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Homework Help: Work of Varying X Component by Integration

  1. Nov 9, 2011 #1
    1. The problem statement, all variables and given/known data
    A 76-kg crate starts from rest at A and is moving at 8 m/s. Using integration, find the work done between A and B of the variable x component of force F= 756 N. (See attached jpeg for the geometry of the problem)

    EDIT: We are doing a variation of the problem shown in the jpeg below, the only difference being the length from A to B (5 m instead of 6 m) and the vertical position of the pulley (5 m instead of 6 m).

    2. Relevant equations
    U1-2= ∫F cos θ dx

    3. The attempt at a solution
    Since F is constant and cos θ is variable, I've tried to relate cos θ in terms of the displacement in x. Since cos θ = Δx/Δ(Length of rope), and because the vertical displacement of the pulley remains constant (5 m), I obtained a value of:

    cos θ = x/(x2 + 25)-1/2

    Whenever I solve this integral I end up with a value equal to the magnitude of the force multiplied by a displacement of 7.5 m in x which, from the geometry of the problem, is impossible. Obviously, I'm just not looking at this correctly. Any hints would be very helpful, and thank you in advance.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Nov 9, 2011 #2

    Doc Al

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    Staff: Mentor

    Looks good. (Except for the minus sign in front of the 1/2. I assume that was a typo.)
    What did you get when you integrated? What were your limits of integration?
  4. Nov 9, 2011 #3
    0m5m (F cos θ) dx

    = F∫0m5m (cos θ) dx

    = F∫0m5m [x(x2 + 25)-1/2] dx

    = F∫0m5m [1 + (x/5)] dx

    = F{∫0m5m dx + 1/5∫0m5m x dx}

    = F{[x + x2/10]5m0m}

    = F{[(5 m) + (5 m)2/10] - [0]}

    = F(7.5 m), where F= 756 N

    = 5670 J

    This just seems to be too high a value to me. Since the x component of F is always decreasing and can never be more than F itself and, furthermore, is not applied for a displacement greater than 5 m, I feel this value should be lower than this. Maybe I'm wrong, but I can't figure out where it is exactly that I'm going wrong.

    Sorry if my notation seems a little sloppy, I've never tried to express an integral via digital media before. I hope it makes sense.
  5. Nov 9, 2011 #4

    Doc Al

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    Staff: Mentor

    OK. Are the limits 0 to 5 or 2 to 7?

    :confused: How did you get this?
  6. Nov 10, 2011 #5
    I wish there was an emoticon for an embarrassed slap to one's own forehead. I can't believe I made such an amateurish move. I guess I somehow convinced myself that (a + b)x was equal to ax + bx and then multiplied both terms by x. I was just trying to simplify my integral before trying to integrate.

    Now I feel better for knowing where I was going wrong, but am left with the task of integrating the square root of a sum of squares. I don't even know where to begin on that one. Maybe some sort of trig substitution?

    My teacher said that it was fairly simple integral and now I feel like a moron for not being able to grasp it. I feel like the answer is right there in front of me and I'm just not seeing it...:confused:

    EDIT: The upper limits are from A to B (in the picture I provided), or from 0m to 5 m so far as I can tell.
  7. Nov 10, 2011 #6

    Doc Al

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    Staff: Mentor

    Nah, nothing that complicated.

    You need to find the antiderivative of [tex]\frac{x}{\sqrt{x^2 + a^2}}[/tex]

    Play around with it a bit. (Or just look it up!)
    In your diagram, B is not directly under C.
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