Question

The force \(\mathbf{F}=2 \mathbf{i}-5 \mathbf{k}\) acts at the point \((3,-1,0) .\) Find the torque of \(\mathbf{F}\) about each of the following lines. (a) \(\quad \mathbf{r}=(2 \mathbf{i}-\mathbf{k})+(3 \mathbf{j}-4 \mathbf{k}) t\). (b) \(\quad \mathbf{r}=\mathbf{i}+4 \mathbf{j}+2 \mathbf{k}+(2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}) t\).

Short answer

The Shortened torque.

Step-by-step solution

- Understanding the Torque

Torque \( \mathbf{\tau} \) is calculated using the cross product of the position vector \( \mathbf{r} \) and the force \( \mathbf{F} \): \[ \mathbf{\tau} = \mathbf{r} \times \mathbf{F}. \]

- Find the Position Vector

The position vector is defined from the reference point to the line along which we measure torque. For part (a), we have \( \mathbf{r} = (2 \mathbf{i} - \mathbf{k}) + t (3 \mathbf{j} - 4 \mathbf{k}) \).

- Substitute and Simplify

Typically, simplify the reference line's parametric equations, for example with parameter \( t = 0 \), our reference point becomes \( 2 \mathbf{i} - \mathbf{k} \). Use this for calculating torque.

- Calculation for Line (a)

Find the displacement vector \( \mathbf{r_{F}} \) from the point of force application (3, -1, 0) to reference point (2, 0, -1): \[ \mathbf{r_{F}} = \mathbf{r_{ref}} - \mathbf{r_{point}} = (2 \mathbf{i} - \mathbf{k}) - (3\mathbf{i} - \mathbf{j}) = - \mathbf{i} + \mathbf{j} - \mathbf{l}. \] Calculate torque: \[ \mathbf{\tau} = \mathbf{r_{F}} \times \mathbf{F} = ( - \mathbf{i} + \mathbf{j} - \mathbf{k}) \times ( 2 \mathbf{i} - 5 \mathbf{k}) = \mathbf{i}(\mathbf{j} - (-5\mathbf{k})) + \mathbf{j}( -2 \mathbf{i} - \mathbf{k}) + \mathbf{k}(4 \mathbf{i} - \mathbf{j}) \]

- Compute Torque for Line (a)

The resulting vector from above multiplication to summarise Adding from Step 4 to torque for line (a)

Key concepts

cross product

In physics and mathematics, the cross product, or vector product, is an essential operation between two vectors in 3D space. Unlike the dot product, which yields a scalar, the cross product results in another vector that is perpendicular to both of the original vectors.

Mathematically, the cross product of vectors \(\textbf{A}\) and \(\textbf{B}\) is denoted as \(\textbf{A} \times \textbf{B}\). Using components, for vectors \(\textbf{A} = [A_x, A_y, A_z]\) and \(\textbf{B} = [B_x, B_y, B_z],\) the cross product is:

\[\begin{equation} \textbf{A} \times \textbf{B} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \ \end{vmatrix} \end{equation}\]

The determinant can be expanded to give:
\(\textbf{A} \times \textbf{B} = \textbf{i}(A_yB_z - A_zB_y) - \textbf{j}(A_xB_z - A_zB_x) + \textbf{k}(A_xB_y - A_yB_x).\)

Notice that the resultant vector from the cross product lies in the direction orthogonal to both original vectors and is useful in torque calculations.

position vector

A position vector describes the location of a point in space relative to an origin. It is often denoted as \(\textbf{r}\) and in Cartesian coordinates it can be expressed as \(\textbf{r} = x\textbf{i} + y\textbf{j} + z\textbf{k}\). This representation allows us to pinpoint exactly where in the 3D space the point lies.

To find a position vector:

• Choose a reference point (often the origin).
• Measure or calculate the coordinates of your target point relative to this reference.

For example, if a point (3, -1, 0) exists in space, the position vector from the origin would be \(\textbf{r} = 3\textbf{i} - \textbf{j}\).

This concept is critical for calculating torque. You take the position vector of the point where force is applied and perform operations like cross products to determine rotational effects.

displacement vector

The displacement vector is a vector that shows the change in position of a point from one location to another.

Steps to calculate a displacement vector:

• Identify initial position: Record the starting point of your vector.
• Identify final position: Record the endpoint of your vector.
• Subtract: Compute the vector difference \((\textbf{final} - \textbf{initial})\).

For instance, given an initial position \(\textbf{r}_{\text{initial}} = (3, -1, 0)\) and a final position \(\textbf{r}_{\text{final}} = (2, 0, -1),\) the displacement vector would be:\[\begin{equation} \textbf{r}_{\text{displacement}} = \textbf{r}_{\text{final}} - \textbf{r}_{\text{initial}} = (2 - 3) \textbf{i} + (0 + 1) \textbf{j} + (-1 - 0) \textbf{k} = -\textbf{i} + \textbf{j} - \textbf{k}. \end{equation}\]
Displacement vectors are crucial in calculating torque because they tell you how the force is applied relative to the reference point in question.

parametric equations

Parametric equations describe a set of related quantities as explicit functions of an independent variable, known as a parameter. In 3D space, these equations describe lines or curves by specifying coordinates in terms of a parameter \(\textbf{t}\).

For example, consider the line:
\(\textbf{r} = \textbf{r}_0 + t\textbf{v},\)
where \(\textbf{r}_0\) is the position vector of a point on the line, \(\textbf{v}\) is a direction vector, and \(\textbf{t}\) is the parameter.

To visualize how parametric equations work, let's break down \(\textbf{r} = (2\textbf{i} - \textbf{k}) + t(3\textbf{j} - 4\textbf{k})\). Here:

• \textbf{r}_0 = (2\textbf{i} - \textbf{k}) : This is the initial point at \(\textbf{t} = 0\).
• \textbf{v} = (3\textbf{j} - 4\textbf{k}) : This vector shows how the line expands with each increment of \(\textbf{t}\).

To find a specific point along this line, substitute \(\textbf{t}\) with a particular value.Parametric equations are fundamental for tracking the path of objects in physics, particularly for torque computations, where you may need to determine points of force application along a line.