Particle Acceleration We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples. Drag Force on a Barge Two tugboats push on a barge at different angles as shown below. The first tugboat exerts a force of 2.7 x 105 N in the x-direction, and the second tugboat exerts a force of 3.6 x 105 N in the y-direction. The mass of the barge is 5.0 x 106 kg and its acceleration is observed to be 7.5 x 10-2 m/s2 in the direction shown. What is the drag force of the water on the barge resisting the motion? (Note: Drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object. Since the barge is flat bottomed, we can assume that the drag force is in the direction opposite of motion of the barge.) a 53.1 Fo F1 - 2.7 x 105 N F2 = 3.6 × 105 N Fo Fret F app 53.1º F. y AX F1 (a) (b) (a) A view from above of two tugboats pushing on a barge. (b) The free-body diagram for the ship contains only forces acting in the plane of the water. It omits the two vertical forces-the weight of the barge and the buoyant force of the water supporting it cancel and are not shown. Note that Fapp is the total applied force of the tugboats. Strategy The directions and magnitudes of acceleration and the applied forces are given in part (a) of the figure. Wo define the total force of the tugboats on the barge as Fapp so that Fapp = F1 + Ë2. The drag of the water Fp is in the direction opposite to the direction of motion of the boat; this force thus works against Fapp, as shown in the free-body diagram in part (b) of the figure above. The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x- and y-axes are in the same direction as F1 and F2. The problem quickly becomes a one-dimensional problem along the direction of Fapp, since friction is in the direction opposite to Fapp. Our strategy is to find the magnitude and direction of the net applied force Fapp and then apply Newton's second law to solve for the drag force Fp -