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Component resolution of g on an inclined plane

  1. Sep 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi, I just had a quick conceptual question.
    I understand that if I have an object moving down a ramp, it is being accelerated only by gravity. My question is this:

    We should be able to experimentally calculate g if we measure the acceleration down the ramp. Then we can use trigonometry to determine a value for g. However, this only works if you take the
    acceleration down the ramp divided by the sin (theta) . I understand where this comes from (redrawing the angle theta in the middle where the object starts from), but why can we not calculate the component of gravity by saying:

    sin (theta) = g / a

    This makes more sense to me looking at the initial angle, because g appears to be the opposite side and the acceleration down the ramp appears to be the hypotenuse.

    Can someone please help clarify?



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 11, 2013 #2

    I think it will be easier if you draw a diagram
     
  4. Sep 11, 2013 #3
    pzqwWWV.png

    I realize that an expression for g should be g = a/sin(theta) by looking at the angle that's formed perpendicular to the plane.

    However, if I just look at angle theta in the bottom left corner, why am I not justified in saying g = sin(theta)*a, since to me it appears that the sine of angle theta is just opposite over hypotenuse, or g/a.
     
  5. Sep 11, 2013 #4
    "However, if I just look at angle theta in the bottom left corner, why am I not justified in saying g = sin(theta)*a, since to me it appears that the sine of angle theta is just opposite over hypotenuse, or g/a."

    Because you've not got the correct triangle set up. The hypoteneuse of a triangle is its longest side. The component of the gravitational acceleration (g) in the down incline direction is what gives you a. Since a component (a) can never exceed the full vector (g) in magnitude, it follows that a cannot be the hypoteneuse.

    Chris.
     
  6. Sep 11, 2013 #5
    The hypotenuse is the longest side, which is (correctly) labelled g in your diagram. The side opposite angle ## \theta ## is (correctly) labelled a. So opposite over hypotenuse is ## \frac{a}{g} ## not ## \frac{g}{a} ##.
     
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