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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 ( [link] ). The first tugboat exerts a force of 2.7 × 10 5 N in the x -direction, and the second tugboat exerts a force of 3.6 × 10 5 N in the y -direction. The mass of the barge is 5.0 × 10 6 kg and its acceleration is observed to be 7.5 × 10 −2 m/s 2 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) A view from above of two tugboats pushing on a barge. One tugboat is pushing with the force F sub 1 equal to two point seven times by ten to the five newtons, shown by a vector arrow acting toward the right in the x direction. Another tugboat is pushing with a force F sub 2 equal to three point six times by ten to the five newtons acting upward in the positive y direction. Acceleration of the barge, a, is shown by a vector arrow directed fifty-three point one degree angle above the x axis. In the free-body diagram,  the mass is represented by a point, F sub 2 is acting upward on the point, F sub 1 is acting toward the right, and F sub D is acting approximately southwest. (b) The vectors F sub 1 and F sub 2 are the sides of a right triangle. The resultant is the hypotenuse of this triangle, vector F sub app, making a fifty-three point one degree angle from the base vector F sub 1. The vector F sub app plus the vector force F sub D, pointing down the incline, is equal to the force vector F sub net, which points up the incline.
(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 F app is the total applied force of the tugboats.

Strategy

The directions and magnitudes of acceleration and the applied forces are given in [link] (a). We define the total force of the tugboats on the barge as F app so that

F app = F 1 + F 2 .

The drag of the water F D is in the direction opposite to the direction of motion of the boat; this force thus works against F app , as shown in the free-body diagram in [link] (b). 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 F 1 and F 2 . The problem quickly becomes a one-dimensional problem along the direction of F app , since friction is in the direction opposite to F app . Our strategy is to find the magnitude and direction of the net applied force F app and then apply Newton’s second law to solve for the drag force F D .

Solution

Since F x and F y are perpendicular, we can find the magnitude and direction of F app directly. First, the resultant magnitude is given by the Pythagorean theorem:

F app = F 1 2 + F 2 2 = ( 2.7 × 10 5 N ) 2 + ( 3.6 × 10 5 N ) 2 = 4.5 × 10 5 N .

The angle is given by

θ = tan −1 ( F 2 F 1 ) = tan −1 ( 3.6 × 10 5 N 2.7 × 10 5 N ) = 53.1 ° .

From Newton’s first law, we know this is the same direction as the acceleration. We also know that F D is in the opposite direction of F app , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F app , but its magnitude is slightly less than F app . The problem is now one-dimensional. From the free-body diagram, we can see that

F net = F app F D .

However, Newton’s second law states that

F net = m a .

Thus,

F app F D = m a .

This can be solved for the magnitude of the drag force of the water F D in terms of known quantities:

F D = F app m a .

Substituting known values gives

F D = ( 4.5 × 10 5 N ) ( 5.0 × 10 6 kg ) ( 7.5 × 10 2 m/s 2 ) = 7.5 × 10 4 N .

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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