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Inclined Plane and Elevator with Acceleration

  1. Aug 5, 2009 #1
    1. The problem statement, all variables and given/known data

    An inclined plane, fixed to the inside of an elevator, makes a 40 angle with the floor. A mass m slides on the plane without friction. What is the acceleration relative to the block if the elevator is a)accelerating upwards .35g b)accelerating downwards .35g? c)constant speed





    2. Relevant equations
    F=ma


    3. The attempt at a solution



    3. The attempt at a solution
    I know there's two accelerations I have to take into the account. The acceleration as the block slides down the incline, and the acceleration of the elevator. Combining these two, I can find the relative acceleration. I chose the reference frame to be titled so that the x-axis is along the incline. The thing I can't figure out is how the motion of the elevator affects the motion of the block. For example, when the elevator is accelerating upwards. Does the block accelerate upwards with it? From the block's force diagram I only have the normal force and mg down. the weight provides the acceleration down the incline but I have no clue what the elevator's acceleration does to the system. Any help would be appreciated.

    So lost though, how does the forty degree angle come into play?
     
  2. jcsd
  3. Aug 5, 2009 #2

    ideasrule

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    Assume there's no acceleration, just gravity. Do you know how to split gravity into a component parallel to the incline and one perpendicular to it? (Hint: draw a vector diagram)

    Now, if you're in an accelerating elevator in vacuum (no gravity), everything seems to accelerate downwards or upwards at the same rate regardless of their mass. That's exactly how gravity behaves: it accelerates all objects at the same rate, regardless of their mass. So acceleration can be considered to be additional gravity. Whereas gravity provides mg of force, acceleration provides ma; whereas gravity accelerates objects at g, acceleration does so at a. This is the equivalence principle, the central tenet of general relativity.
     
  4. Aug 5, 2009 #3
    yeah so mgcos40 right?

    .35mgcos40-mg??? or am I way off? **** this is due in two hours
     
  5. Aug 5, 2009 #4

    ideasrule

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    No, it's mgsin40; mgcos40 is the component perpendicular to the incline.

    Acceleration adds to or subtracts from gravity, net "force"=(mg+ma)sin40 or (mg-ma)sin40, depending on which way the elevator is accelerating. a=F/m, so a=(g+a)sin40 or (g-a)sin40
     
  6. Aug 5, 2009 #5
    Alright, I figured it out!
     
    Last edited: Aug 6, 2009
  7. Aug 6, 2009 #6
    Just as a curious second question, what would the acceleration be if the elevator was falling freely?

    or if it went upward at a constant speed?
     
  8. Aug 6, 2009 #7

    kuruman

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    Use the same equations that you have derived alredy.

    For free fall set a = g
    For constant speed set a = 0

    What do you get?
     
  9. Aug 6, 2009 #8
    yeah I figured free fall would be zero!

    THANKS!DSGHSDKGHDS
     
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