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Homework Help: Work problem - Rope, pulley and brick (applied integration)

  1. Nov 17, 2016 #1
    • Member warned that the homework template is required
    If a brick is pulled across the floor by a rope thruogh a pulley, 1 meter above the ground - and work = W, where [itex]W = 10N [/itex], (in newton).


    Show that the horizontal component of W, which is pulling the brick has the size
    [tex] \frac{10x}{\sqrt{1+x^2}} (*) [/tex]
    Use this to calculate the amount of work needed to move the brick from x = 10 to x = 2.

    This is what I have so far:
    In terms of the expression (*), I'm thinking the 10x has to do with W being equal to 10N, and the W-side (call it Z) being the hypotenuse: [itex] \sqrt{1+x^2} [/itex]
    ex5d.jpg
    Let the angle by the brick be θ:
    [tex] \cos θ=\frac{x}{\sqrt{1+x^2}} [/tex]

    So is the formula found by taking cosθ×10?

    In terms of calculating work, I'm not sure. I first thought taking the definite integral of (*) from 2 to 10?
     
    Last edited by a moderator: Nov 17, 2016
  2. jcsd
  3. Nov 17, 2016 #2

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    You're doing fine. Yes, the formula is good and since the force varies with x you need to integrate. Carry on !
     
  4. Nov 17, 2016 #3
    Can I integrate it as 10*cosine? Or should I use the original expression using integration by parts (and other methods)?
     
  5. Nov 17, 2016 #4

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    Write down the integral you want to calculate (##\ \int F \, dx\ ##). The integrand can be written as ##\ 10 \cos\theta\ ## but then you still need to do something ...
     
  6. Nov 17, 2016 #5
    Clearly, since [itex]10\int_2^{10}cosθ[/itex] gives a negative number... I have no clue about physics, so I might be missing something about the work?

    In the text, the brick is moving from 10 to 2, so maybe i should do:

    [itex]10\int_{10}^{2}cosθ \approx 14.53[/itex]

    How's that?

    Alternatively, I know that [itex]W = F \times s [/itex] or [itex]W = F(s) [/itex], so I need to find some value for displacement
     
    Last edited: Nov 17, 2016
  7. Nov 17, 2016 #6

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    Once you have the right expression for the integral, there isn't much physics going on anymore: it's pure math.

    What do you make of ##dx## ?
     
  8. Nov 18, 2016 #7
    Ah right - above I just used [itex] dθ [/itex] , but It depends on the integrand, so [itex] dx [/itex] means I need the integrand to be in terms of [itex] x [/itex]. So I should use the expression [itex]\frac{x}{\sqrt{1+x^2}} [/itex], thus [tex]10\int_{10}^2 \frac{x}{\sqrt{1+x^2}} dx[/tex]
     
  9. Nov 18, 2016 #8

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    What I meant is that if you want to use theta as integration variable, you should not forget to express ##dx## in terms of ##\theta##. Perhaps keeping ##x## as integration variable is easier...?
     
  10. Nov 18, 2016 #9
    Yes, that was just a typo on my part - I did integrate using dθ with cos (I did get a number above, which I assume is wrong?) and dx using the x-expression, which I haven't calculated yet, looks like a tricky integral, trig substitution maybe?

    Btw about the limits of integration, from the illustration I made, it looks like we are moving in the negative direction, from 10 to 2 (this is the way it was stated in the problem as well). I know under the fundamental theorem you change the sign when flipping the limits of integration.
     
  11. Nov 18, 2016 #10

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    Turns ot both ways are pretty straightforward (goes to show that I'm also not that comfortable with integrals). Good exercise to actually do it both ways ! That also gives you an extra checking opportunity.

    And you know you have the wrong sign if the work comes out negative :smile:
     
  12. Nov 18, 2016 #11
    [tex]10\int_{2}^10 \frac{x}{\sqrt{1+x^2}} dx[/tex] Let [itex]u=1+x^2, du=2x dx[/itex] so we have

    [tex]5\int_{x=2}^{x=10} \frac{1}{\sqrt{u}} du[/tex]

    [tex]10\sqrt{u} = 10\sqrt{1+x^2} [/tex]

    From [itex]x = 2[/itex] to 10

    [tex]\approx 78.14[/tex]
     
  13. Nov 18, 2016 #12

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    Looks good. Now perhaps ##\ \displaystyle \int 10\cos\theta\, dx \ ## with ##\ \theta = \cot x\ ## :rolleyes: ?

    [edit] Sorry, wrong way around o:) : should of course be ##\ \cot\theta = x \ ##
     
    Last edited: Nov 18, 2016
  14. Nov 18, 2016 #13
    Let [itex]θ= cotx , dθ=-csc^2xdx[/itex] so we have

    [tex]-10\int_{2}^{10} cos(cotx)csc^2x dx[/tex]

    Let u = cotx, [itex] du=-csc^2x dx[/itex]

    [tex]-10\int_{x=2}^{x=10} cos(u) dx = -10(sin(cotx))[/tex] from 2 to 10

    Is that negative? No idea how I would calculate [itex] \sin(\cot x) [/itex], it's a composite function but..
     
  15. Nov 18, 2016 #14

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    Does that yield 78.14 ?
     
  16. Nov 18, 2016 #15
    No, so something's wrong. The answer I got was [tex]1.441\times (-10) = -14.41[/tex]
     
  17. Nov 18, 2016 #16

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    Step from ##\ d\theta\ ## to ##\ dx\ ## :$$
    d\theta = -csc^2 x \, dx \Rightarrow \ dx = ... d\theta $$
     
  18. Nov 18, 2016 #17
     
  19. Nov 18, 2016 #18
    [tex] dθ = -csc^2x dx [/tex]
    [tex] dx = \frac{dθ}{-csc^2x} [/tex]
     
    Last edited: Nov 18, 2016
  20. Nov 18, 2016 #19
    I miscalculated something in the post above, the integral of [tex] \sin u du [/tex] Is equal to
    [tex] 14.4[/tex] not negative. I missed a negative sign from the csc^2. But it's still not 78.14.
     
  21. Nov 18, 2016 #20

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    You need to express ##\ \csc^2 x \ ## in terms of ##\theta##
     
  22. Nov 18, 2016 #21
    Not sure how to do that. At the point θ to dθ, I take dθ/dx and get [itex] -\csc^2x dx[/itex]. Isn't it equivalent to when I use u-substitution, and I express the term I differentiate still using x, with the point of cancelling the already existing x-terms? So same here, I replaced θ with x-terms.

    And also, is 78.14 the wrong answer?
     
  23. Nov 18, 2016 #22

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    You have ##\displaystyle \int \cos\theta\, dx\ ## with $$
    x = {\cos \theta\over\sin \theta}\ \Rightarrow \ dx = -{1\over \sin^2 \theta} \;d\theta $$ and you integrate from ##\operatorname{arccot}10## to ##\operatorname{arccot}2##

    Well compare the two answers ... :wink:

    (the value seems quite reasonable: about 10 N times 8 m would be a first guess when you completely ignore the angle)
     
  24. Nov 19, 2016 #23
    Got it! Got the same answer, [itex] 78.138 [/itex]. That was a strange integral lol. I can't thank you enough!:smile:

    However, when the brick is dragged up, the angle could be calculated, because we now the lenght of the sides ([itex]1, 8, 10.1 [/itex]) by pythagorean theorem. Should I take that into account? Since [itex] 10N \times 10.1 = 101 [/itex]. That is the z side is 10.1 long.

    Edit: Nvm, I was only trying to find the horizontal work.
     
    Last edited: Nov 19, 2016
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