Holding onto a tow rope moving parallel to a frictionless ski slope, a 62.5 kg skier is pulled up the slope, which is at an angle of 8.2° with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 2.00 m/s and (b) v = 2.00 m/s as v increases at a rate of 0.105 m/s2? when trying to figure this out, i used the formula, T = mg sin (theta) = (62.5)(9.8)(sin 8.2) = 87.36N then Fn = mg cos (theta) =(62.5)(9.8)(cos 8.2) = 606.2N then the magnitude came to = 612.5N for part B.... I used Fn = m(g+a) = (62.5)(9.8 + .105) =619.1N But both of those answers are incorrect... am i going about this problem the wrong way?? any help would be appreciated.