Problem 24 - Fundamentals of Physics

Question

In unit-vector notation, what is the torque about the origin on a jar of jalapeno peppers located at coordinates \((3.0 \mathrm{~m},-2.0 \mathrm{~m}\), \(4.0 \mathrm{~m})\) due to (a) force \(\vec{F}_{1}=(3.0 \mathrm{~N}) \mathrm{i}-(4.0 \mathrm{~N}) \mathrm{j}+(5.0 \mathrm{~N}) \mathrm{k}\) (b) force \(\vec{F}_{2}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \mathrm{j}-(5.0 \mathrm{~N}) \mathrm{k},\) and (c) the vector sum of \(\vec{F}_{1}\) and \(\vec{F}_{2} ?\) (d) Repeat part (c) for the torque about the point with coordinates \((3.0 \mathrm{~m}, 2.0 \mathrm{~m}, 4.0 \mathrm{~m})\).

Step-by-Step Solution

Answer

The torque for (a) is \(6\hat{i}-3\hat{j}-6\hat{k}\); for (b) is \(-26\hat{i}+27\hat{j}-18\hat{k}\); (c) and (d) both result in zero torque.

Step 1: Understand the Torque Formula

Torque (\( \vec{\tau} \)) is calculated using the formula \( \vec{\tau} = \vec{r} \times \vec{F} \), where \( \vec{r} \) is the position vector from the origin to the point of application, and \( \vec{F} \) is the force vector. It's important to remember that the cross product follows the right-hand rule.

Step 2: Compute Torque for Force \(\vec{F}_1\)

The position vector is \( \vec{r} = 3 \hat{i} - 2 \hat{j} + 4 \hat{k} \). To find the torque from \( \vec{F}_1 = 3 \hat{i} - 4 \hat{j} + 5 \hat{k} \), compute the cross product:\[\vec{\tau}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -2 & 4 \ 3 & -4 & 5 \end{vmatrix}\]Expanding the determinant:\[\vec{\tau}_1 = \hat{i}((-2)(5) - 4(-4)) - \hat{j}(3 \times 5 - 4 \times 3) + \hat{k}(3 \times (-4) - (-2) \times 3) = 6 \hat{i} - 3 \hat{j} - 6 \hat{k}\]

Step 3: Compute Torque for Force \(\vec{F}_2\)

For force \( \vec{F}_2 = -3 \hat{i} - 4 \hat{j} - 5 \hat{k} \), compute the cross product using the same position vector \( \vec{r} \):\[\vec{\tau}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -2 & 4 \ -3 & -4 & -5 \end{vmatrix}\]Expanding the determinant:\[\vec{\tau}_2 = \hat{i}((-2)(-5) - 4(-4)) - \hat{j}(3(-5) - 3 \times 4) + \hat{k}(3 \times (-4) - (-2)(-3)) = -26 \hat{i} + 27 \hat{j} - 18 \hat{k}\]

Step 4: Calculate Torque for Sum of Forces \(\vec{F}_1 + \vec{F}_2\)

Find the sum of the forces: \( \vec{F}_1 + \vec{F}_2 = (3 - 3) \hat{i} + (-4 - 4) \hat{j} + (5 - 5) \hat{k} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \). Since the sum of forces is zero, the torque about the origin is:\[\vec{\tau}_{sum} = \vec{0} \times \vec{r} = \vec{0}\]

Step 5: Recalculate Torque for New Origin and Sum of Forces

For a new origin at \( (3, 2, 4) \), the position vector becomes \( \vec{r}_{new} = 0 \hat{i} + (-4) \hat{j} + 0 \hat{k} \), since the coordinates of the application point are \((3, -2, 4)\). The sum of forces is still \( \vec{F}_{sum} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \), thus the torque remains:\[\vec{\tau}_{new} = \vec{0} \times \vec{r}_{new} = \vec{0}\]

Key Concepts

Unit Vector Notation

In physics and engineering, unit vector notation is a compact way to represent vectors in a Cartesian coordinate system using the unit vectors: \( \hat{i}\), \( \hat{j}\), and \( \hat{k}\). These unit vectors correspond to the \(x\)-axis, \(y\)-axis, and \(z\)-axis, respectively. This notation simplifies calculations and conveys direction as well as magnitude.

For instance, a force vector \( \vec{F} \) can be expressed as \( 3 \hat{i} - 4 \hat{j} + 5 \hat{k} \), where each component represents the quantity of force in that respective direction. Using unit vector notation allows each vector component to be clearly distinguished and manipulated in vector calculations, such as when computing torque, which is done using the cross product of a position vector and a force vector.

Cross Product

The cross product is a crucial vector operation used in physics to find a vector that is perpendicular to two given vectors. It's especially important in calculating torque, as torque involves the position vector and the force vector. The cross product is denoted by the symbol "\( \times \)". If we want to compute the torque (\( \vec{\tau} \)), represented as \( \vec{\tau} = \vec{r} \times \vec{F} \), we need to find the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \).

In three-dimensional space, if \( \vec{r} \) and \( \vec{F} \) are written using unit vector notation, the cross product is calculated by representing these vectors in a matrix-like format. This leads us to determinant expansion, which we'll explain further. The cross product results in a vector, which is always orthogonal to the initial two vectors, aligning with the directions outlined by the right-hand rule.

Determinant Expansion

Determinant expansion is a technique used to solve matrices, like when calculating the cross product of vectors. When given vectors \( \vec{a} \) and \( \vec{b} \) expressed in terms of \( \hat{i}\), \( \hat{j}\), \( \hat{k}\), the cross product \( \vec{a} \times \vec{b} \) is represented by a 3x3 matrix:

• The first row contains the unit vectors \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \).
• The second row holds the components of vector \( \vec{a} \).
• The third row includes the components of vector \( \vec{b} \).

To find the cross product, we perform the determinant expansion of this 3x3 matrix. It involves taking the determinant value of the components associated with each unit vector, following specific mathematical operations.

This method allows us to deduce a new vector properly aligned with the torque's dimensions. Practically, by expanding using \( \hat{i} \), \( \hat{j} \), \( \hat{k} \) each component is calculated by blocking out its row and column, determining the smaller 2x2 matrix's determinant and ensuring a precise output.

Right-Hand Rule

The right-hand rule is a handy method to determine the direction of the resulting vector from a cross product calculation. It's beneficial for visualizing the orientation of torque in physics, which is perpendicular to both the position and force vectors.

To use the right-hand rule, align your right hand as follows:

• Point your fingers in the direction of the first vector \( \vec{r} \).
• Rotate your palm to curl your fingers towards the second vector \( \vec{F} \).
• Your thumb points in the direction of the resulting vector \( \vec{\tau} \) from their cross product.

This method provides an intuitive understanding of vector directions. It ensures consistency in finding torque direction, which is essential in solving physics problems where direction and stability play significant roles.