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Law of conservation of momentum

  1. Oct 24, 2004 #1
    I can't seem to figure out what the damn question is asking :D. Can anyone explain what the question is asking?

    During a chicago storm, winds can whip horizontaly at speeds of 100 km/h. If the air strikes a person at the rate of 40kg/s per square meter and is brought to rest, calculate the force of the wind on a person. Assume the person's area to be 1.50m high and 0.50m wide. Compare to the typical maximum force of friction (μ = 1.0) between the person and the ground, if the person has a mass of 70kg.

    What comes to rest here? is it the air or the person?

    Also on a side note, can anyone explain whether the transfer of force during impulse instantaneous or not?
     
    Last edited: Oct 24, 2004
  2. jcsd
  3. Oct 24, 2004 #2

    Galileo

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    The air is. Consider the person standing still. The rate of change in momentum of the air is equal and opposite to the force on the person.
     
  4. Oct 24, 2004 #3
    Ok this is how I solved the problem. I doubt it's right.

    I see that the force is in units of [tex]kg/sm^2[/tex] and I have an area in square meters, so I multiplied the two of them to get this
    [tex](\frac{40 kg}{m^2s})(1.0m)(0.5m) = 30 \frac{kg}{s}[/tex]

    I plugged that into [tex]\Delta{p} = F\Delta{t} = (\frac{m\Delta{v}}{\Delta{t}} \times \Delta{t}) = m\Delta{v}[/tex] and got [tex]\frac{30kg}{s} \times \frac{27.8m}{s} = 8.3\times10^2N[/tex]
     
  5. Oct 24, 2004 #4
    looks good to me. :-D
     
  6. Oct 24, 2004 #5

    Galileo

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    That's correct. In 1 second 30 kg of air having a speed of 27.8 m/s comes to rest.
    So the change in momentum (which equals the force) is [itex]30 \cdot 27.8 = 8.3 \cdot 10^2N[/itex]
     
    Last edited: Oct 24, 2004
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