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Inclined Plane Clarification.

  1. Jul 30, 2012 #1
    I know that force parallel (the force of the block down the inclined plane) in this case should be mgSin(theta), but why is the derivation in the picture wrong? According to that triangle could it not be mg/(Sin theta)?
     

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  3. Jul 30, 2012 #2
    the corner with an angle of theta is the one with an arm to the center and an arm to the F parallel axis and sine theta is equal to the opposite side divided by the hypotenuse. So indeed sin theta = F parallel / mg so F parallel = mgSin(theta) is correct.
     
  4. Jul 30, 2012 #3
    I agree, sin theta is equal to opposite divided by hypotenuse. But opposite of theta in the picture is mg, and the hypotenuse is F parallel. So that gives us Sin theta = mg/F parallel. Which makes F parallel equal to mg/sin theta....do you see the dilemma?
     
  5. Jul 30, 2012 #4
    In the picture the f parallel line and the geen line make the same angle as the top red line and the green line, correct?
    This second angle mentioned must have an angle of 180-90-theta=90-theta, correct? (Since you know two angles in the triangle).
    so the first angle mentioned must also be 90-theta.
    Now complete the triangle by drawing a line from the tip of the green arrow under a right angle onto F parallel axis. This gives you the proportions in which the forces can be split.
    Now the hypotenuse in in this picture is the green arrow and opposite the F parallel axis is an angle of 180-90-(90-theta)=theta.

    What went wrong is you are using the wrong triangle and there is not just one angle equal to theta. The one drawn just tells you where in the right triangle the angle is theta. You could also have done the exercise is the other non-right angle of the triangle was given.
     
  6. Jul 30, 2012 #5
    Yes, I see how using other triangles gets you to the correct result of mgsin theta for F parallel. I suppose my question is why can one not use the triangle in the image below? It is certainly just as valid of a triangle as any other, and all rules of trigonometry to my knowledge have been followed. There has to be some reason why one cannot use that triangle to get the result.
     

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  7. Jul 30, 2012 #6

    Doc Al

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    As conquest has explained, you're imagining the wrong triangle. If you want to find the components of a vector using a right triangle, then that full vector must be the hypotenuse of that triangle. The other sides are the components. (The components are always smaller than the original vector.)

    Since you want the components of the weight, then it's the weight that must be the hypotenuse of your right triangle.
     
  8. Jul 30, 2012 #7
    Thank you I will look into this a little more. As I understand geometry, as long as you can validly construct the triangle, it cannot be a 'wrong' triangle. Any and all triangles that I could construct validly using the information given should lead to the same result. I know there's an error here - I just have to find it.
     
  9. Jul 30, 2012 #8

    Doc Al

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    You can certainly construct a triangle, as you did, where the weight (mg) is one of the sides. But the hypotenuse will not be the component of the weight parallel to the incline. (That triangle has no physical meaning and its hypotenuse is mislabeled.)

    The problem is not with the geometry of triangles, but with using a right triangle to represent a physical situation.

    Another hint: The component vectors must be perpendicular to each other.
     
  10. Jul 30, 2012 #9

    Doc Al

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  11. Jul 30, 2012 #10
    Yes, I see how the F parallel comes out to be larger than mg if one is allowed to construct that triangle. Perhaps that is the reason? Mg and F parallel are vectors, whereas the bottom of that triangle is not. Maybe since the vectors are in vector space and the bottom of the triangle is not, then that cannot be a valid triangle. After playing with this a bit more, I can see that mixing vectors and real lines leads to many triangles that will give you answers that do not correspond to reality.
     
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