1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the delta x of a bow and arrow pullback

Tags:
  1. Nov 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the delta x of the bow pullback. The weight of the pullback is 45 kg, the speed of the arrow as it shot out was 184mph( i converted that to velocity and got 83 m/s).
    2. Relevant equations
    I'm using K1+U1+W=K2+U2.

    3. The attempt at a solution
    I wasn't sure which formulas to plug into that equation... is it u1+work=k2? or work=k2. Then i didn't know what plugs into u1. work=FxD. please just show me where to start!
     
  2. jcsd
  3. Nov 30, 2014 #2
    So I ended up using work= kinectic energy, because it's the same. I got (444.8xDistance)=1/2(45.36)(82.6^2) then i got a very large distance? was that right?
     
  4. Nov 30, 2014 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The arrow is 45.36kg?!
     
  5. Nov 30, 2014 #4
    No the pull on the bow! Sorry lol... Idk what I should plug in? Does that mean I have to solve for the mass in KE?
     
  6. Nov 30, 2014 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If you don't know the mass of the arrow then you cannot determine the KE achieved. Are you sure you have quoted the whole question, word for word?
     
  7. Nov 30, 2014 #6


    Ya. I think Im missing some pieces.. from what I see towards the end of the video. :D
    136 joules of KE= 1/2m(82.6^2). m=.04kg then i plug that into 444.8xDist.=1/2(.04)(82.06^2)! YES?! The pullback is .3m?!
     
    Last edited: Dec 1, 2014
  8. Dec 1, 2014 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Not quite. The 445 N is the peak draw weight. It will not be constant over the distance. Do you know the formula for PE of a spring?
     
  9. Dec 1, 2014 #8
    It's (1/2)•444.8•Dist.=(1/2)(.04)(82.6^2)
    Yes!!!! Distance=.61meters!!!
     
  10. Dec 1, 2014 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Are you saying that is the correct answer as given by the question setter?
    There is a problem with applying the spring formula to drawing a bow. If we take the arc of the bow as behaving like a spring, the pull on the string does not. To do it more accurately, it would probably be simpler to think in terms of the movement of the ends of the bow, treating the arc as two rods jointed in the middle. When the two rods form an angle ##\pi-2\theta##, the torque at the joint is ##k\theta##. Knowing the draw length, with a bit of trig we can deduce the relationship between the draw weight (the max force applied to the string) and the max torque, and thus determine k. Then the stored energy would be ##\frac 12 k {\theta_{max}}^2##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding the delta x of a bow and arrow pullback
  1. Bows and Arrows (Replies: 3)

  2. Bow and Arrow (Replies: 4)

Loading...