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Galileo's Slope (Math question)

  1. Sep 21, 2006 #1
    Hey guy's just found this forum, seems pretty interesting, but I've been nested in my bedroom trying to figure out one little thing that miffed me off, I did it in A level physics a while back and remember understanding it completely.

    Just to make the question a little easier I'd like to refer you to the following URL as a referance:


    If you look at the vector component map you have the standard right angled triangle that was practically created by Galileo, there is an extra triangle (done in red in the diragram) that is made from a perpendicular line protudeing from the incline and intersecting the vertical line (g).

    Why do they make this "Theoretical" triangle in almost all of the diagrams? Couldnt i theoretically just use the practical triangle? have G as the vertical (Rather than H) A still being the inclines vector and have a horizontal intersection line rather then a perpendicular line to A.

    I tried doing the math for that rather than the theoretical triangle, comes out (if my math is right) as something like:

    Sin*Theta*= G / A​

    And therefore

    Sin*Theta*A = G​

    Which is wrong evidently because all books use A = GSin*Theta*

    Whats wrong with my method (horizontal intersection) over the accepted method (Perpendicular theoretical intersection line) that makes it so bent for math? I know it seems like an annoying little question but it's been nagging me all night

    Adam K
  2. jcsd
  3. Sep 21, 2006 #2
    I think your maths is wrong, [itex]a\sin\theta[/itex] gives the horizontal acceleration, not g. I'm not quite sure why you're using that triangle. The red triangle gives the relationship between a, g and theta, which is what you want.
  4. Sep 22, 2006 #3
    Yeah but theoretically couldnt i just use the ogirional triangle for the acceleration components? Just substitute h for g and use the same layout.

    The maths was an example of the outcome of using the standard practical layout, why is this "imaginary" triangle used is my question, better yet why does it HAVE to be used in order to get the correct number? Something to do with the perpendicular of the incline == the resulting acceleration vectors?
  5. Sep 22, 2006 #4
    The red triangle is equivalent to the original triangle, the angle theta is given by arcsin(h/L)=arcsin(a/g). So in a way you are using the original triangle. You draw the red triangle like that to make the relationship between a,g and theta more obvious.
  6. Sep 22, 2006 #5
    Yeah but the differance is pretty annoying

    Theoretical triangle:
    Opposite (Opposite of theta) = A
    Hypotamuse (Next to theta but opposite the right angle) = G

    Practical triangle
    Opposite = G
    Hypotamuse = A

    If that helps see the differance

    EDIT: So the maths is differant, its not purely asthetics
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