m*g = 220kg * 9.8N/kg = 2156 n. = Wt. of the log.
Fp = 2156*sin26.6 = 965.4 N. = Force
parallel to the incline.
Fn = 2156*Cos26.6 = 1928 N. = Normal. =
Force perpendicular to the incline.
Fk = u*Fn = 0.863 * 1928 = 1664 N. =
Force of kinetic friction.
Fex-Fp-Fk = m*a
Fex-965.4-1664 = 220*0.856
Fex-2629 = 188.3
Fex = 188.3 + 2629 = 2817 N. = Force exerted. = Tension in the rope.
A 220-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 26.6° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.863, and the log has an acceleration of 0.856 m/s2. Find the tension in the rope
I kind of got stuck solving this. Please help if you can.
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