# Finding acceleration given

• alex_todd
In summary, the question is about a box on a board with a coefficient of kinetic friction of 0.4. The board is angled at 36.8699 degrees and the question asks for the coefficient of static friction. The equation gsinθ - ugcosθ = a can be used, but the last sentence mentioning the angle at which the box starts sliding may be causing confusion. The attempt at a solution involved plugging values into the equation, but the resulting answer did not match the teacher's solution. The question also raises the question of what type of friction applies once the box is moving.

## 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)

Hi alex_todd, Welcome to Physics Forums!

alex_todd said:
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?