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Dynamics: Tension (Elevator)

  1. Jan 8, 2015 #1
    1. The problem statement, all variables and given/known data
    An elevator(4850kg) is to be designed so that the maximum acceleration is 0.068m/s2. What are the minumum and maximum forces the motor must exert on the supporting cable?

    2. Relevant equations

    Force = mass(acceleration)
    Forcevertical = Ftens + Fgrav
    3. The attempt at a solution

    Forcevertical = Ftens + Fgrav
    Force = 4850kg(0.068m/s2)
    = 329.8N

    Forcevertical = 329.8N - 47530N
    = -47200N

    I'm not so sure about both answers but i think the first solution is kinda right?

    thanks in Advance!
     
  2. jcsd
  3. Jan 8, 2015 #2
    Not quite. Think about what is happening. It also has to account for the force of gravity. So the minimum amount of force needed would be to support the force of gravity, and the maximum would be to not only support the force of gravity, but to accelerate the elevator upward at 0.068m/s2. So How would you express those forces with mathematics?
     
  4. Jan 8, 2015 #3

    haruspex

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    Please define what Forcevertical represents.
    Which of the forces in the first equation corresponds to "Force" in this equation?
     
  5. Jan 8, 2015 #4
     
  6. Jan 8, 2015 #5

    haruspex

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    No, there are two applied forces (plural) on the y axis, and you have those on the right hand side of the equation. Forcevertical is one force. The word I'm looking for is 'net' force, or 'resultant' force.
    Then that's where you are wrong. In the standard equation F=ma, what does F represent? Yes, I know it's a force, but exactly what force?
     
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