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Homework Help: Force and motion in Newton's laws for elevator

  1. Sep 28, 2011 #1
    An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the cab, originally moving downward at 12 m/s, is brought to rest with constant acceleration in a distance of 42m.

    Taking downwards as positive
    My solution:
    F net = mg - T = ma
    T = m(g - a) (1)

    V^2 - Vi^2 = 2a (y - yi)
    a = (-12)^2 / 2(42) = -1.7 m/s^2

    (1) -----> T = m (g - a) = 1600 (9.8 + 1.7) = 18400 N

    Did I do it right ? cuz my friend got the different answer from me, he basically got T = m (g + a).......
     
  2. jcsd
  3. Sep 28, 2011 #2

    PeterO

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    Homework Helper

    I have seen T = m(g + a) commonly used when the question is find the "magnitude of the force on the lift". In that case, the fact that g is down and a is up, so one of them should be negative, is sort of ignored because we expect that the two magnitudes are simply added - giving the same answer as yours.

    Analysis of a ferris wheel is often written reaction force at top = mg - Fc and reaction force at bottom = mg + Fc.
    This is an implication that we are talking magnitudes only.


    Note that in your line of working that I coloured red, you have a - sign suddenly appearing on the right. Perhaps you just forgot to type is before the 42? ie distance travelled was -42 since the lift was travelling down.
     
  4. Sep 28, 2011 #3
    Thanks, it should be like this

    V^2 - Vi^2 = 2a (y - yi)
    sice V = 0, so we have
    - Vi^2 = 2a (y - yi)
    a = -(Vi^2) / 2(y - yi)
    a = -(12^2) / 2(42) since am taking downwards as positive, so y > yi.
     
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