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).