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    So there is a car going around on a bank. If you want to do stuff, you have to resolve to find R:



    I have in my notes to make sure to ALWAYS resolve vertically and never perpendicular to the slope in this case. So:

     RcosQ = mg + Fr sinQ

    So can someone just explain why this is the case? I don't want to just learn it without understanding.

    Thanks
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    (Original post by 2710)
    So there is a car going around on a bank. If you want to do stuff, you have to resolve to find R:



    I have in my notes to make sure to ALWAYS resolve vertically and never perpendicular to the slope in this case. So:

     RcosQ = mg + Fr sinQ

    So can someone just explain why this is the case? I don't want to just learn it without understanding.

    Thanks
    Because this is circular motion, the horizontal components of R and Fr are providing the centripetal force. This means that the vertical component of R is countering the vertical component of Fr and mg.

    In normal statics questions, the component of mg perpendicular to the slope is equal to R, but in banked circular motion, R > perpendicular component of mg, so you can't work it out that way any more.
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    Because you model the car as moving around the inside of an upside down cone not a inclined plane.
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    i know this a stupid question...but i was doing m3 the other day..they mention slipping..but not whether up or down..so do i assume its slipping down?
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    (Original post by azzkikr616)
    i know this a stupid question...but i was doing m3 the other day..they mention slipping..but not whether up or down..so do i assume its slipping down?
    If you could post the whole question i might be able to help..
 
 
 
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Updated: April 10, 2010
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