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Gr11 Physics Forces

  1. Apr 11, 2005 #1
    Elevator of 2000kg is suspended by a single cable calculate the forces on it while it is at rest.

    F=ma
    well it's at rest so there is 0 acceleration, does that mean the formula is not usable or is it simply 2000N?

    partc asks what the value of all forces is at 4.0 m/s^2, would that just be F=2000(4) then F=8000N
     
  2. jcsd
  3. Apr 11, 2005 #2
    Actually gravity exerts a force of 2000g downward while the roof pulls it up with the same force, and the total net force=0.
     
  4. Apr 11, 2005 #3
    ok well the question is, calculate all the forces acting on the elevator

    c) calculate all the forces acting on the elevator when descending 4.0 m/s^2
     
  5. Apr 11, 2005 #4
    Gravity and normal force which are equal.
     
  6. Apr 11, 2005 #5
    it's actually tension and gravity but I need to find the value in Newtons
     
  7. Apr 11, 2005 #6
    9.81*2000=Fg... Same for tension.
     
  8. Apr 11, 2005 #7
    much appreciated but the value 9.8, I remember it from an example, it has to do with gravity, gravities acceleration is 9.8m/s^2 so assuming thats where its coming from a value of 4.0m/s^2 descending would give me an equation of F=2000(5.8)
     
  9. Apr 11, 2005 #8
    You mean the Elevetor is accelerating 4 m/s^2 upward? Gravity stays the same, 9.81(2000) and then torsion (I call it normal force, it is just a matter of what the force is going trought) will be equal to 2000(9.81+4)
     
  10. Apr 11, 2005 #9
    yes it accelerating but it's going downward so does that make a difference?
     
  11. Apr 11, 2005 #10
    If it accelerates downward then its substracted from gravity so 5.8 upward is right.
     
  12. Apr 11, 2005 #11
    Perfect, thanks for the help.
     
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