up from y=0 to the finishing point x is constant. So, yes, you should get y3/3 in the result.
     
  20. Feb 17, 2015 #19
    Okay then. But for part A, was the answer that I got in joules? Am I supposed to do another calculation to get joules?
     
  21. Feb 17, 2015 #20

    haruspex

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    Why did you multiply it by anything? What quantity do you think the result of the integration is giving you?
     
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