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Who would hit the ground harder?

  1. Dec 1, 2005 #1
    Person A: Falling from the top of a 20' ladder.

    Or

    Person B: Falling with a 20' ladder that is falling over. (from the top)


    Give me your answer and tell me why? This one is driving me crazy!
     
  2. jcsd
  3. Dec 1, 2005 #2

    ranger

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    I'm [only] guessing you could use Newton’s Second Law, F = ma. Of course I am not taking into account air resistance. So Persion B would hit the ground with more force since the combined mass of the person and the ladder would be more. The acceleration remains constant at g and therefore velocity is increasing 9.80m for every second the object is in the air.

    Do correct me if I am wrong.
     
    Last edited: Dec 1, 2005
  4. Dec 1, 2005 #3

    russ_watters

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    The problem may be trying (poorly) to illustrate how the acceleration at the end of the ladder can go above g if you include the weight of the ladder. Regardless, acceleration would start at zero and increase as the ladder supports less and less of the weight of the person. So if you ignore the weight of the ladder, the person's ending speed would be lower. I'm not sure how to work that out mathematically or if you are supposed to ignore the weight of the ladder.

    edit: Hmm - now I'm not sure. Since there is circular motion, there is more than one acceleration - perhaps if you look at it according to potential energy, the end result has to be no difference whether the ladder is there or not, since the ladder does no work on the person.
     
    Last edited: Dec 1, 2005
  5. Dec 1, 2005 #4
    Acceleration will not be at a constant for person B. You need to take in account the ladder. Person B will be supported by the ladder during the fall.
     
    Last edited: Dec 1, 2005
  6. Dec 1, 2005 #5

    berkeman

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    Pretty interesting problem. At first glance, you'd think that the person hits with equal force, since it seems that there is no other place for their initial stored potential energy to go except into their terminal velocity. Yeah, it takes longer to hit the ground for the guy swinging down with the ladder, but he could still have the same terminal velocity.

    But as I was drawing the problem up on a Post-It note to see if I could work out the math quickly, I noticed that there is someplace else where the swinging ladder guy's energy can go. Some energy is transferred sideways into the ground, so less of the ladder guy's initial potential energy will be converted into his terminal velocity, so he will hit the ground a little slower than the free-falling guy. If you sketch out the vertical and horizontal velocity vectors versus time for the ladder guy, you'll see that he has a net area under his horizontal velocity vector curve, and hence net work gets done in the horizontal direction. the horizontal force starts at zero, then increases as the ladder swings down, and then goes back to zero right at impact because the guy's motion is all downward at that instant. Since the horizontal force has a positive net, the ladder and guy did work on the earth sideways during the fall. If you connect the base of the ladder with a hinge on the top of a platform on a frictionless plate, you'd see the sideways motion of the base of the ladder in the direction opposite of the guy's fall.

    I still need to work out how much less speed the ladder guy has at impact, but I'm out of room on this Post-It, and I need to get back to work....:rolleyes:
     
  7. Dec 1, 2005 #6

    Danger

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    You'll have to requisition larger Post-Its if you expect to be of any further use to us. :tongue:
     
  8. Dec 1, 2005 #7

    berkeman

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    :biggrin: :rofl:
     
  9. Dec 1, 2005 #8

    Tide

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    Apply energy conservation including the motion of the person and the ladder. I think the question implies the person is fixed at one end of the ladder and that the base of the ladder is fixed to the ground. The person's speed of impact with the ground will be greater than that of a person free falling from the same height - as long as the mass of the ladder is not zero.
     
  10. Dec 1, 2005 #9

    berkeman

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    You sure? If two objects of different masses fall at the same rate in a vacuum, and if my comments above about some of the ladder guy's potential energy going into horizontal work, then how could the ladder guy hit with a higher velocity than free-fall guy?
     
  11. Dec 1, 2005 #10

    russ_watters

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    Same rate, different mass = higher energy. The question askes about which hits harder, not which is going faster.
     
  12. Dec 1, 2005 #11

    turbo

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    The person at the top of the ladder has lots of potential energy and little kinetic energy. I think it is fair to say that a person clinging to a falling ladder will convert his potential energy to kinetic energy at a slower rate (especially near the top of the arc) than a person falling off the ladder. If it were me, I would hang onto that ladder and ride it down. That way, my rate of acceleration would not approach that of free-fall until the ladder was close to horizontal, and the lower rate of acceleration earlier would result in a slower speed at impact and less injury.
     
  13. Dec 1, 2005 #12

    Tide

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    You have to consider that some parts of the ladder strike the ground with very little speed while parts of the ladder near the oher end strike at much greater speed! I suggest doing the calculation. :)
     
  14. Dec 1, 2005 #13

    berkeman

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    I agree it's time to play with some numbers -- I'm about to head home for the day, so maybe I'll make it my evening project in front of the TV :rolleyes:

    BTW, I don't think the ladder hitting the ground is relavent to the OP's question. The ladder could be weightless, as long as it is rigid enough to hold up the ladder guy as he falls. It converts the vertical downward force of gravity into vertical cos(theta) and a horizontal sine(theta) (maybe) opposing components, where theta is measured between the vertical and the ladder. When the ladder guy is balanced on top, the cos(0) component exactly balances his weight W and holds him up. When the ladder has swung down to 45 degrees, the ladder guy's weight vector points down, but the ladder is pushing radially outward some, which applies force vectors in the vertical and horizontal directions.

