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Pseudo elevator question-not enough info I think

  1. Sep 19, 2008 #1
    A box with a 75 passenger inside is launched straight up into the air by a giant rubber band.
    After the box has left the rubber band but is still moving upward, what is the passenger's apparent weight?

    This is all the information that was given. The question prior to this one wanted to know what the true weight is. I found it to be 735 N which the computer rounded to 740 N. I understand the concept here but can not find anything in the book to help. I thought I would need to be given the velocity or acceleration of the box upwards. I know the accel is g but I am lost. Can I get some help setting this up?
     
  2. jcsd
  3. Sep 19, 2008 #2

    Doc Al

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    You have all the information you need. Hint: The apparent weight is the normal force of elevator pushing up on him. What does that force equal?
     
  4. Sep 19, 2008 #3
    I know that the normal force is equal to weight. I thought that since he is accelerating upwards then his apparent weight would increase. n=w=mg just in the oppisite direction. At rest and constant velocity it would be his true weight. I am missing something and really do not know what it is.
     
  5. Sep 19, 2008 #4

    LowlyPion

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    While the elevator is accelerating upward on the force from the rubber band, that force is translated to the passenger. But what happens within the elevator when there is no further force exerted except the slowing force of gravity on both the passenger and the elevator together as they fly through the air?
     
  6. Sep 19, 2008 #5
    I see what you are saying. He should not be accelerating upwards but being slowed due to gravity. I have tried the answer 740 for this problem thinking that his weight would just be his mass under the force of gravity. The way I understood from class is that the floor is still pushing up on him and therefore his weight would feel heavier. I can picture what is happening in this situation. I see a box that has just left the rubber band and is traveling upwards. I know that he is slowing due to gravity but I still see the box pushing up on him making his apparent weight heavier. I really do not know where to go from here.
     
  7. Sep 19, 2008 #6

    LowlyPion

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    And the box is not being slowed by gravity as well?
     
  8. Sep 19, 2008 #7
    I guess the problem that I am having right now is that w=mg. If the whole problem is that the box is being slowed by gravity, then the acceleration is 9.8. If this is true then the apparent weight would still be 735 N. I know this is not the correct answer as I have considered that and submitted it to be told it is wrong. There would be a difference if you took in to account the initial + acceleration from the box and showed the effect of gravity but I do not know what it is. I see what you are saying about the box being slowed also and I guess I am looking at this from the split second it takes off and the bow is pushing againist the man causing his weight to appear heavier. I am misiing something and have no clue what it is. Thanks forr all the help so far.
     
  9. Sep 20, 2008 #8

    Doc Al

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    Don't confuse apparent weight with the true weight. The true weight is w = mg, and that's not going to change (unless they move far from the earth's surface). The apparent weight is the support (or normal) force that holds you up.

    Some examples: If you are just walking around, not accelerating, then the upward force of the floor (which is your apparent weight) must balance the downward force of gravity (your true weight). What if you are walking toward the end of a diving board. What happens to your apparent weight when you step off the board into the air?

    Back to your passenger in a box. What forces act on him? Obviously his true weight (mg), which acts down. What about the force of the box pushing him up? (That would be his apparent weight.) How much is that force? You know that once the box is no longer being pushed by the rubber band, its acceleration (and the passenger's) is just g downward. So what must the net force be on the passenger? (Consider Newton's 2nd law.) What does that tell you about the upward force of the box?
     
  10. Sep 21, 2008 #9
    Well I finally got the answer. I understand what everyone was saying but I am unsure why. I found his apparent weight to be 0 N or at least that is what the computer accepted as the answer. I know the box is under the effect of gravity and so is the man. I do not see how he feels no weight. I guess it is because the box is not pushing againist him anymore but I feel like he would still feel weight. If anyone would like to explain a little bit more that would be great. Thanks for the help.
     
  11. Sep 21, 2008 #10

    LowlyPion

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    While the two are in flight, if the man was transmitting force to the elevator, and the elevator is in free fall wouldn't that serve to separate the two from contact?

    Incidentally this is how the Vomit Comet that introduces astronauts to momentary weightlessness operates. The plane develops a rapid rate of ascent and then removes power into a freefall allowing the frame of reference of the plane to be accelerating earthward, as its upward speed slows until it is plunging earthward and recovers.
     
    Last edited: Sep 21, 2008
  12. Sep 21, 2008 #11
    He would still technically have weight but because there is nothing pushing up against him, his apparent weight would be 0.

    It's like diving on a diving board into a swimming pool. From the time you jump to the time that you hit the water, your apparent weight is 0. Another example would be if you tied a bathroom scale to your feet and then jumped off a chair. Assuming no air resistance, the scale would read 0.

    The apparent weight requires a normal force which you do not have if both you and the box are free falling at the same rate. This is the reason why the guys in the space shuttle are able to float ; they are technically free-falling towards the earth.
     
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