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Forces again *sigh* please help me , this is not homework

  1. Nov 3, 2009 #1
    forces again.. *sigh* please help me , this is not homework!!

    1. The problem statement, all variables and given/known data
    a man of mass 80 kg ( weight 176 lb), jumps down to a concrete patio from a window ledge only 0.50 m above ground. He neglects to bend his knees on landing, so that his motion is arrested in a distance of about 2.0 cm.

    a) what is the average acceleration of the man from the time his feet first touch the patio to the time he is brought fully to rest?

    b) with what average force does this jump jar his bone structure?



    2. Relevant equations

    f = ma, 3rd law



    3. The attempt at a solution
    I really don't know what to do at all...
    if a man falls down and accelerates a little bit forward after the fall, will the same force from falling transfer to that small time he is moving forward?

    and how will an acceleration be affected if something is falling from the sky and then bounces off the ground? touching the ground will make it go slower wouldnt it

    please help me, this is not homework.. I'm trying to learn physics out of some book and it feels impossible
     
    Last edited: Nov 3, 2009
  2. jcsd
  3. Nov 3, 2009 #2
    Re: forces again.. *sigh* please help me , this is not homework!!

    the force of the man falling is not just weigh is it? the jumping will have some kind of force? so, his force will be F = 176 + ma
     
    Last edited: Nov 3, 2009
  4. Nov 3, 2009 #3
    Re: forces again.. *sigh* please help me , this is not homework!!

    okay, I realized a couple of things.. when the man hits the ground, the force component is cancelled out by the ground. so I'm wondering what keeps a thing in motion after it has hit the ground? why does a coin bounce when dropped to the ground? I thought that it was because the velocity accumulated during the fall becomes large enough, when the force cancels out, you still need something that can set the velocity to zero. you can' t have 100 m/s to 0 instantaneously (I dont think so ).. but then why would crouching as you fall do exactly that?

    please help me im desperate
     
  5. Nov 3, 2009 #4
    Re: forces again.. *sigh* please help me , this is not homework!!

    I can only help you with part a, I'm not so sure about b.

    Given:
    [tex]\Delta{y}=0.5m[/tex]
    [tex]\bar{a}=\frac{\Delta{V}}{\Delta{t}}[/tex]

    To find the man's speed at first impact, use the kinematic equations for uniform acceleration:
    [tex]{V_{imp}}^2={V_o}^2+2g\Delta{y}[/tex]
    Thus,
    (1)[tex]{V_{imp}}=\sqrt{{V_o}^2+2g\Delta{y}}[/tex]

    We now need to find the time it takes the man to decelerate after hitting the ground. Use:
    [tex]{V}^2={V_{imp}}^2+2a\Delta{y}[/tex] where final velocity V=0
    Which translates to
    [tex]a=\frac{V^2-{V_{imp}}^2}{2\Delta{y}}[/tex]

    Using this acceleration, find the time for him to decelerate from Vimp to 0 m/s. Another equation is:
    [tex]V=V_{imp} + at[/tex]
    which becomes:
    (2)[tex]t=\frac{V-V_{imp}}{a}[/tex]

    Finally, use the values obtained in (1) and (2) and divide them to obtain the average acceleration, which is change in velocity divided by change in time.

    Phew! First time using LaTeX. Hope this makes sense!
     
    Last edited: Nov 3, 2009
  6. Nov 3, 2009 #5
    Re: forces again.. *sigh* please help me , this is not homework!!

    thank you sooooo muchh I will read over whatyou have said
     
  7. Nov 3, 2009 #6
    Re: forces again.. *sigh* please help me , this is not homework!!


    will you get a set number if you don't know the velocities?
     
  8. Nov 4, 2009 #7
    Re: forces again.. *sigh* please help me , this is not homework!!

    You know all the "velocities" that you'll need. Granted, you need to infer it from the problem, but it's there.
     
  9. Nov 4, 2009 #8
    Re: forces again.. *sigh* please help me , this is not homework!!

    thanks, so to solve for Vimp, set V = 0 for your "a" equation and then solve for Vimp?
     
