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Inclined ramp inside an accelerating train

  1. Jun 20, 2007 #1
    Hello,
    I have a question for homework. The question reads:

    A body of mass m=4.8 kg is standing on a ramp inclined in an angle of α=12 degrees. The static friction coeficient between the ramp and the body is μ=0.24. The ramp, which is inclined to the left, is in a train, which accelerates to the right. What is the minimal acceleration of the train that will cause the body to slide up the ramp?

    I thought that for the box to slide up force greater than Ff+4.8*9.8*sin(12) should be applied by train (..by friction that is). From there acceleration of train with respect to the box can be calculated. However I unable to figure out how calculate the acceleration with respect to earth.
    Would appreciate any advice.
     
  2. jcsd
  3. Jun 20, 2007 #2

    nrqed

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    just do a free body diagram picturing the situation as seen from outside the train. There will be a normal force, gravity and the static friction force. Since you want to consider the point when it's just about to slide, use the equation for the maximu static friction force with a magnitude [itex] \mu_s N [/itex]. use the fact that the vertical acceleration is zero to find the value of the normal force (along the y direction you will have gravity acting down an dthe y component of the normal force which will involve "N" and the angle of the plane). Then go along the x direction and impose [itex] \sum F_x = m a_x [/itex] to find [itex] a_x [/itex].
     
  4. Jun 21, 2007 #3
    Thanks for the response nrqed.

    Here is my free body diagram.
    [​IMG]

    N = mgcos(O) =>
    fs = μN = 0.24*4.8*9.81*cos(12) = 11.054 N

    Fnet_x = mgsin(12) + 11.054 = 20.844 N

    Fnet_x = ma = 20.844

    ax = 4.343 m/s/s

    I have suspicion that I am not calculating the
    Fnet_x correctly. Is the above equation correct?
     
  5. Jun 21, 2007 #4

    nrqed

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    OOPS EDIT: The friction force should be down since the block is just about to slide UP!

    Btw I meant by "x" an axis being horizontal.

    But the mistake you are doing is to assume that the net force perpendicular to the plane is zero. This is the case when the incline is not accelerating. But it's not the case here because the net force along x is not zero. So you must do it a bit differently than you usually do it with inclined planes.

    You must start by setting the net force along y to zero. so it will be (y component of N) - mg =0. From this you can solve for N. Then you go along x and find a_x.

    Hope this helps.
     
    Last edited: Jun 21, 2007
  6. Jun 21, 2007 #5
    I think I understand it now.
    Thanks again!
     
  7. Jun 21, 2007 #6

    nrqed

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    You are welcome.

    Best luck!
     
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