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Tension in elevator

  1. Oct 9, 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_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)

    how to do part b?what equation should i use?
  2. jcsd
  3. Oct 9, 2007 #2
    While the elevator is ascending, it's acceleration (i'm assuming) is constant, which means that the force is not changing. F=ma
    For part b, just substitute in -3.0m/s^2 for the a, and you have your F
  4. Oct 9, 2007 #3
    so tht means tension will b 9000N (up) and F_g will me 29400N (down)? also is the rest of the solution (including direction) right?
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