  10. Nov 4, 2009 #9
    Re: forces again.. *sigh* please help me , this is not homework!!

    Yes. (If by V you mean Vo):smile:
     
  11. Nov 4, 2009 #10
    Re: forces again.. *sigh* please help me , this is not homework!!

    no, I meant V, to solve for the time it takes to go from Vimpact to V, V = 0
    we don't know Vo
     
  12. Nov 4, 2009 #11
    Re: forces again.. *sigh* please help me , this is not homework!!

    What? To get Vimp, we use
    [tex]
    {V_{imp}}=\sqrt{{V_o}^2+2g\Delta{y}}
    [/tex]

    Vo means initial velocity, aka his velocity when he first walks off/jumps off the ledge.
     
  13. Nov 4, 2009 #12
    Re: forces again.. *sigh* please help me , this is not homework!!

    yes, but what is Vimpact? we can't have a number if we don't know Vo? thanks
     
  14. Nov 4, 2009 #13
    Re: forces again.. *sigh* please help me , this is not homework!!

    We DO know Vo. It's his INITIAL velocity when he starts falling. Imagine that he's a particle-like object that you're holding. What is his speed right when you drop the object?
     
  15. Nov 4, 2009 #14
    Re: forces again.. *sigh* please help me , this is not homework!!

    he jumps though, so we don't know his initial velocity? it isn't like he just walks off the thing
    by the way, thanks for replying to me and everything..I really appreciate it . I'm going to bed now, I'll check on the thread tomorrow morning :)

    also, I think we can solve for Vimp in the 'a' equation, we don't even need Vo
     
  16. Nov 4, 2009 #15
    Re: forces again.. *sigh* please help me , this is not homework!!

    In these problems, if his initial velocity isn't give, assume he just walks off.
     
  17. Nov 4, 2009 #16
    Re: forces again.. *sigh* please help me , this is not homework!!

    ohh okay, but I don;t think we even need Vo, is it right? I think we can find Vimp from just the 'a' equation
     
  18. Nov 4, 2009 #17
    Re: forces again.. *sigh* please help me , this is not homework!!

    You're referring to this, right?
    [tex]
    a=\frac{V^2-{V_{imp}}^2}{2\Delta{y}}
    [/tex]

    Also, no. That only works for after he hits the ground. Read through my explanation again.
     
  19. Nov 4, 2009 #18

    cepheid

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    Re: forces again.. *sigh* please help me , this is not homework!!

    Yeah, that is basically what we are assuming here (i.e. he somehow just starts falling from a height of 0.5 m).

    BlueEight, it doesn't seem strange to you that your solution assumes constant acceleration after impact? Your steps basically amount to:

    1. Assume constant acceleration, and calculate it based on the distance over which it occurs and the change in velocity.

    2. Use the calculated acceleration to find the time over which it occurred.

    3. Divide the change in velocity by the time to get the acceleration.

    If you are allowed to assume a is constant, then steps 2 and 3 are redundant and circular. However, I suspect that assuming the acceleration is constant really makes no sense, because upon impact the force is probably going to vary with time in some complicated way that we have no way of figuring out, which is why we can only speak of the *average* force (and acceleration), which is why that's what the problem asks for.
     
  20. Nov 4, 2009 #19
    Re: forces again.. *sigh* please help me , this is not homework!!

    From my limited knowledge, normal force is what slows down an object after impact. Therefore, according to Newton's Second Law, Normal force N = mg + ma. However, I do not know how the acceleration will be affected after impact, and thus declare this problem unsolvable, having only gone through a quarter of AP Physics.

    But really, we need to somehow find a(t) after impact.
     
  21. Nov 4, 2009 #20

    cepheid

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    Re: forces again.. *sigh* please help me , this is not homework!!

    There is no way to find F(t). Force is the rate of change of momentum, and after impact that is going to depend on all sorts of complicated things like how the object and the ground deform upon impact etc etc. Remember that all contact forces are fundamentally electromagnetic in nature (when you examine things at the scale of the atoms making up the matter in the the two bodies that are in contact). So there is a whole bunch of physics that we are not prepared to really investigate and are basically just glossing over.

    For part b, it seems easy enough to calculate the total work done, and from that, to determine the average force, which equals (total work done)/distance. This would be an average over distance, not time. From that, you could deduce an average acceleration (averaged over distance travelled), but that doesn't seem to be what the problem is asking for, especially since it asks for acceleration *first* (in part a). I'm not sure how to solve part a.
     
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