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Magazine Chapter 6: Problem 7
Let \(\mathbf{F}=2 \mathbf{i}-3 \mathbf{j}+\mathbf{k}\) act at the point (5,1,3) (a) Find the torque of \(\mathbf{F}\) about the point (4,1,0) (b) Find the torque of \(\mathbf{F}\) about the line \(\mathbf{r}=4 \mathbf{i}+\mathbf{j}+(2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}) t\).
Short Answer
Expert verified
Torque around point is \(9 \mathbf{i} + 5 \mathbf{j} - 3 \mathbf{k} \). Torque around the line involves further reduction.
Step by step solution
01
Understanding the Problem
We need to find the torque of the force \(\mathbf{F} = 2 \mathbf{i} - 3 \mathbf{j} + \mathbf{k} \) around two different coordinates. First, we compute the torque around a point and then around a line.
02
- Find Position Vector
To find the torque about point (4,1,0), calculate the position vector \(\mathbf{r} \) from (4,1,0) to (5,1,3). The position vector \(\mathbf{r} \) is: \(\mathbf{r} = (5 - 4)\mathbf{i} + (1 - 1)\mathbf{j} + (3 - 0)\mathbf{k} = \mathbf{i} + 0 \mathbf{j} + 3 \mathbf{k}\)
03
- Calculate Torque about a Point
The torque \(\mathbf{ \tau} \) about a point is given by the cross product \( \mathbf{ \tau } = \mathbf{ r} \times \mathbf{ F} \). Thus,\( \mathbf{ \tau} = (\mathbf{ i} + 0 \mathbf{j} + 3 \mathbf{k}) \times (2 \mathbf{i} - 3 \mathbf{j} + \mathbf{k}) \):
04
- Compute the Cross Product
Compute the cross product:\[\mathbf{ \tau } = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 0 & 3 \2 & -3 & 1 \end{vmatrix} = \mathbf{i}(0 \cdot 1 - (-3) \cdot 3) - \mathbf{j}(1 \cdot 1 - 3 \cdot 2) + \mathbf{k}(1 \cdot (-3) - (0) \cdot 2) = 9 \mathbf{i} + 5 \mathbf{j} - 3 \mathbf{k}\]
05
- Define the Components for Torque about a Line
To find the torque of \(\mathbf{ F} \) about the line \( \mathbf{ r} = 4 \mathbf{ i} + \mathbf{ j} + (2 \mathbf{ i} + \mathbf{ j} - 2 \mathbf{ k}) t \), we first identify the components:- Point on the line: \( \mathbf{r_0} = 4 \mathbf{i} + \mathbf{j} \)- Direction vector: \( \mathbf{d} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \)
06
- Compute Position Vector Relative to the Line
Now find the position vector from point (4,1,0) to point (5,1,3):\( \mathbf{ r } = (5-4) \mathbf{i} + (1-1)\mathbf{j} + (3-0)\mathbf{k} = \mathbf{i}+0\mathbf{ j} + 3 \mathbf{k} \)
07
- Project Position Vector onto Line
Project position vector \(\mathbf{ r} \) onto the direction vector \(\mathbf{d} \) to find the perpendicular component \( \mathbf{ r_\perp} =\mathbf{ r} - \left ( \frac{\mathbf{ r} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \right ) \mathbf{ d}\)
08
- Calculate Torque about the Line
Finally, use the perpendicular component to calculate torque \( \mathbf{ \tau} = \mathbf{ r_\perp} \times \mathbf{ F} \). Thus,\( \mathbf { \tau} = \left ( \mathbf{i} - \frac {1}{3} \mathbf{d} \right) \times (2 \mathbf{i} - 3 \mathbf{j} + \mathbf{k})\) to find the required torque.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Cross Product
The vector cross product is a mathematical operation that takes two vectors and returns a third vector that is perpendicular to the plane of the first two. This operation is essential in physics for calculating quantities like torque. For vectors \(\textbf{A}\) and \(\textbf{B}\), the cross product \(\textbf{A} \times \textbf{B}\) is computed using the determinant of a matrix. The formula looks like this: \[\(\textbf{A} \times \textbf{B}\) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix}\].
Here, \(\textbf{i}, \textbf{j},\) and \(\textbf{k}\) are unit vectors in the x, y, and z directions, respectively. The result is another vector whose x, y, and z components can be determined from the determinant.
Here, \(\textbf{i}, \textbf{j},\) and \(\textbf{k}\) are unit vectors in the x, y, and z directions, respectively. The result is another vector whose x, y, and z components can be determined from the determinant.
Position Vector
A position vector represents the position of a point in space relative to an origin. In the exercise, you need to find the position vector \(\textbf{r}\) from one point (4,1,0) to another point (5,1,3). This is done by subtracting the coordinates component-wise: \[ \textbf{r} = (5 - 4) \textbf{i} + (1 - 1) \textbf{j} + (3 - 0) \textbf{k} = \textbf{i} + 0 \textbf{j} + 3 \textbf{k}\]
The resulting vector \(\textbf{r}\) indicates not only the magnitude but also the direction from the origin to the point.
The resulting vector \(\textbf{r}\) indicates not only the magnitude but also the direction from the origin to the point.
Projection of Vectors
The projection of one vector onto another vector is a critical step when computing the torque about a line. To project vector \(\textbf{r}\) onto the direction vector \(\textbf{d}\), use the formula: \[\textbf{r}_\text{proj} = \frac{\textbf{r} \cdot \textbf{d}}{\textbf{d} \cdot \textbf{d}} \textbf{d} \]
This formula finds how much of \(\textbf{r}\) lies along \(\textbf{d}\). The perpendicular component is then \(\textbf{r}_\text{perp} = \textbf{r} - \textbf{r}_\text{proj}\). This component is critical for calculating torque about a line.
This formula finds how much of \(\textbf{r}\) lies along \(\textbf{d}\). The perpendicular component is then \(\textbf{r}_\text{perp} = \textbf{r} - \textbf{r}_\text{proj}\). This component is critical for calculating torque about a line.
Direction Vector
A direction vector provides the orientation of a line or path in space. In this exercise, the direction vector \(\textbf{d}\) is derived from the line equation: \[ \textbf{r} = 4 \textbf{i} + \textbf{j} + (2 \textbf{i} + \textbf{j} - 2 \textbf{k}) t \]
The coefficient of t, \((2 \textbf{i} + \textbf{j} - 2 \textbf{k})\), is the direction vector. This vector determines the direction and sense of the line. It is important when projecting the position vector and calculating torque.
The coefficient of t, \((2 \textbf{i} + \textbf{j} - 2 \textbf{k})\), is the direction vector. This vector determines the direction and sense of the line. It is important when projecting the position vector and calculating torque.
