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Numerical approach to the spring force of a bowstring (Not Hooke's Law)

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Assume the spring force of a bowstring acting on an arrow does not follow Hooke’s Law because the bow’s material becomes more rigid as it is drawn back. The string’s force as a function of the drawback distance in meters is:
    F(x) = -k1x -k2x|x|
    where positive x is along the eventual direction of flight, x=0 in the bow’s relaxed position, and we write x|x| instead of x2 in order to account for the direction of that force term.

    A: If the force is given in Newtons, what must be the dimensions of constants k1 and k2?

    B: In these units, the values for k1 and k2 are 390 and 215, respectively.
    If the force an archer can apply to pull back the arrow is 310 Newtons, what is the pull back length the archer can maintain as she aims the bow?

    C: Assuming the archer holds the bow handle in a fixed position and assuming all other forces acting on the arrow are negligible to that from the bowstring, what will be the release speed of a 34 gram arrow?

    (NOTE: You will need to perform this calculation numerically, similarly to our free-fall with drag problems in class. You should use a time-step small enough such that choosing a time-step of one tenth that value results in less than a 1% change in the answer of release velocity.)

    D: Using your numerical model, calculate and make a graph of the power as a function of time for the arrow release.

    E: Finally, what is the work done on the arrow by the bowstring? You should numerically integrate the power curve to obtain one answer, and compare this answer to that found through the relation between work and change in kinetic energy.




    2. Relevant equations
    A: F=ma
    B: -


    3. The attempt at a solution
    A: What I did here is assume that F has the dimensions [M]*[L]/[T]², so that both k1x and k2x|x| had these dimensions as well, giving k1 the dimension [M]/[T]² and k2 the dimension M/ ([T]²*[L]

    B: I just filled the numbers in and calculated the answer, being x = -0,60.

    C: This is where I start having trouble. How do I approach this? What I did was make an excel spreadsheet, where I calculated F(x) for every X between -0,60 and 0, using steps of 0,01X. I then calculated W(x) for every F(x), using the fact that W(x) is the integral of F(x) * dx. I then summed all these W(x) values and used -W(x) = 1/2mv², finding v. I'm however not sure that this is the correct way of approaching this issue, as I have no idea how to use this to answer question D, and question E sounds a lot like what I already did. Could someone help me out here? I'd be eternally grateful.
     
  2. jcsd
  3. Oct 3, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    It looks like they're expecting you to integrate by time-steps rather than x-steps. This is borne out by part D, where you want a power versus time plot.

    For a 'primitive' integration, for each "tick" of size Δt you can calculate the acceleration a at the current x position from the force function, then Δv = a*Δt and Δx = Δv*Δt. Update t = t + Δt, v = v + Δv, x = x + Δx at each step. Finish when x [itex]\geq[/itex] 0.
     
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