A 265 kg log is pulled up a ramp by means of a rope that is parallel to the surf

Question

A 265 kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 30.0 degrees with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is .900, and the log has an acceleration of .800 m/s squared. Find the tension in the rope.

I know I have to use Newton's second Law; which would give me these equations:
N-mg-Fsin = 0
T - µfk - Fcos = ma

I'm not sure if I've written the right equations. Please check and explain how to solve the rest. Thanks!
I'm not sure if I've written the right equations. Please check and explain how to solve the rest. Thanks!

Explanation / Answer

From the free body diagram of the log Let us take T be the tension in the rope. Apply Newton's second law    T-f -mg sin? = ma    T - µk mg cos? -mg sin? =ma So the tension in the rope is             T = µk mg cos? +mg sin? +ma                = (0.9)(265 kg)(9.80 m/s2)cos30.0 + (265 kg)(9.80 m/s2)sin30.0 + (265 kg)(0.80 m/s2)                 = 2024.1611 N + 1298.5 N + 212 N                  = 3534.6611 N

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