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 \(\vec{O}\) that is produced.

Short answer

Answer: The torque about point O produced by the applied force is \(\vec{\tau} = \langle -20, 20, 20 \rangle\).

Step-by-step solution

Identify the given values

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

Calculate the cross product of the position vector and applied force

To find the torque, we need to calculate the cross product of the position vector \(\vec{r}\) and applied force \(\vec{F}\). \(\vec{\tau} = \vec{r} \times \vec{F}\)

Compute the cross product components

We can calculate the cross product by finding the determinant of the 3x3 matrix: $\vec{\tau} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -1 & 2\\ 10 & 10 & 0 \end{vmatrix}$

Expand the determinant

Now, let's expand the determinant: $\vec{\tau} = \mathbf{i} \begin{vmatrix} -1 & 2\\ 10 & 0 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 2\\ 10 & 0 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -1\\ 10 & 10\\ \end{vmatrix}$

Find the individual components of the torque vector

Compute the determinants for each part: \(\vec{\tau} = \mathbf{i}((-1)(0) - (2)(10)) - \mathbf{j}((1)(0) - (2)(10)) + \mathbf{k}((1)(10) - (-1)(10))\)

Simplify the torque vector

Simplify the expression to get the final torque vector: \(\vec{\tau} = -20\mathbf{i} + 20\mathbf{j} + 20\mathbf{k}\) The torque about point O produced by the applied force is \(\vec{\tau} = \langle -20, 20, 20 \rangle\).

Key concepts

Cross Product

The cross product is a fundamental operation in vector calculus used to find a vector that is perpendicular to two given vectors. In the context of physics, the cross product helps calculate quantities like torque. Torque measures the rotational effect of a force applied at a point. When you compute the torque by taking the cross product of a position vector \( \mathbf{r} \) with a force vector \( \mathbf{F} \), you get a new vector \( \vec{\tau} \) that describes the magnitude and direction of this rotation.
- To perform a cross product, the two vectors must be in three-dimensional space.- The formula for the cross product of vectors \( \mathbf{A} = \langle A_1, A_2, A_3 \rangle \) and \( \mathbf{B} = \langle B_1, B_2, B_3 \rangle \) is: \[ \mathbf{A} \times \mathbf{B} = (A_2B_3 - A_3B_2)\mathbf{i} - (A_1B_3 - A_3B_1)\mathbf{j} + (A_1B_2 - A_2B_1)\mathbf{k} \]
The resulting vector \( \vec{\tau} \) from the cross product \( \vec{r} \times \vec{F} \) signifies how the force influences the rotation about a specified axis.

Determinants

Determinants are mathematical tools used in various computations including calculating volumes, solving systems of equations, and, importantly for this context, finding the cross product. When it comes to the cross product, determinants help simplify the calculation process by breaking it down into manageable components.
- In the case of the torque problem, to find \( \vec{\tau} \), we use a 3x3 matrix composed of unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), the components of \( \mathbf{r} \), and the components of \( \mathbf{F} \).\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\1 & -1 & 2\10 & 10 & 0\end{vmatrix}\]
This determinant expands through the formula for a 3x3 matrix, helping us compute each component of the resulting torque vector. The scalar coefficients for \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are calculated separately by finding the determinants of smaller 2x2 matrices. This simplification elegantly organizes the calculation, ensuring the right combination of operations is performed.

Vector Calculus

Vector calculus extends traditional calculus to vector fields, making it an invaluable tool in physics and engineering. It allows us to work with quantities that have both magnitude and direction, such as force and velocity. Understanding vector calculus is crucial for analyzing systems where multiple forces interact in three-dimensional space.
In the given exercise:

• The position vector \( \mathbf{r} \) \(i - j + 2k\) shows the position in space.
• The force vector \( \mathbf{F} \) \langle 10, 10, 0 \rangle signifies a force applied at that position.
• The torque vector \( \vec{\tau} \) essentially results from the cross product operation, which gives us a new vector indicating the resultant rotational force about point \( O \).

Through vector calculus, you learn how these vectors interact, allowing you to predict physical dynamics. By understanding the roles different vectors play in a system, you can derive insightful conclusions about motion and forces in the physical world.