Solution Since F2 and Fy are perpendicular, we can find the magnitude and direction of Fapp directly. First, the resultant magnitude is given by the Pythagorean theorem: Papp = F2 + F2 = V(2.7 x 105N)2 + (3.6 x 105 N)2 = 4.5 x 105 N. The angle is given by @ = tan-1 F2 F = tan-1 2.7 × 105 N =53.1°. 3.6 × 105 NY From Newton's first law, we know this is the same direction as the acceleration. We also know that Ip is in the opposite direction of Fapp, since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as Fapp, but its magnitude is slightly less than Fapp. The problem is now one-dimensional. From the free-body diagram, we can see that Fnet = Fopp - FD. However, Newton's second law states that Fnet = ma. Thus, Fapp - FD = ma. This can be solved for the magnitude of the drag force of the water Fp in terms of known quantities: FD = Fapp - ma. Substituting known values gives FD= (4.5×105 N)-(5.0×100kg) (7.5×10-2m/s2)=7.5×104 N. The direction of FD has already been determined to be in the direction opposite to Fapp, or at an angle of 53" south of west.