Problem 21 - Fundamentals of Physics

Question

In unit-vector notation, what is the torque about the origin on a particle located at coordinates \((0,-4.0 \mathrm{~m}, 3.0 \mathrm{~m})\) if that torque is due to (a) force \(\vec{F}_{1}\) with components \(F_{1 x}=2.0 \mathrm{~N}, F_{1 y}=F_{1 z}=0,\) and (b) force \(\vec{F}_{2}\) with components \(F_{2 x}=0, F_{2 y}=2.0 \mathrm{~N}, F_{2 t}=4.0 \mathrm{~N} ?\)

Step-by-Step Solution

Answer

Torque \(\vec{\tau}_1\) is \( 0\hat{i} + 6.0\hat{j} + 8.0\hat{k} \), torque \(\vec{\tau}_2\) is \( -22.0\hat{i} + 0\hat{j} \).

Step 1: Understanding Torque Formula

Torque \(\vec{\tau}\) about the origin is calculated using the cross product of the position vector \(\vec{r}\) and the force vector \(\vec{F}\). The formula is \(\vec{\tau} = \vec{r} \times \vec{F}\), where \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\) and \(\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k}\).

Step 2: Determine Position Vector

The position vector \(\vec{r}\) of the particle is given as \((0, -4.0\ \mathrm{m}, 3.0\ \mathrm{m})\). Therefore, \(\vec{r} = 0\hat{i} - 4.0\hat{j} + 3.0\hat{k}\).

Step 3: Calculate Torque for Force \(\vec{F}_1\)

For force \(\vec{F}_1\), the components are given as \(F_{1x} = 2.0\ \mathrm{N}, F_{1y} = 0, F_{1z} = 0\). Therefore, \(\vec{F}_1 = 2.0\hat{i}\). To find the torque, perform the cross product: \[ \vec{\tau}_1 = \vec{r} \times \vec{F}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -4.0 & 3.0 \ 2.0 & 0 & 0 \end{vmatrix} \]Calculating the determinant, we get:\[ \vec{\tau}_1 = \hat{i}(0 \cdot 0 - 3.0 \cdot 0) - \hat{j}(0 \cdot 0 - 6.0 \cdot 2.0) + \hat{k}(0 \cdot 0 + 8.0 \cdot 2.0) = 0\hat{i} + 6.0\hat{j} + 8.0\hat{k} \]

Step 4: Calculate Torque for Force \(\vec{F}_2\)

For force \(\vec{F}_2\), the components are given as \(F_{2x} = 0, F_{2y} = 2.0\ \mathrm{N}, F_{2z} = 4.0\ \mathrm{N}\). Therefore, \(\vec{F}_2 = 2.0\hat{j} + 4.0\hat{k}\). To find the torque, perform the cross product: \[ \vec{\tau}_2 = \vec{r} \times \vec{F}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -4.0 & 3.0 \ 0 & 2.0 & 4.0 \end{vmatrix} \]Calculating the determinant, we get: \[ \vec{\tau}_2 = \hat{i}((-4.0 \cdot 4.0) - (3.0 \cdot 2.0)) - \hat{j}(0 - 0) + \hat{k}(0\cdot 2.0 - 0\cdot 4.0) = -22.0\hat{i} + 0\hat{j} - 0\hat{k} \]

Step 5: Final Result

The torque due to force \(\vec{F}_1\) in unit vector notation is \( \vec{\tau}_1 = 0\hat{i} + 6.0\hat{j} + 8.0\hat{k} \). The torque due to force \(\vec{F}_2\) in unit vector notation is \( \vec{\tau}_2 = -22.0\hat{i} + 0\hat{j} \).

Key Concepts

Unit Vector Notation

Unit vector notation is a convenient and precise way to express vectors. Vectors can describe quantities like position, velocity, or force, showing both magnitude and direction.
In three dimensions, vectors are represented using unit vectors along the x, y, and z axes, labeled as \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\), respectively. This notation simplifies calculations by breaking a vector into its components.
For example, a position vector given by coordinates \((x, y, z)\) can be expressed in unit vector notation as \(x\hat{i} + y\hat{j} + z\hat{k}\).

• \(\hat{i}\) represents the x-component;
• \(\hat{j}\) represents the y-component;
• \(\hat{k}\) represents the z-component.

Using this notation makes it easier to perform operations like addition, subtraction, and the calculation of cross products, which are essential in physics to compute things like torque.

Cross Product

The cross product is an essential operation in vector algebra, used to find a vector that is perpendicular to two given vectors. It's particularly useful in physics, for example in computing torque.
The cross product of vectors \(\vec{A}\) and \(\vec{B}\) is denoted \(\vec{A} \times \vec{B}\). It results in a vector whose direction follows the right-hand rule—where the thumb, forefinger, and middle finger of the right hand represent the vectors: \(\vec{A}\), \(\vec{B}\), and \(\vec{A} \times \vec{B}\), respectively.
To calculate it, use a determinant:

• Write the unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) in the first row;
• The components of \(\vec{A}\) in the second row;
• The components of \(\vec{B}\) in the third row.

Then, calculate the determinant by expanding along the first row. The result will give you the components of the cross product vector in unit vector notation.
This technique becomes crucial when dealing with forces in a three-dimensional space.

Position Vector

A position vector is a fundamental concept in physics. It describes the position of a point in space relative to an origin. This vector shows where an object is located, which is essential for calculating forces like torque.
The position vector \(\vec{r}\) from the origin \((0, 0, 0)\) to a point \((x, y, z)\) is represented as \(x\hat{i} + y\hat{j} + z\hat{k}\). Each component represents the displacement along its respective axis.
In the exercise here, the position vector is given as \(\vec{r} = 0\hat{i} - 4.0\hat{j} + 3.0\hat{k}\), indicating the particle is 4 meters in the negative y-direction and 3 meters in the positive z-direction.
Understanding this helps when applying the cross product to find the resulting torque. By knowing the position vector, you set the stage for further calculations involving forces acting on particles.

Force Vector Calculation

Force vector calculation is critical when determining the impact of a force acting on a particle. A force vector is expressed in terms of its components along the x, y, and z axes using unit vector notation.
In physics, the force vector \(\vec{F}\) reveals how force is distributed across different directions. For instance, in the exercise, force \(\vec{F}_1\) is \(2.0\hat{i}\), and force \(\vec{F}_2\) is \(2.0\hat{j} + 4.0\hat{k}\).
Graphs or diagrams often illustrate these vectors, helping visualize how forces influence object movement. Knowing the components allows us to compute other vector properties, like torque.
The torque itself is found by taking the cross product of the position vector \(\vec{r}\) and the force vector \(\vec{F}\). Doing this, you find the torque direction and magnitude resulting from given forces.