Finding Friction force impeding boxs motion

[cleo0724]

A 21.0 kg box is released on a 36.0° incline and accelerates down the incline at 0.271 m/s2. Find the friction force impeding its motion.

F=mg*cos(36)

F=21(9.8)*cos(36)
F=205.8*.809
F=166.495

I'm not sure what I'm doing wrong. Please help

 

[rock.freak667]

the component of the weight acts downwards and the friction acts upwards (directions are parallel to the plane)


so if the box moves down then ma=forces down-forces up.


a bit easier now?