91Ó°ÊÓ

The cross product is an essential tool in physics and engineering, especially for calculating torque. It involves determining the product of two vectors, resulting in a third vector that is perpendicular to the plane containing the initial vectors.
In this exercise, we calculate the cross product of a position vector \( \vec{r} \) and a force vector \( \vec{F} \) to find the torque \( \vec{\tau} \). The cross product is typically defined in a three-dimensional space, but it simplifies in two dimensions where results are along the \( \hat{k} \) direction, indicating the torque's perpendicular nature to the xy-plane.
The cross product in two dimensions is represented by the determinant \( | \begin{array}{cc} x & y \ f_x & f_y \end{array} | \), which simplifies to express a scalar value indicating the magnitude of the torque.

The position vector is crucial when calculating torque, as it defines the point of application of force relative to the origin or another reference point.
In our problem, the position vector \( \vec{r} \) represents the coordinates of the force's point of application in the xy-plane, given by \( \vec{r} = (3.00 \, \text{m}) \hat{\imath} + (4.00 \, \text{m}) \hat{\jmath} \). This vector quantifies the position of the force relative to the origin (0,0), providing a way to determine how effectively a force can induce rotational motion.
The position vector plays a critical role when we calculate torque, as it determines the lever arm which directly influences the torque's magnitude.

The force vector shows the direction and magnitude of the given force applied to an object. In this exercise, the force vector \( \vec{F} \) is \( (7.00 \, \text{N}) \hat{\imath} + (-3.00 \, \text{N}) \hat{\jmath} \).
This vector indicates that:

• \(7.00 \, \text{N}\) of force is acting in the positive x-direction.
• \(-3.00 \, \text{N}\) points in the negative y-direction.

Understanding the components of a force vector is imperative for calculating resulting effects like torque, where different vector components interact to produce rotation.
By manipulating force vectors, we can predict and analyze the impact they will have on an object or a system.

Determinants provide us with a mathematical function to extract useful scalar information from a matrix, often used to solve systems of linear equations and calculate vector products like the cross product.
In torque calculations, the determinant simplifies finding the cross product in two dimensions. By organizing the components of our position and force vectors into a 2x2 matrix, we can quickly evaluate the determinant which corresponds to the magnitude of the torque.
In our exercise, the cross product and hence the determinant \( | \begin{array}{cc} 3.00 \, \text{m} & 4.00 \, \text{m} \ 7.00 \, \text{N} & -3.00 \, \text{N} \end{array} | \) is calculated as \( -37.00 \, \text{N}\cdot\text{m} \), indicating both the torque's magnitude and its sense of rotation. Understanding how to evaluate determinants is key in applications like these, ensuring accurate calculations and analyses.