Question

A vector force with components \((1,2,3)\) acts at the point \((3,2,1)\). Find the vector torque about the origin due to this force and find the torque about each coordinate axis.

Short answer

The torque vector is \(\textbf{T} = (4, -8, 4)\). The torque about the x, y, and z axes are 4, -8, and 4, respectively.

Step-by-step solution

- Identify the given vectors

Given the force vector \(\textbf{F} = (1,2,3)\) and the position vector \(\textbf{r} = (3,2,1)\).

- Use the cross product formula to find the torque

The torque \( \textbf{T} \) about the origin is given by the cross product \( \textbf{T} = \textbf{r} \times \textbf{F} \). Compute \( \textbf{r} \times \textbf{F} \): \(\textbf{r} \times \textbf{F} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 3 & 2 & 1 \ 1 & 2 & 3 \ \ \end{vmatrix}\).

- Calculate the determinant

Expand the determinant to find each component of the torque: \( \textbf{T}_x = (2 \times 3 - 1 \times 2) = 4 \), \(\textbf{T}_y = -(3 \times 3 - 1 \times 1) = -8 \), and \(\textbf{T}_z = (3 \times 2 - 2 \times 1) = 4 \). Thus, \( \textbf{T} = (4, -8, 4) \).

- Determine the torque about each coordinate axis

The torque about each coordinate axis corresponds to each component of the torque vector: \(\textbf{T}_x = 4 \), \(\textbf{T}_y = -8 \), and \(\textbf{T}_z = 4 \).

Key concepts

Cross Product

The cross product is a fundamental operation in vector algebra. It combines two vectors to produce a third vector that is perpendicular to both original vectors. In three-dimensional space, if we have vectors \(\textbf{A} = (A_x, A_y, A_z)\) and \(\textbf{B} = (B_x, B_y, B_z)\), the cross product \(\textbf{A} \times \textbf{B}\) results in a vector given by: \[ \textbf{A} \times \textbf{B} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \ \end{vmatrix} \] This expression uses the determinant of a 3x3 matrix where \(\textbf{i}\), \(\textbf{j}\), and \(\textbf{k}\) are unit vectors along the x, y, and z axes, respectively. The resulting vector from the cross product has components that demonstrate rotational effects in a system, which is crucial in the study of torques and rotational forces. The direction of this resultant vector follows the right-hand rule, where if you point your right-hand fingers in the direction of the first vector and curl them toward the second, your thumb points in the direction of the cross product.

Determinant Calculation

Determinants are essential in vector algebra for calculating cross products. In our case, the determinant of the 3x3 matrix gives the components of the torque vector. Given vectors \(\textbf{r} = (3, 2, 1)\) and \(\textbf{F} = (1, 2, 3)\), we find the cross product by computing the following determinant: \[ \textbf{T} = \textbf{r} \times \textbf{F} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 3 & 2 & 1 \ 1 & 2 & 3 \ \end{vmatrix} \] To solve, we expand the determinant by breaking it down into three 2x2 determinants: \[ \textbf{T}_x = \textbf{i} (2 \times 3 - 1 \times 2) \ \textbf{T}_y = -\textbf{j} (3 \times 3 - 1 \times 1) \ \textbf{T}_z = \textbf{k} (3 \times 2 - 2 \times 1) \ \end{vmatrix} \] After calculating, we get: \[ \textbf{T}_x = 4, \ \textbf{T}_y = -8, and \textbf{T}_z = 4 \] Combining them gives the resulting torque vector: \[ \textbf{T} = (4, -8, 4) \]

Force Vector

A force vector represents a force's magnitude and direction in space. In this exercise, we have the force vector \( \textbf{F} = (1, 2, 3) \). This means the force has:

• a component of 1 unit along the x-axis,
• 2 units along the y-axis,
• 3 units along the z-axis.

Each component of the force vector tells us how strong the force is in that particular direction. The direction and point of application of the force are critical for understanding how it influences movement and rotation. When a force acts at a certain point, it can cause an object to rotate around an axis. This rotational effect is the torque, which we calculate using the cross product.

Position Vector

The position vector locates a point in space relative to an origin. For our exercise, the position vector is \( \textbf{r} = (3, 2, 1) \). This vector:

• points 3 units along the x-axis,
• 2 units along the y-axis,
• 1 unit along the z-axis from the origin.

Knowing the exact point where the force acts is necessary for calculating the torque, as torque depends on both the location of the applied force and the force itself. Using the position vector and force vector, we can find the torque vector, providing insight into the object's rotational behavior.