Problem 49 - Trigonometry

Question

Towing a Barge A barge is pulled by two tugboats. The first tugboat is traveling at a speed of 15 knots with heading \(130^{\circ}\), and the second tugboat is traveling at a speed of 16 knots with heading \(190^{\circ}\). Find the resulting speed and direction of the barge.

Step-by-Step Solution

Answer

The resulting speed of the barge is approximately 26.82 knots at a direction of 341.07°.

Step 1: Break down the problem

The problem involves two vectors representing the speeds and directions of the tugboats. We need to find the resultant vector in terms of speed and direction.

Step 2: Convert headings into vector components

For the first tugboat traveling at 15 knots with heading 130°, convert into components: \[ V_{1x} = 15 \cos(130^{\circ}), \quad V_{1y} = 15 \sin(130^{\circ}) \] For the second tugboat traveling at 16 knots with heading 190°, convert into components: \[ V_{2x} = 16 \cos(190^{\circ}), \quad V_{2y} = 16 \sin(190^{\circ}) \]

Step 3: Calculate vector components

Calculate the specific numerical values:\[ V_{1x} = 15 \times (-0.64278) \approx -9.6417, \quad V_{1y} = 15 \times 0.76604 \approx 11.4906 \] \[ V_{2x} = 16 \times (-0.98481) \approx -15.7569, \quad V_{2y} = 16 \times (-0.17365) \approx -2.7784 \]

Step 4: Sum the vectors

Add the components of the two tugboats together: \[ V_{total ext{x}} = V_{1x} + V_{2x} = -9.6417 + (-15.7569) = -25.3986 \] \[ V_{total ext{y}} = V_{1y} + V_{2y} = 11.4906 + (-2.7784) = 8.7122 \]

Step 5: Determine the resulting speed

Use the Pythagorean theorem to determine the resulting speed:\[ V_{resultant} = \sqrt{V_{total ext{x}}^2 + V_{total ext{y}}^2} = \sqrt{(-25.3986)^2 + (8.7122)^2} \approx 26.8184 \text{ knots} \]

Step 6: Determine the resulting direction

Calculate the angle of the resulting vector:\[ \theta = \tan^{-1}\left(\frac{V_{total ext{y}}}{V_{total ext{x}}}\right) = \tan^{-1}\left(\frac{8.7122}{-25.3986}\right) \approx -18.93^{\circ} \] Since the angle is negative, add 360 degrees to find the positive angle in standard position: \[ \theta = 360^{\circ} - 18.93^{\circ} = 341.07^{\circ} \] Adjust based on quadrant understanding if necessary.

Key Concepts

Resultant Vector

When two or more vectors act upon a single object, like in the case of the tugboats pulling a barge, they create what's called a resultant vector. This vector represents the cumulative effect of all acting forces.

Imagine each tugboat is a force, pulling the barge in a specific direction at a specific speed. Each tugboat’s pull, when broken down into its directional components, shows how much of its force acts in the horizontal and vertical directions. By calculating these components, we sum up the horizontal forces and the vertical forces separately to find the total force in each direction.

After finding these total forces or components, the resultant vector is calculated by using them to determine a single vector representing the combination of all forces. This resultant vector tells us both the speed and path the barge will ultimately follow. Essentially, it shows the overall effect of all actions in one concise direction and magnitude.

Pythagorean Theorem

The Pythagorean theorem is a mathematical staple in calculating the resultant vector's magnitude when dealing with vector components. It essentially provides a method to find the resultant speed when you have two component speeds working at right angles to each other.

In the context of vector addition, once you have determined the total horizontal and vertical components through summation, you can envision forming a right triangle. The two sides of this triangle are the summed horizontal and vertical vector components.

The beauty of the Pythagorean theorem is its simplicity:

• It states that for a right triangle, the square of the hypotenuse is the sum of the squares of the other two sides.
• Mathematically, it is expressed as: \[ c^2 = a^2 + b^2 \]
• where 'c' is the hypotenuse or, in our context, the resultant speed.
• The horizontal and vertical components are 'a' and 'b' respectively.
• By applying this theorem, we calculate the length of the resultant vector, offering us the combined speed or magnitude.

In the barge scenario, plugging in the values for the total x and y components gives us the barge's resultant speed, ensuring we clearly understand how fast it moves.

Trigonometric Functions

Trigonometric functions are core to understanding vector problems as they allow us to break down vectors into manageable parts: horizontal and vertical components. These functions, primarily sine and cosine, help translate a vector's angle and magnitude into useful distances that can be added vectorially.

Each of these functions addresses a specific side of a right-angled triangle:

• Sine (\( \sin \)) relates an angle to the ratio of the opposite side to the hypotenuse.
• Cosine (\( \cos \)) deals with the ratio of the adjacent side to the hypotenuse.
• The tangent function helps find the angle of the resultant using: \[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \]

When applied to vectors, cosine will provide the x-component (horizontal), and sine gives the y-component (vertical):

• For any given vector \( V \), with an angle \( \theta \):
• \[ V_x = V \cos(\theta) \]
• \[ V_y = V \sin(\theta) \]

Knowing the angle and magnitude of a vector, these trigonometric functions allow us to calculate exact component values. This step is crucial because these components are eventually added to determine the resultant vector’s direction and speed.