# Finding the acceleration at an angle

"What's the maximum acceleration you can achieve in a a water-slide at a 34 degree angle (If you can't use your arms and legs)"?

This is the free-body-diagram that I drew, assuming $g = 10m/s^2$:

The maximum acceleration is when there is no friction. So when I set friction to zero, the maximum acceleration is just the value of a:

$sin(34^o) = 10/a$

$a = 10 / sin(34^o) = 17.9m/s^2$

However, the book says it's $5.5m/s^2$. What am i doing wrong?

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You should start with a = g cos(90-34), the (90-34) is the angle between g and a.

Then a = 10 cos (56) = 5.5 m/s^2

so you could also go with, a = g sin (34).

I would like to point out that if you used the a as the hypotenuse this would mean that a is the resultant, which is always greater in value than its horizontal and vertical components, which is not the case here. Use g as a hypotenuse (g is the resultant) and you will get correct results.

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That doesn't seem correct. cos(56) = g / a, so a = g / cos 56. –  Peatherfed Apr 7 '13 at 21:06
Draw a line starting from the tip of arrow g on your figure perpendicular on the line a, then you will have your correct triangle. I edited my answer to give you the cause of choosing this triangle over the other. Always check which is the resultant and which is the component. Good Luck –  Force Apr 7 '13 at 21:21
Always use the resultant as the hypotenuse. An easier way to calculate this is, when there's a greater acceleration (g), reduced by being inclined (a), then (a) will always equal to g cos (x) where x is the angle between them. So all you have to do is figure out the angle between the two forces, which is causing which, and go with it. –  Force Apr 7 '13 at 21:35