4.7 Further applications of newton's laws of motion  (Page 2/7)

Since the barge is flat bottomed, the drag of the water ${\mathbf{\text{F}}}_{\text{D}}$ will be in the direction opposite to ${\mathbf{\text{F}}}_{\text{app}}$ , 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. Our strategy is to find the magnitude and direction of the net applied force ${\mathbf{\text{F}}}_{\text{app}}$ , and then apply Newton’s second law to solve for the drag force ${\mathbf{\text{F}}}_{\text{D}}$ .

Solution

Since ${\mathbf{\text{F}}}_x$ and ${\mathbf{\text{F}}}_y$ are perpendicular, the magnitude and direction of ${\mathbf{\text{F}}}_{\text{app}}$ are easily found. First, the resultant magnitude is given by the Pythagorean theorem:

The angle is given by

which we know, because of Newton’s first law, is the same direction as the acceleration. ${\mathbf{\text{F}}}_{\text{D}}$ is in the opposite direction of ${\mathbf{\text{F}}}_{\text{app}}$ , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as ${\mathbf{\text{F}}}_{\text{app}}$ , but its magnitude is slightly less than ${\mathbf{\text{F}}}_{\text{app}}$ . The problem is now one-dimensional. From [link] (b) , we can see that

But Newton’s second law states that

Thus,

This can be solved for the magnitude of the drag force of the water $F_{\text{D}}$ in terms of known quantities:

Substituting known values gives

The direction of ${\mathbf{\text{F}}}_{\text{D}}$ has already been determined to be in the direction opposite to ${\mathbf{\text{F}}}_{\text{app}}$ , or at an angle of $\text{53º}$ south of west.

Discussion

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 $F_{\text{D}}$ is less than 1/600th of the weight of the ship.

In the earlier example of a tightrope walker we noted that the tensions in wires supporting a mass were equal only because the angles on either side were equal. Consider the following example, where the angles are not equal; slightly more trigonometry is involved.

Different tensions at different angles

Consider the traffic light (mass 15.0 kg) suspended from two wires as shown in [link] . Find the tension in each wire, neglecting the masses of the wires.

Strategy

The system of interest is the traffic light, and its free-body diagram is shown in [link] (c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in part (d) of the figure. There are two unknowns in this problem ( $T_1$ and $T_2$ ), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.