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Forces Help

  1. Mar 6, 2006 #1
    [​IMG]

    This is a diagram I drew in hopes of making it easier to understand. I'm supposed to find the Normal Force but I really don't understand the concept. I was wondering if someone could help me through it. Thank you!
     
  2. jcsd
  3. Mar 6, 2006 #2

    Galileo

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    The normal force is a 'reaction force', serving to constrain the object to certain regions. I.e. inclines, tables etc. all exert normal forces on object resting or sliding on them so the object doesn't 'fall through'.

    You only gave an object with a force, a horizontal line and an angle. You have to give more information than that. What is exerting the normal force. Is it resting on something, is there gravity, is there a ceiling?
     
  4. Mar 6, 2006 #3
    It is a box that is lying on a floor.
     
  5. Mar 6, 2006 #4

    Galileo

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    Ok, so there's gravity too. Can you tell me the direction of the normal force? How large should it be to keep the box from falling through the floor?
     
  6. Mar 6, 2006 #5
    uhhmm...I don't know the Normal Force...that's what I'm supposed to find. All I know is that it would be perpendicular to the surface that is contacted, so it would be in the opposite direction of Fg, I believe. I'm just not sure how to calculate it.
     
  7. Mar 6, 2006 #6

    Galileo

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    Right, the normal force is perpendicular to the surface (that's why it's called normal), so that would be upwards. So what's the total force acting this direction? Take into account all forces (there are three).
    What should the resultant force be so the object doesn't fall through the floor? (according to Newtons laws).
     
  8. Mar 6, 2006 #7
    So there's Fapp, Fg and Fn?
     
  9. Mar 6, 2006 #8

    Galileo

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    Yes, those are the three you have to take into account. Now you should know the gravitational force and you can calculate the vertical component of Fapp from the information given.
     
  10. Mar 6, 2006 #9
    If the box is being pulled at a force of 29.5, though, isn't that Fapp anyways?
     
  11. Mar 6, 2006 #10

    Galileo

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    Yes, but it's not directed in the vertical direction, Fg and Fn are. If you add forces you have to take into account the directions.

    So, given the information in the diagram, what is the magnitude and direction of the vertical component of Fapp?
     
  12. Mar 6, 2006 #11
    I'm getting 29.3N but that's not right...
     
  13. Mar 6, 2006 #12

    Galileo

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    Do you mind showing your work? We can't really help if we can't see what you did/thought or did wrong.
     
  14. Mar 6, 2006 #13
    Sorry (I drew it out which is why. gimme a sec to upload it)
     
  15. Mar 6, 2006 #14
  16. Mar 6, 2006 #15
    then I just found x using sine law.
     
  17. Mar 6, 2006 #16
    ??
    The answer on the sheet says 81N...
     
  18. Mar 7, 2006 #17

    Galileo

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    Yes, the normal force is 81 N and that will show if you do it correctly. Take it one step at a time, you seem to lack understanding about the nature of forces and such. I suggest you reread your textbook on this section till it becomes clear.

    I don't know where the 90.16 in your picture comes from. The magnitude of Fapp is 29.5 N and it's direction 18 degrees from the horizontal. So for the vertical component, you'd just take Fapp*sin(18 deg) ok? It's just trigonometry. Now add the gravitational force (mind the direction). What is the resultant of these two?
     
  19. Mar 7, 2006 #18
    Yeah I re-did it because I realized my mistake. Thanks.
     
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