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Homework Help: When will the bolt pass the elevator floor?

  1. Nov 27, 2017 #1
    • Member advised to use the provided formatting template when starting a new thread in a homework forum.
    A 5m tall elevator travels uniformly up an elevator shaft at 8m/s. At some point in time, a bolt falls from the top of the elevator and falls freely down the shaft. You can ignore the effect of air resistance.

    How many seconds after the fall will the bolt pass the bottom of the elevator?



    y=yi + vit + (1/2)at^2


    I tried to use the above equation for both the elevator and the bolt separately. The bolt takes 1.01 seconds to fall 5m and the elevator moves up 5m in 0.625 seconds.
     
  2. jcsd
  3. Nov 27, 2017 #2

    phinds

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    Do you think that will work? How about the fact that the bolt and the bottom of the elevator are not going to meet up at the time when the bolt has dropped 5m because by then the elevator floor will have moved upwards.
     
  4. Nov 27, 2017 #3
    Not really. I am not able to grasp this I guess.
     
  5. Nov 27, 2017 #4

    phinds

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    Well, you had best think about it some, else you will never solve the problem.
     
  6. Nov 27, 2017 #5
    I have been thinking about it all weekend. I know it must take less than 1 second because the floor is moving toward the bolt as it falls. The reason I signed up for this forum is to be given some insight about the problem. Can you explain to me how to solve this?
     
  7. Nov 27, 2017 #6

    phinds

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    What you need is the equation of motion for the bottom of the elevator and the equation of motion for the motion of the bolt and then you solve them for the place where they intersect. The equation of motion for the bottom of the elevator is trivial since it is uniform motion. What is the equation of motion for the bolt?
     
  8. Nov 27, 2017 #7
    0=5 -4.9t^2 ???
     
  9. Nov 27, 2017 #8

    phinds

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    OK, now you need both equations using the same coordinate system.
     
  10. Nov 27, 2017 #9
    This where I am getting confused. So, I need 5=8t and 0=5-4.9t^2 combined as 4.9t^2 + 8t - 10=0? That gives me 0.83 seconds.
     
  11. Nov 27, 2017 #10

    phinds

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    I think you need to draw a diagram and label your variables
     
  12. Nov 28, 2017 #11

    jbriggs444

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    When the bolt falls, does it fall from the elevator's top or from the adjacent wall? What is its initial velocity?
     
  13. Nov 28, 2017 #12
    As it turns out, the bolt has the same initial velocity of the elevator. Rather than noticing this earlier I assumed the initial velocity of the bolt was zero.

    Using the roof of the elevator as my origin my two equations are

    x[bolt] = 8t - 4.9t^2
    and
    x[floor] = -5 + 8t

    Setting them equal to each other eliminates the 8t from both sides and I am left with

    4.9t^2 = 5

    t = 1.01s
     
  14. Nov 29, 2017 #13

    haruspex

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    Ok.
    Are you familiar with using different inertial frames? It is a little simpler if you work in the reference frame of the elevator. The cancellation of the initial velocities is immediate.
     
  15. Dec 4, 2017 #14

    PeterO

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    For future reference to these types of questions, terms like "falls" and "is dropped" can be very misleading, since they imply the object immediately starts moving down. In fact the "breaks free from" or "is released" is a much better descriptor - and you can use that even if the original question doesn't.
    The "frame of reference" mentioned above is good. If you have access to a lift in a tall building, you should try the following.
    Drop (release) an object as you stand in a stationary lift (hold the door open and it won't move) and note how long it takes to reach the floor.
    Now start the lift - either up or down - and wait until it is travelling at constant speed (perhaps going past the 5th floor of 20) and again drop (release) the object and compare the time taken to reach the floor.
     
  16. Apr 17, 2018 #15
    Hello Wrangler_77, did you solve the second question?
    Now assume that the bolt falls from the *wall* of the elevator shaft, and then falls freely without air resistance.

    How many seconds after the fall will the bolt pass the bottom of the elevator?
     
  17. Apr 17, 2018 #16

    phinds

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    He has not been here in 5 months so you're not likely to get an answer.
     
  18. Apr 17, 2018 #17
    yeap... I see it now.... =/ I'm struggling with the second part... I was able to solve the first part:

    I made a draw to understand the "dynamics" of the environment and computed the Xf for the elevator and bolt:

    Xfb = Xib + Vit + 1/2 at^2
    = 0 - 8t + 9.8/2 t^2
    Xfb = 4.9t^2 -8t

    Xfe = 5 - 8t (constant speed)...

    Then, by equaling them:

    5 -8t = 4.9t^2 -8t
    5 = 4.9t^2
    5/4.9 = t^2
    t = sqrt(1.02) = 1.01 ;


    Now... for the b question, I'm really struggling in relating the distance between the bolt and elevator.... I tried to mount the equations based on the bolt frame:

    Xfb = h + 0t + 9.8/2 t^2 (considering initial position at the elevator floor)...

    Xfe = Xi - 8t;

    I'm trying to relate the initial position of the elevator to h of the bolt... But, so far, no good... =/ ...

    Could someone, please, bring me some light to this problem?
     
  19. Apr 17, 2018 #18

    PeterO

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    When the bolt falls, it begins exactly 5m above the floor of the lift - it is level with the roof.
    Draw a couple of position time graphs (at least in your mind). Let the bolt start at +5m, while the floor starts at 0m - then you won't get tied up with negative numbers.
    The bolt follows the usual parabola of a falling object (reaching 0m at 1.01 seconds I think you had already calculated - but exactly 1.0s if you use a g value of 10, as is often done with simple introductory problems.
    The floor proceeds up at a steady 8m/s
    When the lines intersect you have your answer.
    Now use your formulas to find the answer mathematically - complete with 2nd and 3rd decimal places..

    Note: to save the time needed to say the 4 syllable word elevator, we call it a lift.
     
  20. Apr 18, 2018 #19
    Thank you so much for your patience and detailed explanation. I was doing some mass and trying to figure out some distance between the lift and the bolt.... Thinking that, even with the bolt starting with a initial velocity = 0 (the wall of shaft).. I was imagining that at some point, more specifically, at the moment that the bolt meet the roof of the elevator, what would be its velocity... anyway... I was doing such confusion and couldnt realize that what I had to do was just to consider the roof of the lift and the position of the bolt (at the wall of shaft), was my initial position!

    Lastly... I could solve the problem with the same equations and draws I already had:
    Xfe = 5 - 8t (considering down direction positive and bolt initial position as 0); then:

    Xfb= 0 + 0t +1/2 * 9.8 * t^2;

    so: 5-8t = 4.9t^2;
    4.9t^2 + 8t -5 =0

    t' = 0.4824
    t'' = -2.1151

    Since time cant be negative, t= 0.4824s...

    Again.. Thank you for your patience and detailed explanation!
     
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