How Do You Calculate the Coefficient of Friction on a Sloped Surface?

[PhysicsIsRuff]

Homework Statement

A 75kg box slides down a 25.0 degree ramp with an acceleration of 3.60m/s^{2}
a.) Find \muk between the box and the ramp.
b.) What acceleration would a 175kg box have on this ramp?

Homework Equations

Fapplied,y=(?N)(sin[STRIKE]0[/STRIKE])
Fapplied,x=(?N)(cos[STRIKE]0[/STRIKE])
Fg=mg
\sumFy=Fn+Fapplied,y-Fg-0
Fk=\mukFn
\sumFx=Fapplied,x-Fk=ma_{2}

The Attempt at a Solution

Given: ax=3.60m/s
a.) Fg=(75kg)(9.81m/s^{2})=736N
b.) m=175kg

I do not understand how to get Fk or Fn without first finding \mu, so I am stuck for now...

 

[Delphi51]

You can't find the normal force or solve for the coefficient without separating the mg into components parallel and perpendicular to the ramp.

 

[rl.bhat]

The acceleration of the sliding box is given by
ma = mgsinθ - μmgcosθ
Solve for μ.

 

[PhysicsIsRuff]

So in that case, I should change it into
Fapplied,y=736cos25 down
Fapplied,x=736sin25 right
and continue from there? Or are those symbols for something else?

Is this how it should look in graph form?

@rl.bhat ah, so it is that simple... is the acceleration constant or 9.81m/s^2

 

[rl.bhat]

a is given in the problem and it is constant. g is 9.8 m/s^2

 

[PhysicsIsRuff]

So 0 = mgsinθ - μmgcosθ
Ok, thank you both for the help, now to lock this thread...

 

[rl.bhat]

So 0 = mgsinθ - μmgcosθ
Ok, thank you both for the help, now to lock this thread...

It is not correct.
3.6 m/s^2 = mgsinθ - μmgcosθ