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Normal Force acting on tries of a go-cart

  1. Aug 21, 2011 #1
    1. The problem statement, all variables and given/known data
    The total weight of the go-cart and driver is 1068N. The location of their combined center of mass is shown in attachment. The rear wheen exert a 106.7 N horizontal force together on the track. Neglect horizontal force on the front wheels.

    a) What is go carts acceleration? [SOLVED MYSELF]
    b) What are the normal forces exerted on the tires at A and B?

    2. Relevant equations
    Moment around centre of mass.


    3. The attempt at a solution
    NA+NB-W = 0 (where NA and NB is normal force acting on the tires)

    Moment around CM in clockwise direction
    NA*(0.4064)+F*(0.381)-NB*(1,1176) = 0

    solving for NA and NB I get
    My answer; NA = 756,52 N NB = 485 N, which is wrong according to the book.

    I don't know what is wrong. Please help.
     

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  3. Aug 21, 2011 #2

    Doc Al

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    Staff: Mentor

    What orientation (sign) is the torque from each force?
     
  4. Aug 21, 2011 #3
    Not quite sure what you are getting at, but I can take either clockwise or counter clockwise as direction goes. Even if I take it counterclockwise it turns out wrong..
     
  5. Aug 21, 2011 #4

    Doc Al

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    There are three forces acting. Describe the direction of the torque from each, clockwise or counterclockwise. (You're messing up your signs in your equation.)
     
  6. Aug 21, 2011 #5
    Ok so I checked so I belive it should be F+NA-NB as the force is below the centre of mass the cart want to rotate backwards. Doing this I get NA = 788,29 NB= 279,7, this is fairly close to the books answer but it's not correct.
    I'm not hunting for an answer tho, just the correct way of thinking. I thought I understood this type of question, but I seem to get it wrong every time :P
    Please continue to guide me through this! :)
     
  7. Aug 21, 2011 #6

    Doc Al

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    How is this different from what you had before?
    I'm trying to! :smile: Again: Describe the direction of the torque from each force as clockwise or counterclockwise.
     
  8. Aug 21, 2011 #7
    I'm not sure what you mean, im afraid. I have no resource to study this, but i'll try.
    You want me to describe direction of torque from each force as cw, or ccw.

    NA would be CCW, as I stated its because it's below centre of mass.
    NB would be CCW.
    The force F is probably what im having problem with, and it's direction. The force could possibly be CCW aswell. I have an exact answer from a home page I found and it's attached, there they state the force of torque around CM as described in attachment. Its so wierd as I am only +20 N away from the correct answer, and no matter how I choose to change the forces from CCW to CW it wont change just 20N.
    I did the exact problem with a refrigirator and the only general difference is that a) It does not have wheels, 2) NA and NB is on the edges of the refrigiratior on the floor. The only thing not used in this problem is the radii of the wheels, but how can they affect the problem? Im getting a headace...
     

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  9. Aug 21, 2011 #8

    Doc Al

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    Incorrect. You must consider the direction of the force, not just where it acts. If Na were the only force acting, which way would the cart tend to rotate?
    Good.
    No 'possibly' about it--definitely. Again, if F were the only force acting, which way would the cart tend to rotate?
     
  10. Aug 21, 2011 #9
    NA should def be CW as it acts upwards from the floor.

    The cart must rotate CCW. Just as a fast moving car or motorcycle which accelerate fast enough would rotate CCW. So my final answer is that NA is CW and F and NB is CCW in this case.

    What would be the appropriate next step?
     
  11. Aug 21, 2011 #10

    Doc Al

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    Good.
    Good.
    Set up a net torque = 0 equation just like you did before. (Assign one orientation as positive, the other as negative.)
     
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