Dynamics: Incline Slope Derivation Simplification

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Asleky
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Homework Statement


No official problem, just a study guide fill-in-the-blanks with an extended simplification blank. Basically, no values were given, and it is a standard block on a standard slope with a north-east applied force pushing down on the block (not parallel to horizontal or vertical axis).

Homework Equations


Fnet = ma
Fnormal = Fay + Fgy
Fnormal = (Fapplied)(sinØ) + (Fgravity)(cosØ)
Fnet = Fax - Fgx - Ffk
Fnet = (Fapplied)(cosØ) - (Fgravity)(sinØ) - (Uk)(Fnormal)

The Attempt at a Solution


He asked us to substitute the second equation above into the third equation above. I am terrible at simplifying and so this is as far as I've gotten:
ma = (Fapplied)(cosØ) - (Fgravity)(sinØ) -Uk(Fapplied)(sinØ) + (Fgravity)(cosØ)
This is extremely messy and confusing, I am aware, so I am in deep gratitude to anyone taking time to help a scrub physics student like me. :D Thank you so much for any simplifying advice! I am solving for acceleration.
 
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Actually upon further inspection, could I somehow make it (Fapplied)(tanØ) and (Fgravity)(tanØ)? I am confident that the trig law tanØ = sinØ/cosØ will be used. At least I'm on to something :(
 
haruspex said:
The arrangement is not clear. Are you saying that the force is acting down the slope and parallel to it?
According to this equation
the applied force is horizontal.
Yeah that's why I multiplied by sin and cos to make all the components on one axis.
 
haruspex said:
According to this equation ... the applied force is horizontal.
Asleky said:
Yeah that's why I multiplied by sin and cos to make all the components on one axis.
But you wrote in the OP:
Asleky said:
with a north-east applied force pushing down on the block (not parallel to horizontal or vertical axis)
So which is it? Is the applied force horizontal or at an angle?
 
Could you draw a FBD?