# Angle of Child Sliding Down Hill

1. Sep 25, 2007

### shawshank

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-$$\theta$$)] = ma

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

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

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

5. $$\arcsin$$ (a / g) = $$\theta$$ (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. $$\arcsin$$ (1.5 / 9.8) = $$\theta$$

7. $$\theta$$ = 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. Sep 25, 2007

isnt this just a basic triangle? y is gravity, h is 1.5, and find angle with arcsin?

3. Sep 25, 2007

### shawshank

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

4. Sep 25, 2007