Problem 6 - University Physics with Modern Physics

Question

A metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m} .\) (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\overrightarrow{\boldsymbol{F}} ?\)

Step-by-Step Solution

Answer

(a) \( \vec{r} = (3.00 \, \text{m}) \hat{\imath} + (4.00 \, \text{m}) \hat{\jmath} \). (b) Magnitude: 37.00 N⋅m, Direction: into the plane.

Step 1: Find the Position Vector

The position vector \( \vec{r} \) is given by the coordinates of the point where the force is applied, \( x = 3.00 \, \text{m} \) and \( y = 4.00 \, \text{m} \). Therefore, the position vector in terms of the unit vectors \( \hat{\imath} \) and \( \hat{\jmath} \) is \( \vec{r} = (3.00 \, \text{m}) \hat{\imath} + (4.00 \, \text{m}) \hat{\jmath} \).

Step 2: Calculate the Torque

Torque \( \vec{\tau} \) is given by the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \). The cross product in two dimensions is calculated as follows: \( \vec{\tau} = \vec{r} \times \vec{F} = |\begin{array}{cc} 3.00 \, \text{m} & 4.00 \, \text{m} \ 7.00 \, \text{N} & -3.00 \, \text{N} \end{array}| \). This results in \( \vec{\tau} = (3.00)(-3.00) - (4.00)(7.00) \).

Step 3: Evaluate the Determinant

Calculate the determinant to find the torque: \( \vec{\tau} = (3.00)(-3.00) - (4.00)(7.00) = -9.00 - 28.00 = -37.00 \, \text{N}\cdot\text{m} \).

Step 4: Determine Direction of Torque

In the 2-dimensional problem, the direction of torque can be inferred as perpendicular to the plane, along the \( \hat{k} \) direction. Since the torque \( \vec{\tau} \) is negative, this indicates the direction is into the plane (negative \( \hat{k} \)).

Key Concepts

Cross Product

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.

Position Vector

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.

Force Vector

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

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.