Exemplo 4.7 de College Physics, 4.7 Further Applications of Newton’s Laws of Motion

Drag Force on a Barge

Problema e solução em inglês, como no livro original.

Suppose two tugboats push on a barge at different angles, as shown in Figure 4.22. The first tugboat exerts a force of $2.7 \times 10^{5} N$ in the x-direction, and the second tugboat exerts a force of $3.6 \times 10^{5} N$ in the y-direction.

(a) A view from above two tugboats pushing on a barge. One tugboat is pushing with the force F sub x equal to two point seven multiplied by ten to the power five newtons, shown by a vector arrow acting toward the right in the x direction. Another tugboat is pushing with a force F sub y equal to three point six multiplied by ten to the power 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, F sub y is acting on a point upward, F sub x is acting toward the right, and F sub D is acting approximately southwest. (b) A right triangle is made by the vectors F sub x and F sub y. The base vector is shown by the force vector F sub x. and the perpendicular vector is shown by the force vector F sub y. The resultant is the hypotenuse of this triangle, making a fifty-three point one degree angle from the base, shown by the vector force F sub net pointing up the inclination. A vector F sub D points down the incline.

Figure 4.22 (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. Since the applied forces are perpendicular, the x- and y-axes are in the same direction as

F_{x}

and

F_{y}

. The problem quickly becomes a one-dimensional problem along the direction of

F_{app}

, since friction is in the direction opposite to

F_{app}

If the mass of the barge is $5.0 \times 10^{6} kg$ and its acceleration is observed to be $7 . 5 \times 10^{- 2} {m/s}^{2}$ 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.)

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Estratégia
The directions and magnitudes of acceleration and the applied forces are given in Figure 4.22(a). We will define the total force of the tugboats on the barge as $F_{app}$ so that:
$F_{app} = F_{x} + F_{y}$
4.59

Since the barge is flat bottomed, the drag of the water $F_{D}$ will be in the direction opposite to $F_{app}$ , as shown in the free-body diagram in Figure 4.22(b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. 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 $F_{D}$ .
Solução
Since $F_{x}$ and $F_{y}$ are perpendicular, the magnitude and direction of $F_{app}$ are easily found. First, the resultant magnitude is given by the Pythagorean theorem:
$\begin{array}{l} F_{app} & = & \sqrt{F_{x}^{2} + F_{y}^{2}} \\ F_{app} & = & \sqrt{(2.7 \times 10^{5} N)^{2} + (3.6 \times 10^{5} N)^{2}} & = & 4.5 \times 10^{5} N. \end{array}$
4.60

The angle is given by

$\begin{array}{l} θ & = & {tan}^{- 1} (\frac{F_{y}}{F_{x}}) \\ θ & = & {tan}^{- 1} (\frac{3.6 \times 10^{5} N}{2.7 \times 10^{5} N}) = 53º, \end{array}$
4.61

which we know, because of Newton’s first law, is the same direction as the acceleration. $F_{D}$ 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 Figure 4.22(b), we can see that
$F_{net} = F_{app} - F_{D} .$
4.62

But Newton’s second law states that

$F_{net} = ma .$
4.63

Thus,

$F_{app} - F_{D} = ma .$
4.64

This can be solved for the magnitude of the drag force of the water $F_{D}$ in terms of known quantities:
$F_{D} = F_{app} - ma .$
4.65

Substituting known values gives

$F_{D} = (4 . 5 \times 10^{5} N) - (5 . 0 \times 10^{6} kg) (7 . 5 \times 10^{–2} {m/s}^{2}) = 7 . 5 \times 10^{4} N .$
4.66

The direction of $F_{D}$ has already been determined to be in the direction opposite to $F_{app}$ , or at an angle of $53º$ south of west.
Discussão
The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where $F_{D}$ is less than 1/600th of the weight of the ship.

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