Why Does a 0.30 Coefficient Yield a 3.25 m/s² Acceleration Calculation Error?

[helen3743]

A cup of coffee is sitting on a table in an airplane that is flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.30. Suddenly, the plane accelerates, its altitude remaining constant. What is the maximum acceleration that the plane can have without the cup sliding backward on the table?

I got 2.94 m/s^2 as my answer, but the answer is supposed to be 3.25 m/s^2.

This is how I did it:
Fsmax = (funny looking symbol for coefficient of static friction)*FN
FN = mg
Fsmax = (symbol for coefficient of static friction)*mg = ma

amax = ((symbol for coefficient of static friction)*mg)/m
amax = (symbol for coefficient of static friction*g
amax = 0.30 * 9.8 = 2.94 m/s^2

What did I do wrong?
Thanks!

 

[Hootenanny]

Who says the answer is suppose to be 3.25 $m\cdot s^{-2}$? I can't see anything wrong with your working.

 

[Doc Al]

Your method and answer look OK to me.

[Hootenanny's getting faster!]

 

[Da-Force]

This might bring some light.

Force of friction is based on the normal force of the object.

The airplane accelerates, a force ===> in that direction.

The Force of Friction is <===== that direction.

The only way to get 3.25 is when the coefficient is .33 but I remember that static friction exceeds a little bit more then goes down if you look at a graph.