Question

At a point (2m,2m) a force of 10N is acting along positive Z-direction torque of this force about x-axis will be​

Answer 1

Given:

At a point (2m,2m) a force of 10N is acting along positive Z-direction.

To Find:

torque of this force about x-axis will be​?

Solution:

By the definition of torque,

$\[\vec \tau = \vec r \times \vec F\]$

Given point is  (2m,2m) = (2m,2m,0)

$\[\vec r = (2m)\hat i + (2m)\hat j + 0\hat k\]$

Since, force acting along positive Z- direction.

so,$\[\vec F = 10\hat k\]$

Now,

$\[\begin{gathered} \vec \tau = ((2m)\hat i + (2m)\hat j + 0\hat k) \times (10\hat k) \hfill \\ = 20\hat i - 20\hat j \hfill \\ \end{gathered} \]$

Thus, torque about x-axis is

$\[\left| {20\hat i} \right| = 20\]$   .