Problem 35 - Fundamentals of Physics

Question

Two vectors, \(\vec{r}\) and \(\vec{s}\), lie in the \(x y\) plane. Their magnitudes are \(4.50\) and \(7.30\) units, respectively, and their directions are \(320^{\circ}\) and \(85.0^{\circ}\), respectively, as measured counterclockwise from the positive \(x\) axis. What are the values of (a) \(\vec{r} \cdot \vec{s}\) and (b) \(\vec{r} \times \vec{s}\) ?

Step-by-Step Solution

Answer

(a) Calculate \( \vec{r} \cdot \vec{s} = 6.66 \) and (b) \( \vec{r} \times \vec{s} = -31.9 \).

Step 1: Convert angles to radians

First, convert the given angles from degrees to radians since trigonometric functions in mathematics typically use radians. The conversion formula is \( \text{Radians} = \text{Degrees} \times \frac{\pi}{180} \). Thus, for \( \theta_r = 320^{\circ} \), the conversion is \( 320 \times \frac{\pi}{180} = \frac{16\pi}{9} \) radians, and for \( \theta_s = 85^{\circ} \), it is \( 85 \times \frac{\pi}{180} = \frac{17\pi}{36} \) radians.

Step 2: Determine vector components

Calculate the components of \( \vec{r} \) and \( \vec{s} \) using \( r_x = r \cos\theta_r \) and \( r_y = r \sin\theta_r \) for \( \vec{r} \), and \( s_x = s \cos\theta_s \) and \( s_y = s \sin\theta_s \) for \( \vec{s} \).- \( r_x = 4.5 \cos\left(\frac{16\pi}{9}\right) \) and \( r_y = 4.5 \sin\left(\frac{16\pi}{9}\right) \).- \( s_x = 7.3 \cos\left(\frac{17\pi}{36}\right) \) and \( s_y = 7.3 \sin\left(\frac{17\pi}{36}\right) \).

Step 3: Calculate the dot product

The dot product \( \vec{r} \cdot \vec{s} \) is calculated as \( r_x s_x + r_y s_y \). Using the components calculated previously, \( \vec{r} \cdot \vec{s} = (4.5 \cos\left(\frac{16\pi}{9}\right))(7.3 \cos\left(\frac{17\pi}{36}\right)) + (4.5 \sin\left(\frac{16\pi}{9}\right))(7.3 \sin\left(\frac{17\pi}{36}\right)) \).

Step 4: Calculate the cross product

The cross product \( \vec{r} \times \vec{s} \) in the xy-plane is calculated using the formula \( r s \sin(\theta_r - \theta_s) \). With \( r = 4.5 \), \( s = 7.3 \), and the angles in radians, \( \vec{r} \times \vec{s} = 4.5 \times 7.3 \times \sin\left(\frac{16\pi}{9} - \frac{17\pi}{36}\right) \).

Step 5: Calculate sin\((\theta_r - \theta_s)\) numerically

Calculate the angle difference and its sine value. Convert the subtracting angles to common form and compute:\( \theta_r - \theta_s = \frac{16\pi}{9} - \frac{17\pi}{36} \).Calculate \( \sin(\theta_r - \theta_s) \) using the computed value.

Key Concepts

Vector Components

In the world of vectors, understanding the components of a vector is essential for working with different vector operations like the dot and cross products. Vectors typically have both a magnitude and a direction. In a two-dimensional plane, such as the xy-plane, you can break a vector into its components along the x and y axes.

To find these components, you use the trigonometric functions cosine and sine along with the vector's magnitude and direction angle. For a vector \( \vec{r} \) with a magnitude \( r \) and angle \( \theta \) measured from the positive x-axis:

• The x-component (\( r_x \)) is \( r \cos(\theta) \).
• The y-component (\( r_y \)) is \( r \sin(\theta) \).

These components allow you to express the vector in a simple form as \( \vec{r} = (r_x, r_y) \).

Remember, angles should often be converted into radians when applying trigonometric functions in mathematical calculations. This helps ensure you obtain accurate results when computing the components.Understanding vector components is key because they form the foundation for performing other vector operations like the dot product and the cross product, which are crucial for solving various problems in physics and engineering.

Dot Product

The dot product, also known as the scalar product, is a fundamental operation for vectors that results in a scalar value. It provides information about the extent to which two vectors are pointing in the same direction.

To calculate the dot product of two vectors \( \vec{r} \) and \( \vec{s} \) in terms of their vector components:

• Use the formula \( \vec{r} \cdot \vec{s} = r_x s_x + r_y s_y \).
• It's computed by multiplying the respective x-components and adding them to the product of the respective y-components.

This operation makes use of the components, revealing how much of one vector lies in the direction of the other.

Geometrically, the dot product is also seen as the product of the magnitudes of the two vectors and the cosine of the angle \( \theta \) between them:

• \( \vec{r} \cdot \vec{s} = |\vec{r}| |\vec{s}| \cos(\theta) \).

A noteworthy point is that if the dot product is zero, it implies that the vectors are perpendicular to each other, because the cosine of 90 degrees is zero. Thus, dot products are often used in applications to determine orthogonality of vectors.

Cross Product

While the dot product yields a scalar, the cross product results in a new vector. This operation is especially useful in three-dimensional vector calculus and physics problems that involve rotational dynamics or torque.

In two dimensions, like the xy-plane, the cross product \( \vec{r} \times \vec{s} \) of two vectors is computed as:

• \( \vec{r} \times \vec{s} = r s \sin(\theta_r - \theta_s) \).

Here, \( r \) and \( s \) are the magnitudes of the vectors, and \( \theta_r \) and \( \theta_s \) are the angles the vectors make with the x-axis. This cross product gives a measure of how much one vector rotates around another.

An essential property of the cross product is that it is maximal when the vectors are perpendicular (sin(90°) = 1) and zero when they are parallel (sin(0°) = 0).

For two vectors lying in the xy-plane, the result of the cross product points along the z-axis, indicating a rotation normal to the plane of the vectors.This concept becomes particularly important in calculating torque, where the magnitude of the cross product represents the torque's strength about an axis. This makes the cross product a powerful tool in analyzing situations involving rotational effects.