Physics Help! Tension in a rope on a ramp! Formulas needed!?

Question

A 195kg 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 29.1degrees with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.813, and the long has an acceleration of 0.694 m/s^2. Find the tension in the rope.

Answer 1

F=ma
since mass and acceleration of the 'long' is given then:
T'=F=(195kg )(0.694 m/s^2)
T'=135.3 Newtons

Lets forget that we know acceleration
Then
T=Fd + f
( where the force of friction f=uN =u mg cos(29.1) )
T=mg sin(29.1) + u mg cos(29.1)
T=mg(sin(29.1)+ u cos(29.1) )
T=195 x 9.81(sin(29.1)+ .813 cos(29.1) )
T=2289 N

We are ot out of the woods yet

Ttotal=T+T'= 2289 +135.3=
Ttotal=2424.3 N

Answer 2

Sum of the forces, F = mass m times acceleration a. F and a are vectors.

The formula is:

∑F = m a

Sum the forces in the direction of the plane.

T - mg(sin 29.1) {{mass parallel to plane}} - 0.813mg(cos29.1) {{friction}}= m a

T = m (a + 0.486g + 0.813(0.874)g)
T = 195 (0.694 + 4.77 + 6.97)

Answer-------> 2,425 N