Determine the tension in the rope that is holding the 3.8 kg sled in place on a 30 degree incline. The static coefficient of friction between the...

Please refer to the attached free-body diagram.

There are four main forces acting on the sledge: the tension in the rope (that is pulling it along the incline), the mass of the sledge (which can be resolved into two components: one perpendicular to the incline and the other along the incline and in opposite direction to the tension), the static friction and the reaction (normal to the surface of the incline).

Balancing the forces perpendicular...

Please refer to the attached free-body diagram.

There are four main forces acting on the sledge: the tension in the rope (that is pulling it along the incline), the mass of the sledge (which can be resolved into two components: one perpendicular to the incline and the other along the incline and in opposite direction to the tension), the static friction and the reaction (normal to the surface of the incline).

Balancing the forces perpendicular to the incline: N = mg cos30 (equation 1)   

Along the incline: T + `mu`N = mg sin30

substituting the value of N from equation 1, we get:

T = mg sin30 - `mu`mg cos30

here, m = 3.8 kg, g = 9.81 m/s2 and `mu` = 0.45

substituting the values, we get: T = 4.1 N. 

Thus, the tension in the rope holding the sledge is 4.1N.

Hope this helps.