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Tension on elevator cable

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data
    An empty elevator of mass 2.7x10 (3) kg is pulled upward by a cable at an acceleration of 1.2 m/s (2)

    a) What is the tension in the cable?
    b) What would be the tension in the cable if it was accelerating downward at 1.2 m/s (2)?
    c) What would be the tension in the cable if it was moving with a constant velocity of 0.50m/s?


    2. Relevant equations
    So far as I know,
    F=m x a


    3. The attempt at a solution

    I am pretty new to physics and I've been doing alright so far but these elevator tension questions are killing me. I actually have no idea where to go. When I worked it out (probably not right) I found the cable to have more tension in the downwards acceleration rather than upwards. I'm pretty sure I'm wrong I just need someone more experienced to tell me so. I would be really greatful for any help!! Thanks!
     
  2. jcsd
  3. Oct 28, 2009 #2
    Here are a couple hints (I hope they help):

    - If the elevator was in free fall (accelerating downwards only due to gravity), the cable would not be exerting any force on the elevator.
    - Also, if the elevator was just hanging there (not going up or down), the cable would have to exert how much force?
    - If the first two hints helped, think about how much MORE or LESS force is required to get the elevator to accelerate up or down at the rates given
     
  4. Oct 28, 2009 #3
    So when the elevator is stationary the forces would be equal since it is not accelerating. When it's accelerating upwards the cable should have more tension right?

    The way I'm trying to solve it I found the upwards force and the downwards force in a free-body diagram and then adding them. Am I on the right track or am I missing something? I found another equation T=m(g+a) do I need to add both accelerations before calculating the force in a direction to which more than one acceleration is affecting it?
     
  5. Oct 28, 2009 #4
    Correct. In fact, whenever the elevator is not accelerating (whether it's stationary or not), the forces are equal because anet = 0 (HINT for #3).

    Correct again! Now, the key is how much MORE tension? Again, remember Fnet = manet. Since you're now getting the elevator to accelrate upwards with 1.2 m/s2 "more" acceleration, then how much MORE force does it take...

    EDIT:
    Yeah, that's what I was trying to hint at. I didn't see you already posted this. Whoops!
     
    Last edited: Oct 28, 2009
  6. Oct 28, 2009 #5
    Ok I'm starting to get this a little better now but there is one thing that still puzzles me. You said how much "more" acceleration, but the upwards acceleration is going up and gravity goes down so wouldn't the acceleration be (9.8 m/s2 - 1.2 m/s2)? When I do this I get a higher force when the elevator is accelerating DOWNWARDS which just doesn't make sense!
     
  7. Oct 28, 2009 #6
    Sorry for the confusion. To keep the elevator from accelerating upward or downward (basically to keep it at a constant velocity), the tension force MUST be equal to the force of gravity of the elevator. In essence, the cable is providing an upward acceleration of 9.8 m/s2 to balance out the downward acceleration of gravity. So, when the elevator has a net acceleration of 1.2 m/s2 upward, that means it has to supply that much MORE force.
    So, the anet = acable - g.

    Going down, the cable only requires enough force to keep the elevator going at whatever acceleration is needed. For example, if the elevator is to be in free fall (9.8 m/s2), the cable would provide NO force. If the cable were to accelerate downward at 9.7 m/s2, the cable only needs to provide a 0.1 m/s2 upward acceleration (and the force associated with it).
    So, in this sense, anet = g - acable.

    Does that make sense? Sometimes I find it easier to explain this on a board or in person instead of over the internet! Sorry!
     
  8. Oct 28, 2009 #7
    OOOH ok I got it now thanks sooo much! You actually justed saved me :)
     
  9. Oct 28, 2009 #8
    You're welcome. I'm glad I helped!
     
  10. Nov 4, 2009 #9
    a) For when the elevator is accelerating upwards at 1.2m/s^2:
    Tension = m(g+a)
    = 2.7x10^3kg (9.8m/s^2 [upwards] + 1.2m/s^2 [upwards])

    Is this correct? I'm trying to interpret this; g = 9.8m/s^2 [UP] because the Tension is acting opposite to gravities force on the elevator?

    This would make sense because it would give me a greater tension when the elevator is accelerating upwards rather than downwards.

    Another way that i did it prior to finding the tension equation was:
    Fnet = Fgravity + Ftension rearranges to;
    Ftension = Fnet - Fgravity

    Would this equation work as well?
     
    Last edited: Nov 4, 2009
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