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Physics Involving Newton's Laws of Motions (specifically g, and the normal force)

  1. Oct 5, 2004 #1
    An engineer claims to have a new kind of "safety" air-brake for elevators. If the cable should snap, a patented velocity-sensor turns on the brakes once the elevator hits 40 mph. The brakes are powerful enough to stop a 700 lb elevator in 1/20 of a second.

    How much force do the brakes exert to stop the elevator like this? Why isn't the answer 700 lbs? Isn't the force of gravity only 700 lbs?


    I'm not sure where to begin with this one, I understand the concept of the elevator is being held taut by the elevator cord, and that on the sides, there is Ff (f=friction), but I am not really sure what to do with the information being given to me in this problem. Could someone give me a general idea how I should attempt this problem? Thanks in advance.
     
  2. jcsd
  3. Oct 5, 2004 #2

    Pyrrhus

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    The elevator will be moving downwards, and it will have a weight of 700 lbs, so the problem states when the elevator has a speed of 40 mph, then it can stop it, reach a speed of 0 in 1/20 of a second. (Acceleration Equation is ?)

    The answer is not 700 lbs, because the elevator is moving.

    Do a freebody diagram, the brake is supposedly air-brake, so i doubt there will be friction on the sides.
     
  4. Oct 5, 2004 #3
    ok, I somewhat understand the problem a little better, but I did something wrong...

    a=g=9.8 since acceleration is constant? because the cable just snaps, so it's just gravity, and a can be set equal to

    a=Force/mass

    and mass is equal to weight/g which in this case, then 700 lbs/9.8 m/s^2 gives me 71.4285714286, so then, I times both sides by mass, and get

    (9.8)(71.4285714286) = 700 N, so but that was already ruled out, what did I mess up?

    I have another equation where F=(micro)N where (micro)=coefficient of friction and N=normal, but I don't have to coefficient of friction in this problem, so I can't really use that
     
  5. Oct 5, 2004 #4

    Pyrrhus

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    Think about the brake.

    You got the Force the brake will exert to stop the elevator from falling and you got the weight of the elevator, on your diagram, and their moving on the y-axis.
     
  6. Oct 6, 2004 #5
    not really, I'm still lost on this problem... gah, so much work for just one problem... UGH.
     
  7. Oct 6, 2004 #6

    Pyrrhus

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    Think about

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = \frac{d \vec{P}}{dt} [/tex]

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = \frac{\Delta \vec{P}}{\Delta t} [/tex]

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = \frac{m \Delta \vec{v}}{\Delta t} [/tex]

    By the way that's newton second law, and P is lineal momentum.

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = \frac{d \vec{P}}{dt} [/tex]

    If the mass is constant, then it'll be:

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = \frac{m d \vec{v}}{dt} [/tex]

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]
     
    Last edited: Oct 6, 2004
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