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Elevator problem

  1. Apr 25, 2017 #1
    1. The problem statement, all variables and given/known data
    1. John is traveling with an elevator upwards with the speed of 3.0m/s. She drops her bag. How big will the acceleration be right before it hits the elevator floor?
    My answer is 9.82m/s^2 is that correct?

    2. Assuming the bag is 1 meter above the floor with an weight of 1 kg. How long will it take for it to hit the floor and what distance did it travel before hitting the floor?


    2. Relevant equation
    m*a=f? I really don't know.

    3. The attempt at a solution
    1. My answer is 9.82m/s^2 is that correct

    2. My attempt is that they travel the same distance then i can put vt=vt but that's completly wrong. Could anyone help me please?
     
  2. jcsd
  3. Apr 25, 2017 #2

    PeroK

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    Yes.

    What does the problem look like to the people in the elevator?
     
  4. Apr 26, 2017 #3
    It just falls?
    Can anyone show me the solution, or atleast explain it please?
     
  5. Apr 26, 2017 #4

    PeroK

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    You can solve the problem two ways.

    a) Imagine yourself outside the elevator. Initially the bag and the elevator are both travelling upwards. The bag is released and accelerates under gravity, while the floor keeps going upwards.

    You can simply use your equations of motion for the bag and elevator floor separately and calculate when and where their paths intersect.

    This is a good exercise in setting up a system and solving equations.

    b) Imagine yourself first inside the elevator. Is the problem any different from a bag being dropped on the ground? Your answer was correct: no it isn't.

    You can use your equations of motion to calculate how long it takes to fall. Then, you can switch to looking at the problem outside the elevator and use the time you calculated to calculate how far the bag moved.

    This is perhaps a conceptually harder approach, but it is a good exercise in using different reference frames to solve a problem.

    If you are confused by method b) you might want to stick to method a). Alternatively, you could do it both ways and check you get the same answer.
     
  6. Apr 26, 2017 #5
    is it correct to solve this equation to get the time it takes for it to fall down, because the total distance the elevator and the bag travels is 10?

    (9.82x^2)/2+10x=10
    x=0.73
    Is this the correct time the bag travels?

    Or since do i need to have one negative and get
    (9.82x^2)/2-10x=10
    x=2.77

    Or is this completly wrong?? Could anyone type the right equations and explain for me
     
  7. Apr 26, 2017 #6

    PeroK

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    That's not right, I'm afraid. I think you may not understand actually what's happening, so it's not about equations. The equation you need is:

    ##y = y_0 + vt - \frac12 gt^2##

    But, I can see you already know that one.

    Here are questions to help you understand the problem: To someone outside the elevator describe:

    a) The motion of the elevator floor.

    b) The motion of the bag before it is dropped.

    c) The motion of the bag after it is dropped.

    And don't say things like simply "it falls".
     
  8. Apr 26, 2017 #7
    You're right I don't really understand it and i'm totally new to this. I'm doing this on my free time, and just thought about this question :p

    But could you explain y=y0+vt−0.5gt^2 ? What does y stand for? And could you please put the solve the equation?
     
  9. Apr 26, 2017 #8

    PeroK

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    The elevator is going up, so ##y## is a common convention for displacement in the up/down direction. ##y_0## is where it starts (at ##t = 0##), ##v## is the initial velocity (again at ##t=0##) and ##g## is gravity, which is negative in the equation as I decided to take the up direction as positive.

    This isn't the easiest problem, as you will almost certainly need some algebra. As soon as you have two separate equations, you need to do more than just plug in the numbers.
     
  10. Apr 26, 2017 #9
    Okay, thanks!!
    But why can't I think like this.
    I have 10t=y
    and (9.82t^2)/2=y
    Then I put some diffrent times for example 0.3 in both and I get that the elevator moved 3m up, and the elevator 0.4419m down. And then I try until the sum of the both is 10?
    And how do I solve your equation????
     
  11. Apr 26, 2017 #10

    PeroK

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    Where are you getting this critical number 10 from?
     
  12. Apr 26, 2017 #11
    Omg, idk where i got the 10meters. I mean the 1 meters because it's 1 meter above the floor. So the combined movement of the elevator and the bag dropping should be 1 meter?
     
  13. Apr 26, 2017 #12

    PeroK

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    If you are going to try to solve the problem that way, you are still going to have to set up the problem properly.

    As I see it, you have three major issues trying to solve this problem:

    You need to understand the motion of the objects involved
    You need to know how to set up the relevant equations
    You need - shock horror - to learn some algebra!

    We're only supposed to guide you through your homework problems, but you perhaps need some serious tutoring. Are you learning this on your own?
     
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