1. The problem statement, all variables and given/known data A 322-kg boat is sailing 14.5° north of east at a speed of 1.84 m/s. Thirty seconds later, it is sailing 34.2° north of east at a speed of 4.40 m/s. During this time, three forces act on the boat: a 33.6-N force directed 14.5° north of east (due to an auxiliary engine), a 24.3-N force directed 14.5° south of west (resistance due to the water), and FW (due to the wind). Find the magnitude and direction of the force FW. Express the direction as an angle with respect to due east. 2. Relevant equations 3. The attempt at a solution Vo= 1.84cos(14.5) + 1.84sin(14.5) = 2.24 Vf= 4.4cos(34.2) + 4.4sin(34.2) = 6.11 (6.11-2.24)/30 = a = .13 m/s2 Ftotal = 322 x .13 = 41.86 N Fengine= 33.6cos(14.5) + 33.6sin(14.5)= 40.9 Fresistance= -24.3cos(14.5) - 24.3cos(14.5)= -29.6 Ftotal=Fengine+Fresistance+Fw 41.86=40.9+(-29.6)+FW FW= 30.6 But its wrong !! PLEASE help me by explaining with numbers. Ive rechecked my calculations so many times. and I cant get to the second part without the first, so please help. Thanks!