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 of the coordinate axes.

Short answer

Torque vector: \(4\textbf{i} - 8\textbf{j} + 4\textbf{k}\); x-axis: 4, y-axis: -8, z-axis: 4

Step-by-step solution

Understand the Problem

Given a force vector \(\textbf{F} = (1, 2, 3)\) acting at a point \(\textbf{r} = (3, 2, 1)\), the goal is to find the torque \(\textbf{τ}\) about the origin and about each coordinate axis.

Formula for Torque

Torque \(\textbf{τ}\) is calculated using the cross product of the position vector \(\textbf{r}\) and the force vector \(\textbf{F}\). The formula is \[ \textbf{τ} = \textbf{r} \times \textbf{F}.\]

Calculate the Cross Product

The cross product \(\textbf{r} \times \textbf{F}\) is calculated using the determinant of the following matrix: \[\textbf{τ} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 3 & 2 & 1 \ 1 & 2 & 3 \ \ end{vmatrix}\].

Evaluate the Determinant

Solve the determinant to find each component of the torque vector. Using cofactor expansion: \[ \textbf{τ}_x = \textbf{i}(2*3 - 1*2) = \textbf{i}(6 - 2) = 4\textbf{i} \] \[ \textbf{τ}_y = -\textbf{j}(3*3 - 1*1) = -\textbf{j}(9 - 1) = -8\textbf{j} \] \[ \textbf{τ}_z = \textbf{k}(3*2 - 2*1) = \textbf{k}(6 - 2) = 4\textbf{k}.\]

Combine the Components

The torque vector \(\textbf{τ}\) is formed by combining the computed components: \[ \textbf{τ} = 4\textbf{i} - 8\textbf{j} + 4\textbf{k}.\]

Determine Torque about Each Axis

The torque about each axis corresponds to the components of \(\textbf{τ}\). So the torque about the x-axis is 4, about the y-axis is -8, and about the z-axis is 4.

Key concepts

Vector Cross Product

The vector cross product is an essential concept in physics and mathematics. It is used to determine the torque generated when a force is applied at a given position. For two vectors \(\textbf{a}\) and \(\textbf{b}\), their cross product, denoted as \(\textbf{a} \times \textbf{b}\), produces a third vector that is perpendicular to both \(\textbf{a}\) and \(\textbf{b}\). This perpendicular vector's direction is determined by the right-hand rule. The magnitude of the cross product is given by \[ |\textbf{a} \times \textbf{b}| = |\textbf{a}| |\textbf{b}| \text{sin}(\theta), \] where \( \theta \) is the angle between \(\textbf{a}\) and \(\textbf{b}\).

In torque calculations, the cross product helps in finding the direction and magnitude of the rotational effect caused by a force. If you act a force vector \(\textbf{F} = (1,2,3)\) at a point \(\textbf{r} = (3,2,1)\), the torque vector \(\textbf{τ}\) can be found using the formula: \(\textbf{τ} = \textbf{r} \times \textbf{F}\).

Coordinate Axes Torque

Torque about each coordinate axis can be understood by examining the components of the torque vector. The x, y, and z components of \( \textbf{τ} \) show how much rotational effect is present around each axis.

In the exercise solution, after determining the components of the torque vector, we got: \[ \textbf{τ} = 4\textbf{i} - 8\textbf{j} + 4\textbf{k} \]
This gives us:

• 4 units of torque about the x-axis,
• -8 units of torque about the y-axis,
• 4 units of torque about the z-axis.

These components indicate the direction and magnitude of the force's rotational action around each coordinate axis. Positive and negative signs signify the direction based on the right-hand rule.

Determinant Evaluation

Evaluating determinants is a critical step in finding the cross product of two vectors. For vectors \( \textbf{r} = (3,2,1) \) and \( \textbf{F} = (1,2,3) \), we use the determinant to calculate each component of the torque vector.

Set up the matrix: \[ \textbf{τ} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 3 & 2 & 1 \ 1 & 2 & 3 \ \end{vmatrix} \]
Solve the determinant using cofactor expansion:

• For the x-component, \textbf{τ}_x = \textbf{i}(2 \times 3 - 1 \times 2) = 4\textbf{i}\
• For the y-component, \- \textbf{τ}_y = -\textbf{j}(3 \times 3 - 1 \times 1) = -8\textbf{j}\
• For the z-component, \textbf{τ}_z = \textbf{k}(3 \times 2 - 2 \times 1) = 4\textbf{k}\

Combine the components to get the final torque vector: \[ \textbf{τ} = 4\textbf{i} - 8\textbf{j} + 4\textbf{k} \].

Force Vector

The force vector in physics represents both the direction and magnitude of a force. In our exercise, the force vector is \( \textbf{F} = (1, 2, 3) \).

This vector indicates that the force acts in three-dimensional space with components along the x, y, and z axes. These components can have various applications such as calculating torque or analyzing motion.

When dealing with forces and their effects, it’s essential to understand both the magnitude and the direction in which they act. The position vector \( \textbf{r} = (3, 2, 1) \) where this force acts further influences the resulting torque. Combining \( \textbf{r} \) and \( \textbf{F} \) through the cross product generates the torque vector that provides insight into rotational effects.

Using vectors for force and torque calculations helps in simplifying complex problems and offers a clear visualization of how forces interact in a system. Remember: where the force is applied and its direction are equally important for accurately assessing its impact.