Question

A force \(2 \mathrm{i} \wedge+3 \mathrm{j} \wedge\) acts about an axis at a position vector \((\mathrm{j} \wedge+\mathrm{k} \wedge)\) from the axis, then what is the torque due to the force about the axis? (A) \(\sqrt{19} \mathrm{Nm}\) (B) \(\sqrt{13} \mathrm{Nm}\) (C) \(\sqrt{15 \mathrm{Nm}}\) (D) \(\sqrt{17} \mathrm{Nm}\)

Short answer

The torque due to the force about the axis is \(\sqrt{13}\,\text{Nm}\) (B).

Step-by-step solution

Write down the given force and position vectors

The force vector is given by \[ \vec{F} = 2\widehat{i} + 3\widehat{j} \] The position vector is given by \[ \vec{r} = \widehat{j} + \widehat{k} \]

Calculate the vector cross product of the position vector and force vector

The torque vector, \(\vec{\tau}\), is given by the cross product of the position vector \(\vec{r}\) and force vector \(\vec{F}\), so: \[ \vec{\tau} = \vec{r} \times \vec{F} \] We can find the cross product using the determinant method: \[ \vec{\tau} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k}\\ 0 & 1 & 1\\ 2 & 3 & 0\\ \end{vmatrix} \]

Simplify the cross product

Expand the determinant along the first row: \[ \vec{\tau} = \widehat{i} \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} - \widehat{j} \begin{vmatrix} 0 & 1 \\ 2 & 0 \end{vmatrix} + \widehat{k} \begin{vmatrix} 0 & 1 \\ 2 & 3 \end{vmatrix} \] Solve the minors: \[ \vec{\tau} = \widehat{i} (-3) - \widehat{j} (-2) + \widehat{k} (0) \] So, the torque vector is \[ \vec{\tau} = -3\widehat{i} + 2\widehat{j} \]

Calculate the magnitude of the torque vector

The magnitude of the torque vector can be found using the formula: \[ |\vec{\tau}| = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \] So, the torque due to the force about the axis is \(\boxed{\sqrt{13}\,\text{Nm}}\), which corresponds to option (B).

Key concepts

Cross Product

The cross product, also known as the vector product, is a crucial operation in physics and mathematics, specifically when dealing with torque. It's used to find a vector that is perpendicular to two given vectors in three-dimensional space. For torque, it helps determine the rotational effect of a force about an axis.

To perform a cross product, you need to set up a determinant. This involves three unit vectors (\(\widehat{i} \, \widehat{j} \, \widehat{k}\)) and the components of the two vectors being multiplied. In our exercise, the force vector \(\vec{F}\) is \(2\widehat{i} + 3\widehat{j}\) and the position vector \(\vec{r}\) is \(\widehat{j} + \widehat{k}\). Setting up the determinant helps to calculate the resulting vector through expansion of the first row.

Remember, the cross product results in another vector, calculated using the differences between the products of the respective pairs of components. This product is particularly helpful in situations involving an axis of rotation, like those commonly encountered when calculating torque.

Vector Mathematics

Vectors are mathematical entities that have both magnitude and direction, making them essential in physics, particularly when dealing with quantities like force and displacement. In solving problems involving vectors, you often deal with operations such as addition, subtraction, and the cross product.

Understanding the composition of vectors is critical. Vectors are typically expressed in terms of their components along the standard basis vectors (\(\widehat{i}, \widehat{j},\widehat{k}\)). Each component represents the vector's influence in one of the three dimensions of space. For instance, in the exercise, the force is represented as \(2\widehat{i} + 3\widehat{j}\), indicating its exertion in the x and y directions.

Vector mathematics allows for the combination of these components to form new vectors, helping describe physical phenomena. Mastery of vectors includes understanding operations like the dot product (for computing scalar projections), and more importantly here, the cross product, which results in a vector perpendicular to the plane of the two initial vectors.

Determinants in Physics

Determinants are mathematical tools often used in linear algebra to solve systems of equations, among other applications. In physics, determinants are instrumental when calculating cross products, which is the case when finding torque.

In a simple cross product operation involving two vectors, arranging these vectors into a 3x3 matrix with unit vectors (\(\widehat{i} \, \widehat{j}\, \widehat{k})\) on the top row, is crucial.
For example, given vectors \((0, 1, 1)\) and \( (2, 3, 0)\), the determinant helps to find the resulting vector. This involves breaking down the matrix into smaller 2x2 matrices called minors and calculating the solutions to these minors.

The minors themselves are solved by multiplying diagonals and reversing signs according to the rules of determinant calculation (e.g., \det\(\begin{pmatrix} a & b \ c & d \end{pmatrix}) = ad - bc\). This process yields the vector components after simplifying, which in our example were (\(-3, 2, 0\)).
Understanding determinants allows for the proper application of these foundational operations to calculate quantities like torque effectively.