Question

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

Short answer

Line (a): \(\mathbf{\tau}_a = -2 \mathbf{i} - 5 \mathbf{j} - 3 \mathbf{k} \)Line (b): \(\mathbf{\tau}_b = 25 \mathbf{i} - 6 \mathbf{j} + 10 \mathbf{k} \).

Step-by-step solution

- Define the torque formula

The torque of a force \(\mathbf{F}\) about a point \(\mathbf{r}_0\) is given by the vector cross product: \[\mathbf{\tau} = \mathbf{r}_0 \times \mathbf{F}\].

- Find a point on the lines

For line (a): Use \(t=0\) to get a point on the line: \mathbf{r} = 2 \mathbf{i} + 0 \mathbf{j} - 1 \mathbf{k}\. For line (b): Use \(t=0\) to get a point on the line: \mathbf{r} = \mathbf{i} + 4 \mathbf{j} + 2 \mathbf{k}\.

- Compute the displacement vector

For both points found: \[\mathbf{r}_0 = (3 \mathbf{i} - (-1) \mathbf{j} - 0 \mathbf{k}) - (2 \mathbf{i} + 0 \mathbf{j} - 1 \mathbf{k}) = \mathbf{i} + \mathbf{j} + \mathbf{k}\] for line (a). ewline For line (b): \[\mathbf{r}_0 = (3 \mathbf{i} - (-1) \mathbf{j} - 0 \mathbf{k}) - (\mathbf{i} + 4 \mathbf{j} + 2 \mathbf{k}) = 2 \mathbf{i} - 5 \mathbf{j} - 2 \mathbf{k}\].

- Calculate the cross product

Using \(\mathbf{r}_0\) and \(\mathbf{F}\): ewline Line (a): \[\mathbf{\tau}_a = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \mathbf{r}_{0x} & \mathbf{r}_{0y} & \mathbf{r}_{0z} \ \mathbf{F}_x & \mathbf{F}_y & \mathbf{F}_z \end{vmatrix} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 1 \ 2 & 0 & -5 \end{vmatrix} = -5\mathbf{j} - 2 \mathbf{k} - 2 \mathbf{i} \]. ewline Line (b): \[\mathbf{\tau}_b = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -5 & -2 \ 2 & 0 & -5 \end{vmatrix} = 25\mathbf{i} - 6\mathbf{j} + 10\mathbf{k} \].

Key concepts

Vector Cross Product

To understand torque calculation, you'll first need to be familiar with the vector cross product. A vector cross product, denoted as \(\textbf{A} \times \textbf{B}\) for vectors \(\textbf{A}\) and \(\textbf{B}\), results in a new vector that is perpendicular to both \(\textbf{A}\) and \(\textbf{B}\). This is computed using the determinant of a matrix formed by the unit vectors \(\textbf{i}\), \(\textbf{j}\), and \(\textbf{k}\), and the components of \(\textbf{A}\) and \(\textbf{B}\):

• Form a matrix with first row as \(\textbf{i}, \textbf{j}, \textbf{k}\)
• The second row contains the components of \(\textbf{A}\)
• The third row contains the components of \(\textbf{B}\)

For example, if \(\textbf{A} = \begin{pmatrix}A_x \ A_y \ A_z\text{\} \textbf{B} = \begin{pmatrix}B_x \ B_y \ B_z\text{\}\)\text), the cross product is obtained by:
\begin{pmatrix}\textbf{i} & \textbf{j} & \textbf{k} \ A_x & A_y & A_z \ B_x & B_y & B_z\text{\text)\text). Expanding this determinant gives you the resulting vector for torque. This concept is essential in mechanics for analyzing rotational forces and understanding torque.

Torque About a Point

Torque is a measure of the rotational force acting on a body about a specific point. It’s given by the cross product of the position vector \(\textbf{r_0}\) (from the point to where the force is applied) and the force vector \(\textbf{F}\): \(\textbf{τ} = \textbf{r_0} \times \textbf{F}\).

• Calculate the position vector \(\textbf{r_0}\) between two points.
• Use the cross product formula to find the torque.

For instance, if a force \(\textbf{F}= 2\textbf{i} - 5\textbf{k}\) acts at a point \((3, -1, 0)\), and you need to find its torque about a line passing through a point, first, find any point on the line by setting \(\text{t} = 0\).
Let's take this line: \(\textbf{r} = 2\textbf{i} - \textbf{k} + (3\textbf{j} - 4\textbf{k}) t_y\). At \(\text{t} = 0\), the point is \(\textbf{r} = 2\textbf{i} - \textbf{k}\). The position vector \(\textbf{r_0}\) from \((2, 0, -1)\) to \((3, -1, 0)\) is obtained by subtracting coordinates:
\(\textbf{r_0} = (3 - 2)\textbf{i} + (-1 - 0)\textbf{j} + (0 - (-1))\textbf{k} = \textbf{i} - \textbf{j} + \textbf{k}\).
Next, calculate the cross product \(\textbf{r_0} \times \textbf{F}\) to find the torque vector.

Line Equations in 3D

Understanding the line equations in 3D can simplify solving mechanics problems involving torque. A line in 3D space is represented as \(\textbf{r} = \textbf{r_0} + \textbf{d} t\), where \(\textbf{r_0}\) is a point on the line and \(\textbf{d}\) is the direction vector.

• \(\textbf{r}\) is the position vector of any point on the line
• \(\textbf{r_0}\) is the starting point vector
• \(\textbf{d}\) is the direction vector multiplied by parameter \(\text{t}\)

Consider two lines:
(a) \(\textbf{r} = (2\textbf{i} - \textbf{k}) + (3\textbf{j} - 4\textbf{k}) t_y\). At \(\text{t} = 0\), the point is \(\textbf{r} = 2\textbf{i} - \textbf{k}\).
(b) \(\textbf{r} = \textbf{i} + 4\textbf{j} + 2\textbf{k} + (2\textbf{i} + \textbf{j} - 2\textbf{k}) t\). At \(\text{t} = 0\), the point is \(\textbf{r} = \textbf{i} + 4\textbf{j} + 2\textbf{k}\).
In both cases, these line equations help you identify specific points and direction vectors crucial for computing vectors and forces in 3D. By substituting \(\text{t} = 0\), you simplify the calculations for torque.