Question

A particle is located at \(\overrightarrow{\mathbf{r}}=(0.54 \mathrm{~m}) \hat{\mathbf{i}}+(-0.36 \mathrm{~m}) \hat{\mathbf{j}}+\) \((0.85 \mathrm{~m}) \hat{\mathbf{k}}\). A constant force of magnitude \(2.6 \mathrm{~N}\) acts on the particle. Find the components of the torque about the origin when the force acts in \((a)\) the positive \(x\) direction and \((b)\) the negative \(z\) direction.

Short answer

The components of the torque when force acts in the positive x direction are -2.21 Nm in the j direction and 0.936 Nm in the k direction. When force acts in negative z direction, the components are 0.936 Nm in the i direction and -1.404 Nm in j direction.

Step-by-step solution

Scenerio (a) where force is in positive x direction

The force vector for a force acting in the positive x direction is given by \(\vec{F} = 2.6 \, \hat{i}\). The torque around the origin is then given by: \(\vec{\tau} = \vec{r} \times \vec{F} = (0.54 \, \hat{i} - 0.36 \, \hat{j} + 0.85 \, \hat{k}) \times 2.6 \, \hat{i} \). Performing the cross product operation gives: \(\vec{\tau} = (0.36*2.6) \, \hat{k} - (0.85*2.6) \, \hat{j}\)

Calculating the components of Torque in Scenerio (a)

Carrying out the calculations results in \(\vec{\tau} = -2.21 \, \hat{j} + 0.936 \, \hat{k} \, N*M\). These are the components of the torque when the force acts in the positive x direction.

Scenerio (b) where force is in negative z direction

The force vector for a force acting in the negative z direction is given by \(\vec{F} = -2.6 \, \hat{k}\). The torque around the origin is then given by: \(\vec{\tau} = \vec{r} \times \vec{F} = (0.54 \, \hat{i} - 0.36 \, \hat{j} + 0.85 \, \hat{k}) \times (-2.6 \, \hat{k})\). Performing the cross product operation gives: \(\vec{\tau} = (-0.54*2.6) \, \hat{j} + (0.36*2.6) \, \hat{i}\).

Calculating the components of Torque in Scenerio (b)

Carrying out the calculations results in \(\vec{\tau} = 0.936 \, \hat{i} - 1.404 \, \hat{j} \, N*M\). These are the components of the torque when the force acts in the negative z direction.

Key concepts

Cross Product

The cross product, also known as the vector product, is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to the plane in which the original two vectors lie. The magnitude of the cross product vector represents the area of the parallelogram with sides represented by the original vectors.

When calculating torque, which is a measure of the force causing an object to rotate, the cross product becomes crucial as it follows the formula \( \vec{\tau} = \vec{r} \times \vec{F} \). Here, \( \vec{r} \) is the position vector from the pivot point to the point of force application, and \( \vec{F} \) is the force vector. The resulting torque vector \( \vec{\tau} \) tells us not only the magnitude of the torque but also its direction, which indicates the rotation axis and sense (clockwise or counterclockwise).

When you're performing a cross product operation, you're basically determining how much of the force vector \( \vec{F} \) is effectively contributing to rotation around a point that's not lying along the direction of the force. The sin component of the angle between \( \vec{r} \) and \( \vec{F} \) contributes to the magnitude of the torque vector.

Force Vectors

Force vectors represent forces as directional quantities with both magnitude and direction. In calculations, vectors are depicted by arrows, where the length represents the magnitude (how strong the force is) and the arrowhead points in the direction the force is applied.

When discussing torque, forces must be considered as vectors because the rotation effect they generate depends not only on the force's magnitude but also on the line of action and the point of application relative to the fulcrum or pivot point. Therefore, to solve torque problems, each force vector must be broken down into its components along the x, y, and z axes.

In the provided exercise, force vectors are specified in different scenarios to show how their direction affects the resulting torque. Understanding how to resolve a force into its vector components is essential to solve many physics problems that involve forces.

Moment of Force

The moment of force, also known as torque, is the rotational equivalent of linear force. It is a measure of the tendency of a force to rotate an object about some axis. Just as force causes an object to accelerate in linear motion, torque causes an object to acquire angular acceleration.

The magnitude of torque depends on three factors: the magnitude of the force applied (\( F \)), the distance from the rotation axis to where the force is applied (\( r \)), and the angle \( \theta \) between the force vector and the lever arm, which is the straight line drawn from the rotation axis to the point where the force is applied. The vector equation for calculating the magnitude of torque is \( \tau = r \cdot F \cdot \sin(\theta) \), where \( \tau \) is the magnitude of torque, and \( r \), \( F \) are the magnitudes of the position vector and force vector, respectively.

The direction of torque is given by the right-hand rule: if you curl the fingers of your right hand in the direction of rotation (from the position vector to the force vector), your thumb will point in the direction of the torque vector. This rule helps in visualizing and solving problems involving torque, ensuring a clear understanding of the axis and direction of rotation.