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Net acceleration, free body diagrams, velocity, force normal

  1. Mar 29, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-3-29_20-22-38.png

    2. Relevant equations
    relevant equations are provided with each question below

    3. The attempt at a solution
    please check my solutions to ensure they are correct. thanks.
    a)

    upload_2016-3-29_20-21-57.png
    b)

    Calculate net force:

    Fnet = FA + Ff + Fg

    Fnet = (1.20 x 104 N [up]) + (1.40 x 103 N [down]) + (10094 N [down])

    Fnet = (1.20 x 104 N [up]) – (1.40 x 103 N [up] + 10094 N [up])

    Fnet = 506 N [up]

    Calculate net acceleration:

    Fnet = m a

    a.net = Fnet /m

    = 506 N [up] / 1030 kg

    = 0.49 m/s2 [up]

    c)
    upload_2016-3-29_20-22-12.png


    d)

    Relevant Equation to solve for the force normal:

    FN = mg + ma

    FN = (35 kg) (9.8 N.kg [down]) + (35 kg) (0.49 m/s2 [up])

    FN = 360.15 N [up]

    The force normal acting on this passenger is approximately 360 N [up].

    e)

    Given information:

    v1 = 0

    a = 0.49 m/s2 [up]

    t = 12.0 s

    Unknown:

    v2 =?

    Equation:

    v2 = v1 + a x total time

    = 0 + (0.49 m/s2) x (12.0 s)

    = 5.88 m/s [up]

    The velocity is approximately 5.9 m/s [up]
     

    Attached Files:

  2. jcsd
  3. Mar 29, 2016 #2

    Simon Bridge

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    You always do nice diagrams... you mostly did good, I'll add some notes for tweaks and questions to help you think about things.

    (a) thats not a fbd ... just draw a box, forces should be shown coming from their point of action.
    Where two or more arrows must overlap, you can "explode" the diagram, say, using a dotted line to indicate where the forces act.

    (b)You were careful to indicate direction for forces... good.
    A shortcut is to write which direction is positive... ie "+ve = upwards" ... then you can just convert.
    The 10^4 didnt come out.

    It is best practise to do the algebra first... ie. F-W-f=(W/g)a : where W is weight, F is the applied force, f is friction, and g is the acceleration die to gravity. Then a = g(F-W-f)/W ... plug numbers in as magnitudes since direction is already in the realtion.

    (c) you may have too many forces... the fbd should isolate the mass in question, so do not draw the elevator in. What are the forces just on the passenger?
    Note: the passenger accelerates at the same rate as the whole elevator, so you can check your answer against what you got before.

    (d) this equation should come from a free body diagram

    (e) works fine... you can check by sketching the v-t graph.

    Well done.
     
  4. Mar 29, 2016 #3
    for c) are you saying the diagram is incorrect, regarding the force normal, gravity force, and force applied. the person accelerates at the same rate as the elevator (Fapplied), there is also gravity acting downwards on the person, and then the normal force that supports the weight of the person. You mentioned that i have too many forces, what does this imply?
     
  5. Mar 30, 2016 #4

    Simon Bridge

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    By your numbers: a = (FA+FN+Fg)/m ... so what is this value using:
    FA=12000N, FN=360.15N, Fg= -343N, and m=35kg (your figures)?
    Compare this value with that obtained for the entire elevator in part (a)

    Consider: Where on the person does the "force applied" act?
     
    Last edited: Mar 30, 2016
  6. Mar 30, 2016 #5
    are you referring to the direction that the force applied acts on the person. So wouldnt that be up, because the elevator is going up.
     
  7. Mar 31, 2016 #6
    so for c) is this how the fbd goes:
    upload_2016-3-31_14-30-30.png i used the same forces, but just took out the elevator to isolate the passenger ?
     
  8. Apr 1, 2016 #7

    Simon Bridge

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    Please answer guiding questions - vis:
    By your numbers: a = (FA+FN+Fg)/m ... so what is this value using:
    FA=12000N, FN=360.15N, Fg= -343N, and m=35kg (your figures)?
    Compare this value with that obtained for the entire elevator in part (a)
    No... "up" would be a direction. I have asked about a location.
    eg. If I pull on your hand, the location where the force of me pulling on you is acting is "your hand".

    For the lift: the applied force comes via a cable attached to the roof - so the location is "on the roof".
    However - there is no cable attached to the person.

    Isolating the passenger is good - but you should draw the arrows as forces that are attached to the passenger ... i.e. if the passenger had a rope pulling on each arm, you'd show an arrow coming from each arm. The person's weight acts from their centre of mass, for eg. and points down.
    There must be an upwards force so where does it act? Does it pull the passenger up by the hair for example?
     
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