    Fun problem. I'll bet somebody answers it quantitatively by the morning. Cheers, all. -Mike-
     
  15. Dec 1, 2005 #14

    pervect

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    I could use a double check of this, but I'm getting a (oops) faster impact speed, with the assumption that the mass of the ladder is uniformly distributed along its length.

    Let the mass of the person be m
    Let the mass of the ladder be u*m
    Let the length of the ladder be L

    Then the potential energy of the system is
    [tex]
    V = m g L (1+\frac{u}{2}) sin \theta
    [/tex]

    The kinetic energy of the system is

    [tex]
    T = \frac{1}{2} m (L \dot{\theta})^2 +\frac{1}{2} \int_{x=0}^L \frac{m u}{L} (x \dot{\theta})^2 dx = \frac{1}{2} m v^2 + \frac{1}{6} u m v^2 = \frac{1}{2} m (1+\frac{1}{3} u) v^2
    [/tex]

    If the ladder starts out with [itex]\theta=90[/itex] and winds up with [itex]\theta=0[/itex], we have

    [tex] v^2 = 2 g L \frac{1 + \frac{u}{2}} {1 + \frac{u}{3}}
    [/tex]
     
    Last edited: Dec 1, 2005
  16. Dec 1, 2005 #15

    Tide

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    Pervect,

    That is correct so that as long as the ladder has any mass he'll hit the ground harder.
     
  17. Dec 2, 2005 #16

    pervect

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    I agree, I also get a faster impact (I had to correct some arithmetical errors as well as some Latex issues, hence the 'oops').

    My original post is probably a bit terse, let me fill in some of the coordinates and discussion


    P
    .
    .
    .
    .
    .[itex]\theta[/itex]
    ---------

    the P is the person of mass M on the ladder. He is at a height L above the ground with the ladder vertical as shown.

    The bottom of the ladder is assumed to be a fixed pivot point.

    [itex]\theta[/itex] is the angle of the ladder with respect to the ground, 90 degrees when the ladder is vertical

    [itex]\dot{\theta}[/itex] is [tex]\frac{d\theta}{dt}[/itex], the rate at which the angle [itex]\theta[/itex] is changing.

    The velocity of the person at the end of the ladder is v, and has a magnitude of [itex]r \dot{\theta}[/itex]

    Note that with the ladder vertical, the center of mass of the ladder is at height L/2 above the ground. This is the origin of the factor of 1/2 in the first section of the post for the potential energy of the system.

    Different parts of the ladder have different velocities. The total kinetic energy is the intergal of .5*v(x)^2*dm, where v(x) is the velocity of the ladder element which is a function of position (x), and dm is the mass of the element of the ladder at position x.

    v(x) is just [itex]x \dot{\theta}[/itex], where x is the distance from the bottom of the ladder to the mass element.

    The mass of the ladder is u*M =
    [tex]\int dm = \int_{0}^{L} \frac{u M}{L} dx[/tex]

    thus dm = (u M / L) dx

    Putting this all together gives the factor of u/3 in the bottom of the equation.
     
    Last edited: Dec 2, 2005
  18. Dec 2, 2005 #17

    berkeman

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    That's really interesting. So the ladder mass helps to add some velocity to the final impact for the ladder guy, and if the ladder is massless, the impact velocity is the same as for free-fall. Cool.

    I also think I see where my argument about some horizontal work being done wasn't quite right for the situation given. Since the earth is so heavy compared to the ladder and guy, the earth moves very little sideways from the sideways force component, so the work done (and energy lost) is negligible. But if you put the ladder base on a plane with moderate friction, so that the base of the ladder displaces a few meters sideways during the fall, then the work done and energy lost would count, I would think. So with a massless ladder in this scenario with the friction plate and displacement of the bottom of the ladder, I think the guy would have a lower impact velocity compared to the free-fall guy.

    Cool problem.
     
  19. Dec 2, 2005 #18
    Hu..no need to think complicity here. Think the basics. Your question seems to be dealing with different masses falling and which one would fall harder if they feel from the same distance? If i'm correct, then just think of a simple analogy..this should help. Imagine dropping a car and a ant from 20ft above..which would hit harder on the ground? It should be clear to see now.
     
  20. Dec 2, 2005 #19

    Tide

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    Presumably, Person A and Person B have the same mass and we're asked which of them hits the ground "harder." I think it's reasonable to infer the intent of the question is to ascertain which person hits the ground with greater speed.
     
  21. Dec 3, 2005 #20

    Art

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    A person free falling travels the 20 metres to the ground sooner than the person travelling the longer route (of around 31 metres) by clinging to the ladder as it scribes an arc through the air and so discounting the weight of the ladder the free falling person hits the ground harder as his vertical velocity is greater.
     
    Last edited: Dec 3, 2005
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