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Homework Help: Hard Physics Question

  1. Aug 18, 2004 #1
    A lift ascends with an upward acceleration of 1.25ms^-2. At the instant its upward speed is 2.50ms^-1. A loose bolt falls from the ceiling of the lift, 2.75m from the floor.
    (a) the time of flight of the bolt from the ceiling to floor
    (b) the distance it has fallen relative to the lift shaft

    Please help me. I have tried this question a few times and don't understand why i am getting it wrong, i think the aswer in the back of the book may be wrong. Can you please post your method and the answer please.

    :smile: Thanx for your time and your brain cells.
  2. jcsd
  3. Aug 18, 2004 #2
    Oh to the floor of the lift! I thought the question meant from the ceiling of the lift to the floor of wherever it's elevating from.
  4. Aug 18, 2004 #3


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    Those are not the correct answers!

    While the bolt is falling the elevator continues to accelerate so we must account for that. Inside the elevator the bolt falls 2.75m (the distance to the floor).

    Let [tex] e_d [/tex] be this distance the elevator moves while the bolt is falling.
    Let x be the distance the bolt falls compared to the elevator shaft.

    We must have [tex] e_d + x = 2.75 [/tex]

    can you come with expressions for the above quantities in terms of the time of the fall?
  5. Aug 18, 2004 #4
    The bolt does not fall 2.75m, it falls a lesser distance because the elevator is also travaling towards the bolt.
  6. Aug 18, 2004 #5
    The easiest way to calculate part a) is to think of the elevator as a frame with a different value of g for gravitation. Just add to the regular g (~9,8 ms^2) the acceleration of the elevator.

    part b) is easy then. Just calculate the distance traveled by the elevator minus the distance the bolt fell.

    Good luck!
  7. Aug 18, 2004 #6
    By using the formula for uniform acceleration: [tex] s(t)=v_0 t + \frac{1}{2}at^2[/tex]

    I got: a) 0,7055s b)-0,675m (so it fell 0,675m relative to the shaft...)
    Last edited: Aug 18, 2004
  8. Aug 18, 2004 #7

    Doc Al

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    As Integral stated, it falls 2.75m with respect to the elevator car. See Integral's equation for x to find out how far it falls with respect to the elevator shaft.
  9. Aug 18, 2004 #8

    Doc Al

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    da_willem's solution is perfectly correct. In case you don't see what he's doing, you can always solve it this way: Write the equation of motion for (1) the bolt and (2) the bottom of the elevator car, both with respect to the elevator shaft. They both have the form:
    [tex]y = y_0 + v_0 t + 1/2at^2[/tex]

    which is just the formula for uniformly accelerated motion. For the bolt: [itex]y_0 = 0[/itex], [itex]v_0 = 2.5 m/s[/itex], and [itex]a = - 9.8 m/s^2[/itex]. I'll let you figure out what values must be used for the bottom of the elevator car.

    Then just set the equations equal and solve. You'll get da_willem's answers.
  10. Aug 18, 2004 #9
    [tex]s = v_0 t + \frac{1}{2}(a_1+a_2)t^2[/tex]

    [tex]2.75 = 2.5 t + \frac{1}{2} (1.25 + 9.8) t^2[/tex]

    [tex]5.525 t^2 + 2.5 t - 2.75 = 0[/tex]

    [tex]t = \frac{-2.5 + \sqrt{2.5^2 + 60.775}}{11.05}[/tex]

    [tex]t = 0.51465[/tex]


    [tex]h = \frac{1}{2}at^2[/tex]

    [tex]h = \frac{1}{2} (9.8) (0.51465^2)[/tex]

    [tex]h = 1.2978[/tex]



    [tex]a) t\simeq{0.5}s[/tex]

    [tex]b) h\simeq{1.3}m[/tex]
    Last edited: Aug 18, 2004
  11. Aug 18, 2004 #10
    You're solving in the elevator frame, [tex]v_0[/tex] is zero there...
  12. Aug 18, 2004 #11
    I didn't take the elevator as reference.
    Last edited: Aug 18, 2004
  13. Aug 18, 2004 #12
    Where does the other acceleration ([tex]a_1[/tex]) come from then?
  14. Aug 18, 2004 #13
    elevator acceleration = 1.25
    bolt acceleration = -9.8
    Last edited: Aug 18, 2004
  15. Aug 18, 2004 #14

    Doc Al

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    It looks to me that you are using the falling bolt as your reference. I presume that "s" is the position of bottom of the elevator with respect to the falling bolt, in which case [itex]s_0 = -2.75[/itex] and [itex]v_0 = 0[/itex]. So:

    [tex]s = -2.75 + \frac{1}{2} (1.25 + 9.8) t^2[/tex]

    Solve for when the floor hits the bolt (s = 0) and you'll get da_willem's answers.
  16. Aug 18, 2004 #15


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    I solved it in the frame of the elevator, with g' = 11.05 m/s^2.

    a) T = sqrt (2h/g') = 0.706 s

    b) d = h - s(elev) =2.75 - (2.5*.706) - (.5*1.25*0.498) = 0.675 m

    These are the same as da_willem's answers.
  17. Aug 18, 2004 #16
    My mistake!
    Of course the bolt was travelling at the same elevator speed. So, the term [tex]2.5t[/tex] doesn't exist.
    All we have is
    [tex]2.75 = \frac{1}{2} (1.25 + 9.8) t^2[/tex]
    that yields the same da_willem's answer. :smile:
  18. Aug 21, 2004 #17
    Thanx everybody. I knew my teacher was wrong. I think he put in the Term "2.5t" same mistake as Rogerio had at first. Can some1 set it out with full working for me so i can print it out for my teacher.
    Last edited: Aug 21, 2004
  19. Aug 21, 2004 #18

    Doc Al

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    Why don't you really impress your teacher and work it out for yourself? Seriously, you've been given several ways of solving this. Pick one and work out all the steps. (We'll make sure you do it right. :smile: )

    For example, you can start with what I gave in post #8.
  20. Aug 21, 2004 #19
    Thanx all, you have been a great help and have showed me how to work it out.
    I've wrote down the process to show my teacher.
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