Question

what is the torque of the force f=(2i-3j+4k)F, acting at the point r=(3i+2j+3k)m ABOUT THE ORIGIN (in N-m)

Answer 1

Torque of a Force $\vec{F}$ acting on a point with position vector $\vec{r}$ is given by:

$\boxed{\vec{\tau}=\vec{r}\times \vec{F}}$
So, we can find Torque by finding the cross product of $\vec{r}$ and $\vec{F}$

We have:

$\vec{r} = 3 \hat{\imath} + 2\hat{\jmath} + 3\hat{k} \, \, m \\ \\ \vec{F} = 2 \hat{\imath} - 3\hat{\jmath} + 4\hat{k} \, \, N$
So, torque will be:

$\vec{\tau} = \left|\begin{array}{ccc}\hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{array}\right| \\ \\ \\ \implies \vec{\tau} = \hat{\imath} ((2)(4)-(-3)(3)) - \hat{\jmath} ((3)(4)-(2)(3)) + \hat{k} ((3)(-3)-(2)(2)) \\ \\ \\ \implies \vec{\tau} = \hat{\imath} (8+9) - \hat{\jmath}(12-6) + \hat{k} (-9-4) \\ \\ \\ \implies \boxed{\vec{\tau}=17\hat{\imath}-6\hat{\jmath}-13\hat{k} \, \, \, N \, m}$

This is the torque of the force acting about Origin.

Answer 2

$\textbf{Your answer is --}$

Let, the torque of the given force acting on a point be T.

So, vectorT = vector(r)×vector( F)

we know that torque is a vector quantity and we take the cross product of F and r in the formula .

we have ,

F = (2i^ - 3j^ + 4k)F

r = (3i^ + 2j^ +3k^)m

Now , we take the cross product of F and r

F × r = (2i^ - 3j^ + 4k) × (3i^ +2j^ +3k^)

= i^( -3×3-2×4) -j^ (2×3-3×4) +k^(2×2-3×-3)

= i^(-9-8) -j^(6-12) +k^(4+9)

= -17i^ + 6j^ + 13k^

Hence,

T = ( -17i^ + 8j^ + 13k^)Nm

$\textbf{HOPE IT HELPS YOU }$