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Accelertion Due to gravity on a ramp equation

  1. Mar 3, 2009 #1
    I'm new to this forum so please don't mind my being a newbie. I found this equation at school, .5((90-theta)/90)*G=Ag where theta is the angle on the base (not the right angle)


    |\
    | \ <ramp
    |_ \<Theta
    ^
    right angle

    and G is the acceleration due and Ag is acceleration in the direction of the ramp. i found this bu knowing if it's a 90 (horizontal/flat) degree angle then there is no acceleration and if it's 0 (vertical) degrees then there's full acceleration. so every degree in between 0 and 90 it adds (1/2)1/90*G to the acceleration to the acceleration in the direction of the ramp.
     
  2. jcsd
  3. Mar 3, 2009 #2

    Nabeshin

    User Avatar
    Science Advisor

    Your equation is incorrect.

    While it may seem tempting at times to contrive some relationship between two data points an extrapolate a linear relationship (what you've done), this type of analysis is often incorrect (and difficult to prove).

    The correct equation results from the fact that the gravitational force parallel to the incline is given by:
    [tex]F_{//}=mgsin(\theta)[/tex]

    So acceleration is given by:
    [tex]a_{//}=gsin(\theta)[/tex]
     
  4. Mar 4, 2009 #3
    you have to break it up into two components. Thats the idea and thats what makes it a triangle and for triangles we have trigonometry. So how much "down" are you going if you are going that much "side"? well Fg=mg but thats only if you drop something directly down to the ground so if its on an angle and its the angle of the slope to the gound then you need to do (mg)sin(theta). So this gives you the force, now you have to find the acceleration and f=ma so f/m=a so just divide your previous answer by m and you are left with a. Hey what do you know? Newton's observations were correct, mass is canceled and irrelevant because the object accelerates at a certain rate despite it's mass. woohoo!!! i took a few extra steps to show you how the idea works :P. Oh by the way, a sin function is nonlinear therefore your linear equation i.e. (1/2)1/90*G simply cant be right. i mean i could be missing something but the only time lines and curvy lines are related and interchangeable is in calculus and this is high school physics so unless your prof or someone used calculus to get this relationship i just don't think it makes sense. The rate of change changes depending on theta so you cant treat it as if it doesn't and expect to be right.
     
  5. Mar 4, 2009 #4
    i decided to change the equation to theta/90*g=Ag... My physics teacher told me that it was sheer luck that my data came close to the same. i was using a ball down a ramp and apparently the ball rolling can cause changes in the accelaration. so i'm gonna change the ball to a cart and see what went wrong so i can figure out that the equation is actually G*sin(theta)=Ag.
     
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