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Student in Elevator

  1. Oct 6, 2007 #1
    1. The problem statement, all variables and given/known data
    A 545N physics student stands on a bathroom scale in an 847kg (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 427N.

    Find the magnitude of the acceleration of the elevator.
    Find the direction of the acceleration of the elevator.
    What is the acceleration if the scale reads 635N?
    If the scale reads zero, should the student worry? Explain.
    What is the tension in the cable in part A?
    What is the tension in the cable in part D?

    2. Relevant equations

    F=ma
    w=mg

    3. The attempt at a solution

    I first find the mass of the student which is 55.61 kg.

    Ok from the problem, we know that the elevator is 847kg including the mass of the student, thus one can conclude that the elevator has a mass of 791.39kg.

    Anyways,

    As the elevator moves, the scale reads 427N.

    so it would be 427N = 847 (a) = .50m/s^2?

    Correct thus far, or where did I go on the wrong track.
     
  2. jcsd
  3. Oct 6, 2007 #2
    Well what direction is the elevator doing? up or down?

    what forces are acting on the elevator? and if the elevator is accelerating will the person inside also experience the acceleration?
     
  4. Oct 6, 2007 #3

    learningphysics

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    no that's not correct. The 427N isn't being exerted on the entire elevator... the 427N is the normal force on the student inside the elevator (the bathroom scale gives the normal force... also called the "apparent weight").

    find the acceleration of the student... that's the same as the acceleration of the elevator.

    What are the two forces acting on the student?
     
  5. Oct 6, 2007 #4
    the two forces acting on the student would be the weight of 545N, and the normal force of 545N when the elevator is at rest.

    so i should be...
    0 = 55.61a
    a=0?

    but when the elevator moves it should be

    118 = 55.61a
    a= 2.12

    so it should be moving up right? since accel is + ..
     
    Last edited: Oct 6, 2007
  6. Oct 6, 2007 #5

    learningphysics

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    Well you should have -118 = 55.61a..

    gravity is downwards hence negative:

    -545 + 427 = 55.61a
    a = -2.12.

    so it is accelerating downwards.
     
  7. Oct 6, 2007 #6
    ok, since acceleration is -2.12 the elevator is moving downwards. then.

    If the scale reads zero, should the student worry? Explain.

    This would mean that the all forces are equal at this point, and that the elevator has comes to a stop...that acceleration is 0.
     
    Last edited: Oct 6, 2007
  8. Oct 6, 2007 #7

    learningphysics

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    yes, since it starts from rest it is moving downwards... in this case it is moving downwards...

    However, it is generally possible for it to be moving upwards and have the acceleration downwards... in that case it would be moving upwards and slowing down...

    But in this particular case, it is moving downwards and speeding up...
     
  9. Oct 6, 2007 #8

    learningphysics

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    No, if the scale reads 0 then the normal force is 0... what is the student's acceleration then? what does this acceleration mean?
     
  10. Oct 6, 2007 #9
    the acceleration should be 7.68

    meaning that when elevator reaches this point , that the weight of the person is the same as the normal force, it cannot be greater, as then it would seem that the person body is going to slam into the floor. lol
     
  11. Oct 6, 2007 #10

    learningphysics

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    yup, he's going to slam into the floor. but how did you get 7.68?
     
  12. Oct 6, 2007 #11
    I got the acceleration, by doing the following:

    427= 55.61kg (a) = 7.68
     
  13. Oct 6, 2007 #12
    Are you familiar with pseudo forces?
     
  14. Oct 6, 2007 #13
    unfortunately not. :(
     
  15. Oct 6, 2007 #14
    Whenever we are in an accelerating reference frame. We cannot apply Newtons laws of motion.

    But we can make th newtons laws of motion hold true by introducing the pseudo force

    Pseudo force always acts in the direction opposite to the applied force.For example when a bus comes suddenly to rest then we feel a force in the forward direction. This is the pseudo force as the force is applied by the brakes in the backward direction.

    Hence when the lift is going up with acc a, we cannot apply newtons laws of motion, as it is a non inertial reference fame.But after applying pseudo force we can introduce newtons laws of motion.Here the pseudo force will act downwards and hence the weight of the man would be m(a+g)

    Similarly if the lift is going downwards by a then the pseudo force would be upwards and hence the weight of the man would be m(g-a)
     
  16. Oct 6, 2007 #15
    so, you are saying that it would be this:

    w=ma
    w=55.61(2.12+9.8)
    w = 662.87

    F=ma
    545 - 662.87
    117.87 = 55.61a
    a= 2.12

    if this is correct then, T should be 10096.24...
     
    Last edited: Oct 6, 2007
  17. Oct 6, 2007 #16
    When the scale reads 427N it definetly means that the lift is coming down so

    427 = 55.61(9.8 - a1)

    Solve it and you will get a.

    Now when it is 635 it is going up for sure

    635 = 55.61(9.8 + a2)

    now if the scale reads zero than it is for sure coming down

    0 = 55.61(9.8 - a). Here a should be equal to 9.8 as LHS is zero.And this is in fact nothing but free fall.
     
  18. Oct 6, 2007 #17

    learningphysics

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    but the normal force is 0... hence the only force is the weight...

    -mg = ma

    which leaves a = -g

    or using your numbers:

    -545 = 55.61a

    a = -9.8m/s^2.

    So the student is in freefall.
     
  19. Oct 6, 2007 #18
    so if the student is in freefall when the scale is 0, then that must mean that that there is no tension.
     
  20. Oct 6, 2007 #19
    Of course.When the lift is having a free fall the tension is zero.
     
  21. Oct 6, 2007 #20

    learningphysics

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    yes, for part D tension is 0.

    it is impossible for the student to accelerate downwards faster than the elevator... so the elevator must be accelerating downwards at -9.8m/s^2 also... hence the net force on the elevator is -mg... so no tension... so the cable must be cut or something...

    You are right that the tension in part A is 10096.24N
     
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