Problem 63 - Master Resource Book in JEE Main Physics

Question

Here, \(\mathbf{F}=-F \hat{\mathbf{k}}\) $$ \begin{aligned} \mathbf{r} &=(\hat{\mathbf{i}}-\hat{\mathbf{j}}), \tau=? \\\ \text { As } & \tau=\mathbf{r} \times \mathbf{F}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(-F \hat{\mathbf{k}}) \end{aligned} $$ \(=-F(\hat{\mathbf{i}} \times \hat{\mathbf{k}}-\hat{\mathbf{j}} \times \hat{\mathbf{k}})\) \(=-F(-\hat{\mathbf{j}}-\hat{\mathbf{i}})\) \(=F(\hat{\mathbf{i}}+\hat{\mathbf{j}})\)

Step-by-Step Solution

Answer

The torque \( \tau = F(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \).

Step 1: Understand the Problem

We are given a force vector \( \mathbf{F} = -F \hat{\mathbf{k}} \) and a position vector \( \mathbf{r} = \hat{\mathbf{i}} - \hat{\mathbf{j}} \). The task is to find the torque \( \tau \), which is defined as the cross product \( \tau = \mathbf{r} \times \mathbf{F} \).

Step 2: Set Up the Cross Product Equation

Using the cross product definition, \( \tau = \mathbf{r} \times \mathbf{F} = (\hat{\mathbf{i}} - \hat{\mathbf{j}}) \times (-F \hat{\mathbf{k}}) \). This sets the stage for finding \( \tau \) by simplifying the expression.

Step 3: Expand the Cross Product

Apply the distributive property to calculate each individual cross product: \( (\hat{\mathbf{i}} - \hat{\mathbf{j}}) \times (-F \hat{\mathbf{k}}) = -F (\hat{\mathbf{i}} \times \hat{\mathbf{k}}) + F (\hat{\mathbf{j}} \times \hat{\mathbf{k}}) \).

Step 4: Use the Right-Hand Rule for Basic Cross Products

Utilize the properties of basic unit vector cross products:- \( \hat{\mathbf{i}} \times \hat{\mathbf{k}} = -\hat{\mathbf{j}} \)- \( \hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}} \).Using these, the expression becomes \(-F(-\hat{\mathbf{j}}) + F(\hat{\mathbf{i}}) \).

Step 5: Simplify the Expression

Further simplify the expression: \( -F(-\hat{\mathbf{j}}) + F(\hat{\mathbf{i}}) = F\hat{\mathbf{j}} + F\hat{\mathbf{i}} = F(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \).

Step 6: Conclusion: Write the Final Expression for Torque

Thus, the torque \( \tau \) is \( \tau = F(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \).

Key Concepts

Cross Product in Vectors

The cross product is a mathematical operation that is often used when dealing with vectors in three-dimensional space. It results in another vector that is perpendicular to the plane formed by the two original vectors. This operation is crucial in physics for calculations involving torque, magnetic force, and angular momentum.
To calculate the cross product, consider two vectors \( \mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}} \) and \( \mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}} + B_z \hat{\mathbf{k}} \). The cross product \( \mathbf{A} \times \mathbf{B} \) is given by the determinant of a matrix involving these unit vectors:

• Calculate each component: \( C_x = A_yB_z - A_zB_y \)
• \( C_y = A_zB_x - A_xB_z \)
• \( C_z = A_xB_y - A_yB_x \)

The resulting vector is \( \mathbf{C} = C_x\hat{\mathbf{i}} + C_y\hat{\mathbf{j}} + C_z\hat{\mathbf{k}} \).
In the given problem, the torque \( \tau \) is calculated as the cross product of the position vector \( \mathbf{r} \) and the force vector \( \mathbf{F} \). This cross product ensures that the torque vector is perpendicular to the plane created by \( \mathbf{r} \) and \( \mathbf{F} \).

Unit Vectors in Physics

Unit vectors are the building blocks of vector representation in physics, indicating direction without conveying magnitude. Typically, they are denoted with a hat (\( \hat{} \)), such as \( \hat{\mathbf{i}} \), \( \hat{\mathbf{j}} \), and \( \hat{\mathbf{k}} \), representing the x, y, and z axes, respectively.
In vector notation:

• \( \hat{\mathbf{i}} \) indicates direction in the x-axis
• \( \hat{\mathbf{j}} \) indicates direction in the y-axis
• \( \hat{\mathbf{k}} \) indicates direction in the z-axis

These vectors are essential for breaking down vectors into their components and applying vector operations, such as addition and cross products.
For the forces and position vectors in the exercise:

• The force vector \( \mathbf{F} = -F \hat{\mathbf{k}} \) indicates a force acting in the negative z-direction.
• The position vector \( \mathbf{r} = \hat{\mathbf{i}} - \hat{\mathbf{j}} \) represents a vector that lies in the x-y plane, pointing diagonally.

Understanding these unit vectors allows us to perform vector operations like the cross product, determining the direction of resultant vectors like torque.

Right-Hand Rule

The right-hand rule is an essential tool in physics used to determine the direction of vectors resulting from cross products, such as torque or angular momentum. It's a simple, physical aid that helps visualize the perpendicular direction of the resulting vector.
To apply the right-hand rule:

• Point your fingers in the direction of the first vector (\( \mathbf{r} \)).
• Then curl them towards the second vector (\( \mathbf{F} \)).
• Your thumb now points in the direction of the cross product. This thumb direction gives the direction of the torque vector in this problem.

In the exercise, the cross products \( \hat{\mathbf{i}} \times \hat{\mathbf{k}} = -\hat{\mathbf{j}} \) and \( \hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}} \) were evaluated using this rule.
With practice, the right-hand rule becomes a quick and intuitive method to ensure that torque and force calculations lead to physically believable results, indicating correct directional relationships between vectors.