Problem 83 - University Physics with Modern Physics

Question

The scalar product of vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is \(+\)48.0 m\(^2\). Vector \(\overrightarrow{A}\) has magnitude 9.00 m and direction 28.0\(^{\circ}\) west of south. If vector \(\overrightarrow{B}\) has direction 39.0\(^{\circ}\) south of east, what is the magnitude of \(\overrightarrow{B}\)?

Step-by-Step Solution

Answer

Vector \(\overrightarrow{B}\) has a magnitude of approximately 27.9 meters.

Step 1: Understanding the Dot Product

The dot product of two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \) is given by \( \overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos{\theta} \), where \(\theta\) is the angle between the two vectors. From the problem, we know this dot product equals 48 m\(^2\).

Step 2: Calculating the Angle Between the Vectors

The angle given in the problem are with respect to cardinal directions. Vector \( \overrightarrow{A} \) is 28.0\(^{\circ}\) west of south, which translates to 28.0\(^{\circ}\) + 90.0\(^{\circ}\) = 118.0\(^{\circ}\) from the east direction. Vector \( \overrightarrow{B} \) is 39.0\(^{\circ}\) south of east. The angle \(\theta\) between the vectors is \(118.0^{\circ} - 39.0^{\circ} = 79.0^{\circ}\).

Step 3: Apply Dot Product Formula

Substituting the known values into the dot product formula, we have:\[ 48.0 = 9.00 \times |\overrightarrow{B}| \times \cos(79.0^{\circ}) \]

Step 4: Solving for Magnitude of Vector B

First, calculate \(\cos(79.0^{\circ})\) using a calculator to find it approximately equal to 0.1908. Then solve the equation:\[ 48.0 = 9.00 \times |\overrightarrow{B}| \times 0.1908 \]\[ |\overrightarrow{B}| = \frac{48.0}{9.00 \times 0.1908} \]\[ |\overrightarrow{B}| \approx 27.9 \text{ meters} \]

Step 5: Verification

Verify your calculations to ensure accuracy. You can double-check the computations of \(\cos(79.0^{\circ})\) and the division to confirm that \(|\overrightarrow{B}| \approx 27.9 \text{ meters}\) is consistent throughout the steps.

Key Concepts

Magnitude of a Vector

The magnitude of a vector can be thought of as the length or size of the vector. It's like measuring a straight-line distance from one point to another. This is important because, in physics, vectors have both a magnitude and a direction.

When you need to find the magnitude of a vector, you often use the formula \(|\overrightarrow{V}| = \sqrt{V_x^2 + V_y^2}\). This applies to vectors in a two-dimensional space where \(V_x\) and \(V_y\) are the vector's components along the x and y axes. However, if you know the direction and other relationships between vectors (as in the dot product), you may not need this formula directly.

In our example, you calculated the magnitude of vector \(\overrightarrow{B}\) using the scalar (dot) product and angles involved. It shows how a vector's magnitude can be derived from its relationship with other vectors rather than its components.

Angle Between Vectors

The angle between vectors is key to understanding their relationship. It tells us how the vectors are oriented towards each other. In trigonometry, this angle is fundamental because it helps calculate many vector operations, like the dot product.

Here, you express the angle using the directions provided in the problem. Vectors can be described with angles relative to cardinal directions (north, south, east, west), and converting these can clarify the angle between the vectors. In this exercise, you found the relevant angle using simple subtraction of these converted angles to reach the value of \(79.0^{\circ}\).

Understanding these angles becomes crucial when applying trigonometric functions, as they determine functions like \(\cos(\theta)\), extensively used in vector mathematics to find products or projections.

Trigonometry in Physics

Trigonometry connects deeply with physics because it provides tools to manipulate and understand vectors, which represent forces, velocities, and displacements in the physical world.

• Through trigonometry, you can decompose vectors into components to better analyze them.
• Using functions like cosine and sine, you quantify relationships between vector magnitudes and angles.

In the given problem, trigonometry is pivotal for cracking the dot product equation. The cosine of the angle (\(\cos(79.0^{\circ})\)) is essential because it's the trigonometric function that relates the scalar product to the magnitudes of the vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\).

Utilizing trigonometric identities and functions allows us to handle more complex interactions between vectors, predicting outcomes in fields like engineering or physics efficiently. By understanding these concepts, physics problems become simpler and more intuitive to resolve.