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Quiz#4, question 8

  1. Aug 24, 2014 #1
    A physics student in a hot air balloon ascends vertically at constant speed. Consider the following four forces that arise in this situation:
    F1 = the weight of the hot air balloon F3 = the force of the student pulling on the earth
    F2 = the weight of the student F4 = the force of the hot air balloon pulling on the student



    8. Which one of the following relationships concerning the forces or their magnitudes is true?
    (a) F4 > F2 (c) F4 > F1 (e) F3 = –F4
    (b) F1 < F2 (d) F2 = –F4

    Ok I know the answer is (d) because the forces form a pair someone told me but I don't see they form a pair, do they? and why do we have -F4 (I thought the normal force is positive as the balloon is pulling student up the air)
     
    Last edited: Aug 24, 2014
  2. jcsd
  3. Aug 24, 2014 #2

    AlephZero

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    We can't see the diagram, so we don't know what any of this means.
     
  4. Aug 24, 2014 #3
    thanks! I edited the question! :)
     
  5. Aug 24, 2014 #4

    Nathanael

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    Yes, the normal force is upwards, but his weight is downwards (so they have opposite directions)

    His weight pushes down on the hot air balloon with some force, and the hot air balloon pushes up on him with an equal and opposite force (the normal force)
     
  6. Aug 26, 2014 #5
    thanks! (ill get back to this question later, as i am working on another one but I am seeing your responses)
     
  7. Aug 26, 2014 #6

    jbriggs444

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    Since there is no defined surface, there is no "normal" force and no need to characterize any force as such.

    This skips a step.

    In the physics classroom a person's "weight" is the downward force of gravity on their body. It is not the downward force that they then exert on another object.

    The upward force of the balloon on the person (F4) and the downward force of gravity on the person (F2) are not a third law pair. But they do not have to be. This is not a third-law exercise. What about Newton's second law?
     
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