Question

The moment of the force, vector F = 4i cap + 5j cap - 6k cap at (2, 0, - 3), about the point (2, –2, –2), is given by:-

Answer 1

F= 4i +5k_6j r=[2i_3j] -[2i_2j_2k] where r=2j_K   r×f= take matrix for both force and r distance. =_7i_4j_8k.

Answer 2

The force is given as $\vec F=4\hat i+5\hat j-6\hat k$

The force is acting at the point (2, 0,-3).

The point under consideration is (2,-2, -2). The moment of force  is calculated around this point.

Hence, the position vector of (2,0,-3) with respect to (2,-2,-2) will be -

                                    $(2-2)\hat i+(0-[-2])\hat j+([-3]-[-2])\hat k$

                                    $=\ 0\hat i+2\hat j-1\hat k$

The moment of the force is calculated as -

 $\vec\tau\ =\ \vec r\times \vec F$

                        = $(0\hat i+2\hat j-1\hat k)\times (4\hati+5\hat j-6\hat k)$

                       = $(2\times [-6]-[-1]\times 5)\hat i-(0\times [-6]-[-1]\times 4)\hat j+(0\times 5-4\times 2)\hat k$

                       = $\ -7\hat i-4\hat j-8\hat k$

Hence, the moment of the force is $-7\hat i-4\hat j-8\hat k$