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Normal Forces Between Two People on Top of each other

  1. Apr 9, 2015 #1
    1. The problem statement, all variables and given/known data
    During an extreme stunt, a 45 kg cheerleader stands on the shoulders of a 55 kg cheerleader, while riding in a transparent elevator with an upward acceleration of 2 m/s2. Find the normal force between the cheerleaders and between the elevator and the bottom cheerleader.

    2. Relevant equations
    F=ma
    ∑F=ma

    3. The attempt at a solution
    I drew a free body diagram first to identify all forces.

    For the first part I came up with this equation:
    ∑F= Fn - mg = ma
    ∑F= Fn = mg + ma
    ∑F= Fn = m(g + a)
    45 kg ( 9.81 + 2 )
    = 531.45 N​

    For the second part I came up with this equation:
    ∑F= Fn - mg = ma
    Fn = mg + ma
    Fn = m (g + a)
    100(11.8)
    M of cheerleader 1 and 2: 45 kg + 55 kg =100
    = 1181 N ​
    Did I interpret the problem correctly? I always get confused when it comes to normal forces between two objects...
     
  2. jcsd
  3. Apr 9, 2015 #2
    That seems correct to me.
     
  4. Apr 9, 2015 #3

    SammyS

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    For the person on the bottom, you may want to use two normal forces, the one from the top person, which you already know is equal in magnitude and opposite direction to the normal force on the top person. The other is the one you are already using, coming from below. This allows you to isolate the forces acting soly on the bottom person.
     
  5. Apr 9, 2015 #4

    collinsmark

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    I think the others have helped you with the answer. :wink:

    On a different note, I don't mean to be nit-picky, but I'd advise being careful of your notation.

    The above is good. :smile: That's the correct interpretation of Newton's second law, as applied to this problem.
    But from here on down (for this part of the problem) you've revised Newton's second law. The way you are using it, you have changed its meaning.

    The [itex] \Sigma \vec F [/itex] in "[itex] \Sigma \vec F = m \vec a [/itex]" actually means something. It is not put there to say the equation is balanced or the equation makes sense, rather it is there because it is a part of the equation itself. Let's break it down:

    [tex] \sum_i \vec F_i = m \vec a [/tex]

    It means, "The (vector) sum of all forces acting on a body is equal to the body's mass times its acceleration."

    Another way to say "The sum of all forces" is

    [tex] \sum_i \vec F_i \Longleftrightarrow \vec F_1 + \vec F_2 + \vec F_3 + \cdots + \vec F_n [/tex]

    (where [itex] \vec F_n [/itex] here isn't necessary the "normal" force, it's just the nth force.) But once you start moving some (but not all) of the forces to the other side of the equation where the [itex] m \vec a [/itex] is, the [itex] \Sigma \vec F [/itex] no longer applies.
     
    Last edited: Apr 9, 2015
  6. Apr 11, 2015 #5
    I am a bit confused on this part. I don't understand why two normal forces are used and why the top person's normal forces is equal in magnitude and opposite direction to the normal forces on the top person. Is that both referring to the top person?

    For the person on the bottom, the combined weight of person in contact with the elevator and the person on top and respecting the upwards acceleration, does that give me the normal force acting on the bottom cheerleader?

    Is the normal force experienced by the cheerleader on top 531.45N?


    I see my mistake. My textbook gave me this equation for apparent weight in an elevator accelerating upwards: Wa - W = ma
    So basically I need to leave out the ΣF for correct notation.
     
    Last edited: Apr 11, 2015
  7. Apr 11, 2015 #6

    SammyS

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    That's simply Newton's Third Law.

    The force that the top person exerts on the bottom person is equal (in magnitude) and opposite (in direction) to

    the force that the bottom person exerts on the top person.
     
  8. Apr 11, 2015 #7
    Oh, ok that makes sense.
     
  9. Apr 11, 2015 #8
    Does this mean that I have to subtract 531.45N from 1181N to get the normal force acting on the bottom person?
     
  10. Apr 11, 2015 #9

    SammyS

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    I wouldn't state it that way. Rather there are two forces in addition to gravity which act on the bottom person. One is the normal force the floor exerts upward, 1181Newtons. The other is the normal force exerted downward by the top person, 531.45Newtons.

    I would not lump these together as one single normal force.

    This way, when you account for gravity acting on the bottom person you only use that person's weight.
     
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