Question

Answer the following questions about torque. Let \(\mathbf{r}=\overrightarrow{O P}=\mathbf{i}-\mathbf{j}+2 \mathbf{k} .\) A force \(\mathbf{F}=\langle 10,10,0\rangle\) is applied at \(P .\) Find the torque about \(O\) that is produced.

Short answer

Question: Find the torque about point O produced by the force F applied at point P given the position vector \(\mathbf{r} = \mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and the force vector \(\mathbf{F} = \langle 10, 10, 0 \rangle\). Answer: The torque about point O is given by the torque vector \(\boldsymbol{\tau} = \langle -20, 20, 20 \rangle\).

Step-by-step solution

Identify the given vectors

We are given the position vector \(\mathbf{r} = \mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and the force vector \(\mathbf{F} = \langle 10, 10, 0 \rangle\).

Compute the cross product

To find the torque, we need to calculate the cross product of the position vector and the force vector: \(\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}\) Before computing the cross product, we can write down the vectors in component form: - \(\mathbf{r} = \langle 1, -1, 2 \rangle\) - \(\mathbf{F} = \langle 10, 10, 0 \rangle\) Now, we can calculate the cross product: $\boldsymbol{\tau} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -1 & 2\\ 10 & 10 & 0 \end{vmatrix}$

Calculate the torque components

Expanding the determinant using the cofactor method, we get: $\boldsymbol{\tau} = \mathbf{i}\begin{vmatrix} -1 & 2\\ 10 & 0 \end{vmatrix} -\mathbf{j}\begin{vmatrix} 1 & 2\\ 10 & 0 \end{vmatrix} + \mathbf{k}\begin{vmatrix} 1 & -1\\ 10 & 10 \end{vmatrix}$ \(\boldsymbol{\tau} = \mathbf{i}((-1)\cdot0 - 2\cdot10) - \mathbf{j}((1\cdot 0 - 2\cdot 10)) + \mathbf{k}((1\cdot 10) - (-1\cdot 10))\) \(\boldsymbol{\tau} = \mathbf{i}(-20) - \mathbf{j}(-20) + \mathbf{k}(10 +10)\) \(\boldsymbol{\tau} = -20\mathbf{i} + 20\mathbf{j} + 20\mathbf{k}\) So, the torque about point O produced by the force F applied at point P is: \(\boldsymbol{\tau} = \langle -20, 20, 20 \rangle\)

Key concepts

Understanding the Cross Product

The cross product (denoted as \( \mathbf{a} \times \mathbf{b} \)) is a mathematical operation that helps us find a vector that is perpendicular to two given vectors \( \mathbf{a} \) and \( \mathbf{b} \). In the context of torque, this is critical because it tells us the direction in which the torque acts. The magnitude of the cross product gives us the amount of torque or rotational force. When you compute the cross product of two vectors:

• It results in a vector that is orthogonal (perpendicular) to both original vectors.
• The direction of this vector follows the right-hand rule.
• The magnitude is proportional to the sine of the angle between the two vectors and the product of their magnitudes.

In physical terms, if you apply a force at an angle to a lever, the cross product helps us calculate the rotational effect or torque, giving us both direction and magnitude.

Determinant Expansion Technique

The determinant expansion is a technique we use to compute the cross product using a symbolic matrix. It's a structured way of finding the magnitude and direction by breaking it down into smaller calculations.To expand the determinant, follow these steps:

• Set up a 3x3 matrix where the first row contains the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \)
• The second row contains the components of the first vector.
• The third row contains the components of the second vector.

For our exercise, the matrix was set up as follows: \[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & -1 & 2 \10 & 10 & 0\end{vmatrix}\]To expand this matrix determinant:

• Calculate the determinant of 2x2 matrices formed by excluding one row and one column (cofactor expansion).
• Apply alternating signs starting from positive with \( \mathbf{i} \).
• Combine the calculated values with the corresponding unit vectors.

This method gives us the resulting torque vector components clearly.

Breaking Down Vector Components

Vectors have components that represent their influence in different directions. Each vector in our problem can be broken down into its components along the standard coordinate axes \( x, y, \) and \( z \). Understanding each component helps in calculating precise vector operations.For the position vector \( \mathbf{r} = \mathbf{i} - \mathbf{j} + 2\mathbf{k} \):

• \( x \)-component: 1 (along \( \mathbf{i} \))
• \( y \)-component: -1 (along \( \mathbf{j} \))
• \( z \)-component: 2 (along \( \mathbf{k} \))

Similarly, the force vector \( \mathbf{F} = \langle 10, 10, 0 \rangle \) has:

• \( x \)-component: 10
• \( y \)-component: 10
• \( z \)-component: 0

Understanding these individual components:

• Helps in vector addition, subtraction, and operations such as cross product computation.
• Provides clarity on how each part of the vector contributes to the overall vector operation outcome.
• Ensures directional precision which is key when determining influences like torque in a rotational system.