Make a sketch showing a situation in which the torque due to a single force about some location is \(20\,\,{\rm{N}} \cdot {\rm{m}}\) in the positive \(z\) direction, whereas about another location the torque is \(10\,\,{\rm{N}} \cdot {\rm{m}}\) in the negative \(z\) direction.

The force that can cause an object to twist along an axis is measured as torque. In linear kinematics, force is what causes an object to accelerate.

The relation describes the torque operating on an object at a specific point.

\(\overrightarrow \tau = \overrightarrow r \times \overrightarrow F \)……. (1)

Here,

\(\overrightarrow r \to \)Position vector of the given location

\(\overrightarrow F \to \)Force vector acting at the location

The force and position vectors to create a torque at a given place are depicted in the figure:

Let unit vectors along X, Y, X axes are \(\widehat x,\widehat y,\widehat z\)respectively. In order to get torque acting about some location is, \(\tau = 20N \cdot m\)along the positive \(z\)direction, it is essential to consider, the position vector of that location is along the positive \(X\)axis\(\overrightarrow r = (2m)\widehat x\)and force vector acting along the positive \(Y\)axis, \(\overrightarrow F = (10N)\widehat y\)

Using equation (1), torque acting about some location is given by,

\(\begin{aligned}{}\overrightarrow \tau &= \overrightarrow r \times \overrightarrow F \\ &= (2m)\widehat x \times (10N)\widehat y\\ &= (2N \cdot m)(\widehat x \times \widehat y)\\ &= (2N \cdot m)\widehat z\,\,\,\,\,\,\,\,\,\,\,\,\,\,as\,(\widehat x \times \widehat y) = \widehat z)\end{aligned}\)

In order to get torque acting about another location is, \(\tau = 10N \cdot m\)along the negative \(z\)direction, it is essential to consider, the position vector of that location is along the negative \(X\)-axis \(\overrightarrow r = (1m)( - \widehat x)\)and same force vector acting along the positive \(Y\)axis, \(\overrightarrow F = (10N)\widehat y\).

Using equation (1), torque acting about another location is given by,

\(\begin{aligned}{}\overrightarrow \tau & = \overrightarrow r \times \overrightarrow F \\ &= (1m)( - \widehat x) \times (10N)\widehat y\\ &= (10N \cdot m)( - \widehat x \times \widehat y)\\ &= (10N \cdot m)( - \widehat z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,As( - \widehat x \times \widehat y) &= - \widehat z)\end{aligned}\)

Torque produced on positive and negative direction is\(2\,\,{\rm{N}}\,{\rm{m}}\) and \( - 10\,\,{\rm{N}}\,{\rm{m}}\) respectively.