Question 16 3 pts F Fig. 6-23 Part 4 of 4. The loaded penguin sled weighing 120 N rests on a plane inclined at angle 0 = 30 to the horizontal (Fig. 6-23). Between the sled and the plane, now consider that the sled is now in contact with the inclined plane with a coefficient of static friction Us = 0.65 and a coefficient of kinetic friction MK = 0.45, The force F is a tension force moving in the +x direction. IF the sled now is moving upward in +x, what is the minimum Force F to keep it moving at a constant velocity? Hint: Use the appropriate form of this equation, using the type of friction present: [ FX = F - mg sin 0 - MFN = max OF = 120 N OF = 67.5 N OF = 60 N OF = 47.5 N OF = 106.7 NQuestion 15 3 pts V Fig. 6-23 Part 3 of 4. The loaded penguin sled weighing 120 N rests on a plane inclined at angle 0 = 30 to the horizontal (Fig. 6-23). Between the sled and the plane, now consider that the sled is now in contact with the inclined plane with a coefficient of static friction Us = 0.65 and a coefficient of kinetic friction MK = 0.45, The force F is a tension force moving in the +x direction. If the sled is first in static equilibrium, calculate the normal force FN and find the minimum value of the tension force F to overcome the static friction, substituting the values in. Draw a FBD and use this as the equation [ FX = F - mg sin 0 - Us FN = 0 OF = 106.7 N OF = 60 N OF = 67.5 N OF = 47.5 N OF = 127.5 N