Question

The force \(\mathbf{F}=\mathbf{i}+3 \mathbf{j}+2 \mathbf{k}\) acts at the point (1,1,1). (a) Find the torque of the force about the point (2,-1,5) . Careful! The vector \(\mathbf{r}\) goes from (2,-1,5) to (1,1,1). (b) Find the torque of the force about the line \(\mathbf{r}=2 \mathbf{i}-\mathbf{j}+5 \mathbf{k}+(\mathbf{i}-\mathbf{j}+2 \mathbf{k}) t\). Note that the line goes through the point (2,-1,5).

Short answer

The torque about the point is \(16 \mathbf{i} - 2 \mathbf{j} - 5 \mathbf{k}\), and the torque about the line is \(\frac{4}{3} \mathbf{i} - \frac{4}{3} \mathbf{j} + \frac{8}{3} \mathbf{k}\).

Step-by-step solution

- Calculate \(\mathbf{r}\)

Find the vector \(\mathbf{r}\) that goes from point (2, -1, 5) to point (1, 1, 1). Subtract the coordinates:\[\mathbf{r} = (1 - 2) \mathbf{i} + (1 - (-1)) \mathbf{j} + (1 - 5) \mathbf{k} = -\mathbf{i} + 2 \mathbf{j} - 4 \mathbf{k}\]

- Calculate the Cross Product for Torque (a)

Use the cross product \(\mathbf{r} \times \mathbf{F}\) to find the torque about the point (2, -1, 5):\[\mathbf{r} \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 2 & -4 \ 1 & 3 & 2 \end{vmatrix} \]= \[\mathbf{i}((2)(2) - (-4)(3)) - \mathbf{j}((-1)(2) - (-4)(1)) + \mathbf{k}((-1)(3) - (2)(1))\]= \[\mathbf{i}(4 + 12) - \mathbf{j}(-2 + 4) + \mathbf{k}(-3 - 2)\]= \[16 \mathbf{i} - 2 \mathbf{j} - 5 \mathbf{k}\]So the torque is \[16\mathbf{i} - 2\mathbf{j} - 5\mathbf{k}\].

- Convert the Line Equation

Write the parametric form of the line given by \(\mathbf{r} = 2 \mathbf{i} - \mathbf{j} + 5 \mathbf{k} + (\mathbf{i} - \mathbf{j} + 2 \mathbf{k}) t \). When \( t = 0 \), the point on the line is (2, -1, 5).

- Calculate Direction Vector of the Line

Identify the direction vector from the line equation: \(\mathbf{d} = \mathbf{i} - \mathbf{j} + 2\mathbf{k} \).

- Calculate Torque with Respect to the Line (b)

The torque about a line can be found using the projection of the torque vector \(\mathbf{T} = \mathbf{r} \times \mathbf{F} = 16 \mathbf{i} - 2 \mathbf{j} - 5 \mathbf{k}\) onto the direction vector \(\mathbf{d} \). Calculate this as follows:\(\mathbf{T}_{\mathbf{d}} = \frac{\mathbf{T} \, . \, \mathbf{d}}{\|\mathbf{d}\|^2} \mathbf{d} \)Calculate the dot product \(\mathbf{T} . \mathbf{d} \):= \( (16)(1) + (-2)(-1) + (-5)(2) = 16 + 2 - 10 = 8 \)Calculate \(\|\mathbf{d}\|^2 \):= \( 1^2 + (-1)^2 + 2^2 = 1 + 1 + 4 = 6 \)Substitute these into the projection equation:\(\mathbf{T}_{\mathbf{d}} = \frac{8}{6} \mathbf{d} = \frac{4}{3} (\mathbf{i} - \mathbf{j} + 2 \mathbf{k})\)Therefore, the torque about the line is \(\frac{4}{3} \mathbf{i} - \frac{4}{3} \mathbf{j} + \frac{8}{3} \mathbf{k} \).

Key concepts

Cross Product

The cross product, also known as the vector product, is a crucial concept in physics, especially when dealing with torque. The cross product of two vectors, \(\mathbf{a} \times \mathbf{b}\), results in a third vector which is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). This is useful when you need to find a vector that is normal to a plane defined by two vectors.

To compute the cross product, you can use the determinant of a 3x3 matrix composed of the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) along with the components of the vectors being multiplied. For example, for vectors \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \(\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), the cross product is calculated as follows:

\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]

The result is a vector given by:

\[(a_2 b_3 - a_3 b_2)\mathbf{i} - (a_1 b_3 - a_3 b_1)\mathbf{j} + (a_1 b_2 - a_2 b_1)\mathbf{k} \]

Vector Calculations

Vectors are fundamental in physics for representing quantities that have both magnitude and direction. In vector calculations, you can add, subtract, and multiply vectors to solve various problems. The force given in this exercise, \( \mathbf{F} = \mathbf{i} + 3 \mathbf{j} + 2 \mathbf{k} \), is a vector that shows its influence in three-dimensional space.

To find the torque, which is another vector quantity representing rotational force, you need to compute the cross product of two vectors. Specifically, you need the vector \(\mathbf{r}\) that connects the point of application to the point about which you are finding the torque. Given two points (1,1,1) and (2,-1,5), vector \(\mathbf{r}\) is found by subtracting the coordinates:

\[ \mathbf{r} = (1 - 2) \mathbf{i} + (1 - (-1)) \mathbf{j} + (1 - 5) \mathbf{k} = -\mathbf{i} + 2 \mathbf{j} - 4 \mathbf{k} \]

This vector \(\mathbf{r}\) is then used in the cross product \(\mathbf{r} \times \mathbf{F}\) to find the torque.

Parametric Equations

Parametric equations are used to represent lines in a three-dimensional space. In the exercise, the line equation is given in its parametric form:

\[ \mathbf{r} = 2 \mathbf{i} - \mathbf{j} + 5 \mathbf{k} + (\mathbf{i} - \mathbf{j} + 2 \mathbf{k}) t \]

This indicates that the line passes through the point (2, -1, 5) and has a direction vector \(\mathbf{d} = \mathbf{i} - \mathbf{j} + 2 \mathbf{k} \). Parametric equations make it easier to describe the coordinates of any point on that line as the value of the parameter \(t\) changes.

When you need to find the torque about a line, taking the parametric form helps identify both the point of reference and the direction vector, which are crucial for subsequent calculations.

Direction Vectors

Direction vectors indicate the direction of a line or pathway in space. In calculations of torque about a line, direction vectors are particularly important. For the given line

\[\mathbf{r}=2 \mathbf{i}-\mathbf{j}+5 \mathbf{k}+ (\mathbf{i}-\mathbf{j}+2 \mathbf{k}) t\]

the direction vector \(\mathbf{d} \) is

\[\mathbf{d} = \mathbf{i} - \mathbf{j} + 2 \mathbf{k}\]

The direction vector helps determine the orientation of the line in three-dimensional space. When calculating the torque about a line, you project the torque vector onto this direction vector, helping to understand how the rotational force aligns with or deviates from the direction of the line.

The projection is calculated with the formula:

\[\mathbf{T}_{\mathbf{d}} = \frac{ \mathbf{T} \cdot \mathbf{d} }{ \|\mathbf{d}\|^2} \mathbf{d}\]

This process helps isolate the component of the torque that aligns with the direction of the line.