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How Much Force is Needed?-Part Three.

  1. Jan 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Bighorn sheep have extremely tough foreheads. A 200 pound bighorn sheep is running at 25 mi/h. (Converted to 91 kg.) He slams into his opponent and comes to a stop in 1 cm.(Converted to 0.01 m.) What force was exerted on his forehead?


    2. Relevant equations

    a=v[itex]^{2}[/itex]-v[itex]_{0}[/itex][itex]^{2}[/itex]/2Δx
    F=ma
    3. The attempt at a solution
    Using the following equation I got an deceleration of -6050 m/s[itex]^{2}[/itex] Using F=ma, I got an answer of -550550 Newtons. This seems wrong, as the deceleration of a car going 100 mi/h is less than that. Where did I go wrong?




    Thanks! :smile:
     
  2. jcsd
  3. Jan 9, 2014 #2

    mfb

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    Cars don't stop within 1cm.
    WolframAlpha confirms your answer (with a small deviation - rounding error?), assuming the whole Bighorn is as tough as its forehead.
     
  4. Jan 9, 2014 #3
    Oh, I see! That's similar to what I was thinking, thank you! Was the rounding error my calculator's, or WolframAlpha's?
     
    Last edited: Jan 9, 2014
  5. Jan 9, 2014 #4

    mfb

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    WolframAlpha is very precise - if a digit is displayed, you can be nearly sure it is right. Even if there are hundreds of them (click on "more digits" a few times).
     
  6. Jan 9, 2014 #5
    Ah, okay. :) I wasn't sure which you meant. Silly calculator, I put the entire answer in that it gave me. XD Since my physics teacher most likely would use a calculator to check my answer, which answer do you think I should go with?

    Thanks again!
     
  7. Jan 9, 2014 #6
    I suggest trying to solve problems on your own and see how things work as you progress through the calculations. If you plug it into an online calculator, sure you will get your answer, but is it really only the answer you are interested in?
     
  8. Jan 9, 2014 #7
    I didn't use an online calculator, mfb did to check my answer, and we found a slight rounding error. :) That's why I was asking about which answer to use, because there was a one-digit difference and I wasn't sure which was better.
     
  9. Jan 10, 2014 #8

    mfb

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    The difference is roughly 2%. Either you used an imprecise conversion factor for the units or you rounded very imprecise. The latter reason would be problematic, the first reason is probably not an issue. Either way, don't blame the calculator, it does calculate several more digits than you need.
     
  10. Jan 10, 2014 #9
    Haha, I didn't mean to imply that it was the calculator's fault, I was saying that I didn't do any kind of rounding. Maybe I'll try the unit converting again, but I think I did use an online calculator for that because I already have a pretty good grasp on it, and it saves a bit of time. I'll try it manually and compare.

    Thanks!
     
  11. Jan 10, 2014 #10

    haruspex

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    I can't let this go without pointing out that the question is flawed. (This same flaw in textbook problems crops up on the forum several times a year.)
    By asking for 'the force', the question assumes the force is constant over the 1cm. That is clearly not the case. Since nothing breaks, it might be pretty much like a spring, with the force increasing linearly with extent of compression (but with a much lower constant during decompression). The peak force will therefore be about twice that computed by energy/distance.
    Most questions like this try to get around the problem by asking for the average force, but that doesn't work either. There are two ways we might try to define average force here:
    Average over distance: ∫F.ds / ∫ds = ΔE/Δs
    Average over time: ∫F.dt / ∫dt = mΔv/Δt
    If the force is not constant, they will generally produce different answers. Which to use?
    We already have a well-defined concept of average acceleration: Δv/Δt. So for consistency we should define average force as mΔv/Δt. But we're not told Δt.
    A reasonable punt at the average force here can be made by assuming SHM during the compression phase. This leads to 2mv2/(πs), i.e. (ΔE/Δs)(4/π).
    A fully realistic model might be (imperfectly) elastic compression for some short distance, but with the force becoming capped at some point. E.g. there might be some hydraulic damping going on as fluid is forced out through membranes.
     
  12. Jan 10, 2014 #11
    Wolfram|Alpha is the supreme overlord of the universe.

    If Wolfram|Alpha were to be incorrect, the Universe would alter in a way that would make Wolfram|Alpha correct in the first place.

    Never question it again.
     
  13. Jan 11, 2014 #12
    Oops...I did blame it on the calculator. I didn't mean it, I promise! XDXD
     
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