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Angle of Child Sliding Down Hill

  1. Sep 25, 2007 #1
    1. The problem statement, all variables and given/known data

    A child on a toboggan slides down a hill with an acceleration of magnitude 1.5 m/s^2. If friction is negligible, what is the angle between the hill and the horizontal?

    a = 1.5 m/s^2
    g = 9.8 m/s^2

    3. The attempt at a solution

    Ok, there is no mass given so it it's a little confusing but I tried it anyways.


    First of all here is my diagram of the situation. I flipped my free body diagram so that Fn is y+ (the dotted black line of x is perpendicular to Fn). The dotted red line represents what the X and Y line if Fg was perpendicular to the X. So I rotated my FBD and therefore had to rotate my angle.

    so now according to the FBD (used black dotted line as origins in the solution).

    1. mg[cos(90-[tex]\theta[/tex])] = ma

    2. g[cos(90-[tex]\theta[/tex])] = a (factored the mass out)

    3. cos(90-[tex]\theta[/tex]) = a/g (brought g to the other side)

    4. cos90 cos[tex]\theta[/tex] + sin 90 sin [tex]\theta[/tex] = a/g (Angle subtraction)

    5. [tex]\arcsin[/tex] (a / g) = [tex]\theta[/tex] (cos 90 is 0 so we can get rid of that first term, and sin 90 is simply 1 so it simplifies to this)

    6. [tex]\arcsin[/tex] (1.5 / 9.8) = [tex]\theta[/tex]

    7. [tex]\theta[/tex] = 8.8 degrees

    Now I can't really check my answer, and I may have used that whole angle subtraction out of place so can someone just check for me. Thanks
  2. jcsd
  3. Sep 25, 2007 #2
    isnt this just a basic triangle? y is gravity, h is 1.5, and find angle with arcsin?
  4. Sep 25, 2007 #3
    I guess I took the longer stop, I just looked at the diagram and realized if I took sin theta, it would've been same as taking cos 90 - theta. Stupid :D
  5. Sep 25, 2007 #4
    i dunno, im noob at this too. i did take trig so i guess thats the first step that popped in my mind
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