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*A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 160 N at an angle of 35 degrees above the horizontal. The box has a mass of 25.8 kg and the coefficient of friction between box and floor is 0.254. Find the acceleration.*

Alright, so here is what I did: F(N)+160*sin(35)-mg=0, F(N)=161.06777 N

F(f)=uN=0.254-161.06777=40.9112

F=160*cos(35)-40.9112=90.15311=ma

a=91.15311/25.8=3.4943 m/(s^2)

And that's correct ... now the second part is just that the student starts moving the box up a 13.2 degree incline, keeping the same 160 N force directed at 35 degrees.

Here is what I tried first:

F=160*cos(35)-mgsin(13.2)-40.9112=32.416=ma

a=32.416/25.8=1.256 m/(s^2) ... WRONG

So then I thought that since I rotated the diagram by 13.2 degrees, it should have affected the 35 degrees:

F(N)+160*sin(35-13.2)-mg=0, F(N)=-193.42 ... this doesn't seem right ... try it anyway ...

F(f)=uN=0.254--193.42=-49.1289

F=160*cos(35-13.2)-mgsin(13.2)+49.1289=106.696=ma

a=106.696/25.8=4.1355 m/(s^2) ... WRONG AGAIN

I've drawn my free body diagrams over and over and I just don't see what to do in the second part where it's inclined 13.2 degrees. I know must be setting it up all incorrectly. Any hints, please? Thanks in advance!