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Chapter 9: Problem 27

What is the torque of the force \(\mathbf{F}=(2 \mathbf{i}-3 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}) \mathrm{N}\) acting at the point \(\mathbf{r}=(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \mathrm{m}\) about the origin? (a) \(-17 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}\) (b) \(-6 \hat{i}+6 \hat{j}-12 \hat{k}\) (c) \(17 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-13 \hat{\mathbf{k}}\) (d) \(6 \hat{i}-6 \hat{j}+12 \hat{k}\)

Short Answer

Expert verified
Option (c): \( 17 \hat{\mathbf{i}} - 6 \hat{\mathbf{j}} - 13 \hat{\mathbf{k}} \).

Step by step solution

01

Understanding Torque

Torque, often represented as \( \mathbf{\tau} \), is the rotational analog of force. It is defined as the cross product of the position vector \( \mathbf{r} \) and the force vector \( \mathbf{F} \), given by \( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \). In this exercise, we are given \( \mathbf{r} = 3 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}} \text{ m} \) and \( \mathbf{F} = 2 \hat{\mathbf{i}} - 3 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}} \text{ N} \). We need to compute \( \mathbf{\tau} \).
02

Setting Up the Cross Product

The cross product \( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \) is computed using the determinant of a matrix: \[ \mathbf{\tau} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ 3 & 2 & 3 \ 2 & -3 & 4 \end{vmatrix} \]This matrix is constructed using the unit vectors in the first row, the components of \( \mathbf{r} \) in the second row, and the components of \( \mathbf{F} \) in the third row.
03

Calculating the Determinants

Expand the determinant to calculate \( \mathbf{\tau} \):\[ \begin{align*} \tau = & \hat{\mathbf{i}} \left((2)(4) - (3)(-3)\right) \ & - \hat{\mathbf{j}} \left((3)(4) - (3)(2)\right) \ & + \hat{\mathbf{k}} \left((3)(-3) - (2)(2)\right) \end{align*} \]Simplifying, we calculate the expressions inside the determinant.
04

Solving Each Component

Calculate each component of the torque:- \( \hat{\mathbf{i}} \) component: \( (2)(4) + (3)(3) = 8 + 9 = 17 \)- \( \hat{\mathbf{j}} \) component: \( (3)(4) - (3)(2) = 12 - 6 = 6 \)- \( \hat{\mathbf{k}} \) component: \( (3)(-3) - (2)(2) = -9 - 4 = -13 \)So, \( \mathbf{\tau} = 17 \hat{\mathbf{i}} - 6 \hat{\mathbf{j}} - 13 \hat{\mathbf{k}} \).
05

Checking Against the Options

By comparing our calculated torque \( \mathbf{\tau} = 17 \hat{\mathbf{i}} - 6 \hat{\mathbf{j}} - 13 \hat{\mathbf{k}} \) with the given options, we find that option (c) \( 17 \hat{\mathbf{i}} - 6 \hat{\mathbf{j}} - 13 \hat{\mathbf{k}} \) matches our result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross Product
The cross product is a fundamental operation in vector calculus, especially when dealing with three-dimensional vectors. It is used to determine a vector that is perpendicular to the plane formed by two input vectors. In the context of physics, particularly when solving problems related to torque, the cross product is essential.
  • The cross product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is a vector \( \mathbf{C} \) and can be denoted as \( \mathbf{A} \times \mathbf{B} = \mathbf{C} \).
  • Mathematically, it is computed using the determinant of a matrix that includes unit vectors and the components of the vectors involved.
  • The magnitude of the cross product can be determined by \( |\mathbf{A} \, \mathbf{B}| \sin(\theta) \), where \( \theta \) is the angle between \( \mathbf{A} \) and \( \mathbf{B} \).
  • The direction of the resultant vector follows the right-hand rule, which is vital in understanding rotational directions in physical systems.
In the given problem, the cross product helps find the torque generated by a force vector \( \mathbf{F} \) applied at a position vector \( \mathbf{r} \). The result is a vector perpendicular to both \( \mathbf{r} \) and \( \mathbf{F} \), reflecting the axis about which the torque acts.
Vector Calculus
Vector calculus deals with operations on vectors and is a powerful tool in understanding physical phenomena, including torque. It involves a variety of operations, such as the cross product, which is crucial in this exercise.
  • Vectors have both magnitude and direction, which makes them ideal for representing physical quantities like force and displacement.
  • Vector calculus allows the manipulation of these quantities to find results that are not immediately apparent from their scalar counterparts.
  • The operations of addition, subtraction, dot product, and cross product enable complex problem solving in physics and engineering.
  • For instance, the use of determinants in vector calculus helps compute the cross product, as seen in this torque problem."
Engaging with vector calculus equips students with the skills needed to tackle problems across various fields, ranging from physics to engineering disciplines.
Physics Problem Solving
Physics problem-solving involves a structured approach to understanding and tackling questions systematically, using mathematical and analytical methods.
  • Understanding the question is the first step, which involves recognizing the physical quantities involved and the relationships between them.
  • Applying the relevant mathematical operations, such as the cross product for torque, is crucial for finding the right solution.
  • Checking the solution is essential, ensuring that the results align with physical intuition and any given choices, like the options provided in the exercise.
    • When faced with physics problems like those involving torque, students must also familiarize themselves with the meaning of vector components and their significance in calculations. Over time, practicing these steps in problem-solving builds confidence and competence in handling more complex physics scenarios.

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