Question

Find the torque of a force F=a(hati+2hatj+3hatk) N about a point O. The position vector of point of application of force about O is r=(2hati+3hatj-hatk) m.

Answer 1

Answer:

Sorry I don't know anything about this

Answer 2

The value of Torque = a(11 i  -7 j + k)

Step by step explanation:

Given:

Force F =$a(\hat i + 2\hat j + 3\hat k)$ N                                        

Position vector r = $(2\hat i +3\hat j - 1\hat k)$ m

To find

Torque= τ

$\textbf{\large Formula of torque = r} \times \textbf{\large F}$                   (equation 1)

So,Torque is the vector product of position vector (r) and force vector (F).

Putting the values of r and F ,

We can find the value of torque by using matrix:

Torque= τ =

                         $\left[\begin{array}{ccc}\hat i&\hat j&\hat k\\2&3&-1\\a&2a&3a\end{array}\right]$                       (equation 2)

So,

Torque= τ = $\hat i(9a + 2a) -\hat j(6a + a) +\hat k(4a - 3a)$

   

                 =$11a \ \hat i \ -7a\hat j \ +a\hat k$

$\textbf{\Large So, the value of Torque = a(11 i -7 j + k)}$