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Elevator tension physics problem

  1. Oct 10, 2007 #1
    A stationary elevator and its contents have a combined mass of 3000Kg. The elevator is suspended by a single cable. (Assume three significant digits.)
    Draw a free-body diagram of the elevator and calculate the values of all the forces that are acting on it when at rest.
    If the elevator is ascending at a speed of 3.0 m/s, what are the values of the forces acting at this point?
    If the elevator is descending at 3.0 m/s2, what are the values of all the forces acting at this point?

    For a i did---F_g=mg = 3000*9.8=29400N (Down)
    F_g=〖-F〗_T
    F_T=29400 N (Up)

    For c i did---Net Force downwards=mg-T
    downward acceleration= Net force/m= (mg-T)/m
    a = g-T/m
    T= (g-a)m
    = (9.8-3)3000= 20400N(Up)
    Net Force acting upwards= mg= 3000*9.8= 29400N(Down)

    Is this right? How to do part b?what equation should i use?
     
  2. jcsd
  3. Oct 10, 2007 #2

    Hootenanny

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    It's best to stick with a sign convention, so lets say positive is up and negative is down, ok? So that would make;

    Fg = -mg = -29400 ; T = -Fg = 29400

    So, now we have the forces at rest. Now, the net force at all times can be written as Fnet = T+Fg. So you have your basic equations set up, all you have to do it equate them with Newton's Second law.
     
  4. Oct 10, 2007 #3
    But i part b, they have given the speed, not the acceleration. How do use that in Ma=T+F_g?
     
  5. Oct 10, 2007 #4

    Hootenanny

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    Indeed, if an object is travelling at constant speed, what is it's acceleration? What does this mean the net force acting on the object is?
     
  6. Oct 10, 2007 #5
    if the elevator is travelling at a constant speed then the a is 0. That means the T and F_g will be equal, so the net force will b 0. Right?
     
  7. Oct 10, 2007 #6

    Hootenanny

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    Spot on. And for (c)...?
     
  8. Oct 10, 2007 #7
    for c: net force= T+F_g
    and we know a=f/m...so a= T+F_g/m
    hence then T will be equal to m(a-g)= -20400N. hence tension will be 20400N down, and F_g will be 29400 N Up. Right?
     
  9. Oct 10, 2007 #8

    Hootenanny

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    Careful, your numbers are right but check your sign. How do you propose the tension acts downwards and gravity acts upwards?

    You were correct in your first post, I just wanted to make sure you understood the importance of defining a definite coordinate system. :smile:

    As an aside, an easier method would be just to consider the force required to accelerate the elevator at 3 m/s^2. :smile:
     
    Last edited: Oct 10, 2007
  10. Oct 10, 2007 #9
    Thank you so much...:)
     
  11. Oct 10, 2007 #10
    i had another question..
    A rock of mass 200g is attached to a 0.75m long string and swung in a vertical plane.
    A) what is the speed that the rock must travel so that its weight at the top of the swing is 0?
    B) what is the tension in teh string at the bottom of the swing?
    C) if the rock is now moved to a horizontal swinging position, what is the angle of the string to the horizontal?
    Here is what i could do:
    I drew a vertical circle with 2 positions a and b. A is at the top and B is at the bottom. so at A the F_g is down and T is up. at B,T is down and F_g is up. Right? then what?
     
  12. Oct 10, 2007 #11

    Hootenanny

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    :bugeye:
     
  13. Oct 10, 2007 #12
    i dint understand..so tension is down for A? and T is up for B and F_g is down for B?..k then how do i proceed?
     
  14. Oct 10, 2007 #13

    Hootenanny

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    Two points to remember;

    (1) The tension always acts towards the centre of the string.

    (2) Gravity always acts towards the centre of the earth.

    Next, I would draw a FBD for each case individually.
     
  15. Oct 10, 2007 #14
    so then at A net force will b mg-T, and at b net force will b T-mg. right? but mg has to b 0 since for part A, weight is 0. that means net force at A will b equal to -T. and we knoe tht centripital force=Mv^2/r. So that will be -T=-mg=0.2*v^2/0.75. when i solve this i get v=2.71m/s. is that right?
     
  16. Oct 10, 2007 #15

    Hootenanny

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    No, at A both forces act in the same direction, i.e. down. Like I said previously, the tension in a string always acts towards the centre of the string.
     
  17. Oct 10, 2007 #16
    but then the net force at A will still be -T.so the v will still b 2.71m/s
     
  18. Oct 10, 2007 #17

    Hootenanny

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    Nope I'm afraid it isn't. The question asked for the tension when the speed is zero...
     
  19. Oct 10, 2007 #18
    no we have to find the speed when the weight is 0. dosent that mean that we hve to find the speed when mg=o.that means that only tension force is actin at A. isint that right..?
     
  20. Oct 10, 2007 #19

    Hootenanny

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    :redface: My bad, I misread the question, you are indeed correct, your solution is right.
     
  21. Oct 10, 2007 #20
    so at B, the tension will b 1.96 N?
     
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