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Forces (Just need it checked)

  1. Dec 9, 2011 #1
    Question:
    A crate of oranges slides down an inclined plane without friction. If it is released from rest and reaches a speed of 5.832 m/s after sliding a distance of 2.29m, what is the angle of inclination of the plane with respect to the horizontal?


    I first set up a free body diagram then I discovered that finding the acceleration in the x direction would be the best idea.

    I have the question down to a=7.42 m/s^2 which is correct for the acceleration, but I am getting stuck on where to go next. The formula I used is:

    v^2= Vi^2+2a(x-xi) and that gave me the 7.42 m/s^2. The book says the angle is 41 degrees, but I don't know how to take the acceleration and determine the angle just from the acceleration. Anyone have some pointers?
     
  2. jcsd
  3. Dec 9, 2011 #2

    ehild

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    Think: What forces act on the crate? What force accelerates the crate down the slope?

    ehild
     
  4. Dec 9, 2011 #3
    Thenacceleration along the slope is a component of g acting at an angle.... that should help. I reccomend drawing a nice big diagram of the situation with forces on it and looking for a nice right angled trianlge....
     
  5. Dec 9, 2011 #4
    I did this but the problem never gave a mass. The angle is my desired result. Gravity is acting on the y direction F=mg. So I found acceleration in the x direction so...can I find the angle with the acceleration known even if the mass is not given?
     
  6. Dec 9, 2011 #5
    You have 2 sides to a triangle - g of 9.81 in a vertical sense, and the component along the slope you calculated earlier. Could some trig be applied here?
     
  7. Dec 10, 2011 #6

    ehild

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    Of course you can. You got a, the acceleration in the x direction. That corresponds to a force F=ma along the slope, red arrow in the picture, and that force is the parallel component of gravity (G, in blue). The arrows make a right triangle (yellow) with an angle α, equal to the inclination angle of the slope. You need only remember the definition of sine, cosine, tangent....

    ehild
     

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  8. Dec 10, 2011 #7
    I get the angle as 49.3 degrees to 3 s.f. Correct?
     
  9. Dec 10, 2011 #8
    Possibly you have found the wrong the angle in the triangle there...
     
  10. Dec 10, 2011 #9

    ehild

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    It has to be correct. The acceleration along the slope due to gravity is gsin(α). The acceleration is 7.426 ms-2, that corresponds to 49.2°, instead of 41.

    ehild
     
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