Acceleration of box on inclined plane at sliding angle

[alex_todd]

Homework Statement

he question is:
A box is sitting on a board. The coefficient of kinetic friction between the box and the board is 0.4. One side of the board is raised so that the board is 36.8699 degrees from horizontal. This is the angle that the box starts sliding.There are two parts to the question, I figured out part A (What is the coefficient of static friction between the box and the board?) by setting Mk equal to Tan and got .75

The part that throws me off about this equation is the last sentence " This is the angle that the box starts sliding"
I'm pretty sure that if it weren't for that, I would use the equation gsinθ - ugcosθ = a however I tried that and didn't get the right answer

Homework Equations

mgsinθ - umgcosθ = ma
gsinθ - ugcosθ = a
mk=tan theta

The Attempt at a Solution

I tried plugging the values into the gsinθ - ugcosθ = a equation, I assumed I could just get rid of the mass variables because they'll all cancel out anyways
gsinθ - ugcosθ = a = 9.8sin(36.8999 degrees) - .75(9.8)cos(36.8999 degrees)
the answer I got doesn't my teachers solution of 2.744
I'm think the part that's throwing me off is "this is the angle that the box starts sliding)

 

[gneill]

Hi alex_todd, Welcome to Physics Forums!

I tried plugging the values into the gsinθ - ugcosθ = a equation, I assumed I could just get rid of the mass variables because they'll all cancel out anyways
gsinθ - ugcosθ = a = 9.8sin(36.8999 degrees) - .75(9.8)cos(36.8999 degrees)
the answer I got doesn't my teachers solution of 2.744
I'm think the part that's throwing me off is "this is the angle that the box starts sliding)

Once the box is moving, what type of friction applies?