Box on an inclined surface with Force of Friction and angled applied Force.

[Wara]

Incline Surface force

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[Mathoholic!]

In that case:

$\vec{F}$_f=-μ.N$\hat{x'}$

Being μ the coefficient of kinetic friction and N the normal reaction of the ramp.
As the applied force makes 30º with the ramp, it's y component is going to contribute to a decrease in friction, and so being F_y=N=sin(30)F, you have:

$\vec{F}$_f=-μ.sin(30)F$\hat{x'}$

 

[Wara]

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[HallsofIvy]

You are given the acceleration up the incline so you can calculate $F_x$. Set up your force equations for the x and y components of all forces. Knowing the horizontal force will allow you to solve for the magnitude of $F_a$ and you can get $F_a$ from that.

 

[Wara]

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[Mathoholic!]

F_net=Fcos(30)-mgsen(20)-μN , with N=mgcos(20)-Fsen(30)
ma=Fcos(30)-mgsen(20)-μmgcos(20)+μFsen(30)
F=mg(sen(20)+μcos(20)+(a/g))/(cos(30)+μsen(30))