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Elevator Acceleration Help needed

  1. Nov 7, 2006 #1
    I got a list of problems that will be on some of our exams soon, and i dont know how to do a few of them, heres one.

    a 60kg passenger standing on a scale in an elevator notes that the scale's maximum reading is 800 newtons when the elevator accelerates upward. the scales minimum reading is 400 newtons.
    You must draw a free-body-diagram of the passenger. You must use the free-body-diagram to determine the elevators acceleration when the scales reading is at the maximum and the minimum.



    I know how to draw the diagram, just now sure how to solve the problem.
    My geuss is he weighs 60kg, which is 588.6 N after multiplying kg*9.81.
    800-588.6=211.4
    a=F/m = 211.4 N/60 2Kg = 3.52 m/s^2 down
    gravity is expierienced, so total acceleration= 3.52m/s^2 +9.81 m/s^2 = 13.33 m/s^2 down...Thats the total acceleration of the elevator when the scale is its max. Does that look anywhere near correct? Or am i doing this completely wrong? Please help, i have a few more problems too, i really need to know these for the exam i will be taking.
     
  2. jcsd
  3. Nov 7, 2006 #2

    PhanthomJay

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    When the scale reads 800, the net force is UP (800 up -588down = 211 UP) So the accelration is up. Forget about total aceleration! The net force is 211 up, so the acceleration works out to 3.52m/s/s UP. That's newton's law. Now try the 2nd part when the scale reads lower.
     
  4. Nov 7, 2006 #3
    Before i try to find the 2nd part, when the elevator is going down, let me make sure i understand what your saying about my answer...

    When the scale is reading 800 Newtons, the elevators acceleration is 3.52 M/S^2 up. Is that correct? It seems like too little work was done to prove that...
     
  5. Nov 7, 2006 #4

    PhanthomJay

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    Yes, you are right. That is the person's acceleration with respect to the ground. But since the person stays in contact with the floor, I should have noted that the acceleration of the elevator with respect to the ground must be the same!
     
  6. Nov 7, 2006 #5
    My geuss is he weighs 60kg, which is 588.6 N after multiplying kg*9.81.
    588.6-400=188.6
    a=F/m = 188.6 N/60 2Kg = 3.14 m/s^2 down, so it the answer would be -3.14 m/s/s. Right? So the total answer would be that when the scale is at its MAX at 800 N, the elevators acceleration is 3.52M/S/S and when the scale is at its min. the elevators acceleration is -3.145M/S/S?
     
  7. Nov 7, 2006 #6

    PhanthomJay

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    Yes, that is correct. The person accelerates upward at 3.52m/s/s and feels heavier, and then decelerates (accelerates down) at 3.14m/s/s when coming to a stop, and feels lighter during this deceleration phase. These are also the same magnitudes and directions for the elevator's accelerations.
     
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