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2 motion questions

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    1. Boat is towing two water skiers, each skier has it's own rope that's at 40 degrees the boats axis and the tension is 450N, how much force does the boat have to have to keep them all moving?

    2. The plane is traveling at 200 km/h at the bottom of the loop, and the scale (the pilot happens to be sitting on one) reads 4x normal weight/Fg. What is the radius of the loop.


    2. Relevant equations
    1. I don't really know...

    2. I don't really know...


    3. The attempt at a solution
    My teacher put these on an assignment that's due tomorrow, so I really need some info fast.

    I'm thinking she missed out on some variable... Please correct me if I'm wrong.
     
    Last edited: Oct 13, 2008
  2. jcsd
  3. Oct 13, 2008 #2

    Hootenanny

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    "I don't really know" isn't an attempted solution. There are no missing variables, the questions are complete and can be answered as set.
     
    Last edited: Oct 13, 2008
  4. Oct 13, 2008 #3
    See, the equations I got in my lessons are of circular motion (v=2piR/T) and a pulley equation a=ft/m.

    I don't know how to fit them in. Could you please at least direct me in the right direction?
     
  5. Oct 13, 2008 #4

    Hootenanny

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    Well both those equations are irrelevant for the first question since the skiers aren't undergoing circular motion.

    For the second question (which is circular motion), do you have an equation connecting centripetal acceleration/force and velocity?
     
  6. Oct 13, 2008 #5
    I have many equations for circular motion: v=2piR/T a=2piv/T a=4piR/T^2 a=v^2/R

    Also centripetal force: Fcent=mv^2/R Fcent=m4pi^2R/T^2 Fcent=m2piv/T

    Basically, for circular motion, I don't possess the period and for the centripetal force, I don't have mass and period.
     
  7. Oct 13, 2008 #6

    Hootenanny

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    This equation looks appropriate for the second question.
     
  8. Oct 13, 2008 #7
    I was looking at that too, but I need a mass variable, which the questions is lacking. Am I overlooking something?
     
  9. Oct 13, 2008 #8

    Hootenanny

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    What is the net force acting on the pilot at the bottom of the loop?
     
  10. Oct 13, 2008 #9
    Right, it's 4x Fg.

    What I'm understanding is this:
    Fg=mg
    Fg=m(9.8m/s^2)

    Then Fg would be multiplied by 4, which will give me the Fcent. Then just plug it into the equation to get R.
     
  11. Oct 13, 2008 #10

    Hootenanny

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    You're close, but not quite correct. 4mg would be the reading on the scales, which is the normal reaction force of the scales, but this is not the same as the net force acting on the pilot. You need to find the net force acting on the pilot and use this as your centripetal force.

    Do you follow?
     
  12. Oct 13, 2008 #11
    I know that at the bottom of a "loop", Fcent is Ft-Fg. Is this what you are getting at?
     
  13. Oct 13, 2008 #12

    Hootenanny

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    Looks good to me :approve:
     
  14. Oct 13, 2008 #13
    Very well. Still, I'm in need of a mass, which the question lacks...

    As for my first question, could you steer me as well? My substitute teacher gave us this and I think she messed up on a couple of lessons.
     
  15. Oct 13, 2008 #14

    Hootenanny

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    Are you sure you need to know the mass? :wink:
    If you assume that the skiers are travelling at a constant velocity, what can you say about the net force acting on them?
     
  16. Oct 13, 2008 #15
    Mass: not following you there, I'm afraid.

    1st question: No acceleration means no net force. Fnet=Ft? So Ft would be 0.
     
  17. Oct 13, 2008 #16

    Hootenanny

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    Try writing out the equation that I quoted before.
    Correct and the same goes for the boat.
     
  18. Oct 13, 2008 #17
    Alright, first is done. :P

    Second:
    Fcent=mv^2/R
    Ft-Fg=m*(55.55m/s)^2/R
    Please clarify on the next step.
     
  19. Oct 13, 2008 #18

    Hootenanny

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    Well what is Ft-Fg?
     
  20. Oct 13, 2008 #19
    Ft-Fg=Fcent=Fnet
     
  21. Oct 13, 2008 #20
    Please respond... this thing is due tomorrow.
     
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