Question

A box (mass=34kg) is set on an incline (angle=36.5 degrees), attached to a string. The box is let go slowly with a tension of 205N on the string, and starts sliding down with an acceleration of 0.12m/s^2. What was the coefficient of kinetic friction between the box and the ramp?

*Fk is force of kinetic friction.
Work:
In my Free Body Diagram, I oriented +x down the ramp and +y orthagonal to the ramp. Fn was going up, Ft and Fk were going left, and Fg went down towards the floor, with x and y components Fgx and Fgy (m*g*sin(36.5deg) and m*g*cos(36.5deg)).
Fk is kinetic friction; Fk=Mu_k*Fn=Mu_k*m*g*cos(36.5deg)

Fnetx = ma_x = 34kg*0.12m/s^2 = Fgx-Fk-Ft = 34 *(9.8m/s^2) * sin(36.5 degrees) - Mu_k*34kg*(9.8m/s^2)*cos(36.5 degrees) - 205N

Fnety=ma_y= 0 (because it's only sliding along the x axis) = Fn - Fgy

I keep getting a negative answer. I wind up with 10.89N=-Mu_k*267.85N, which solves to a negative number. As far as I know, the coefficient of friction shouldn't be negative.
What's going on?

Answer 1

For reference, I get -0.0406 for Mu_k. Which, y'know, shouldn't be negative, probably. At least considering that acceleration is to the right and therefore friction should be to the left (on the FBD).