Question

a force F=4i-5j+3k is acting in point r=i+2j+3k . the torque about r=3i-2j-3k is

Answer 1

The torque acting on the particle due to this force is $(42i+30j-6k)\ N-m$.

Explanation:

It is given that,

Force, $F=(4i-5j+3k)\ N$

initial position, $r_1=(i+2j+3k)\ m$

Final position, $r_2=(3i-2j-3k)\ m$

To find,

Torque

Solution,

The torque acting on the object is given by the below relation as :

$\tau=F\times r$

r is the position of the particle

$r=r_2-r_1$

$r=(3i-2j-3k)-(i+2j+3k)$

$r=(2i-4j-6k)\ m$

$\tau=(4i-5j+3k)\times (2i-4j-6k)$

$\tau=\begin{pmatrix}42&30&-6\end{pmatrix}$

$\tau=(42i+30j-6k)\ N-m$

So, the torque acting on the particle due to this force is $(42i+30j-6k)\ N-m$. Hence, this is the required solution.

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Torque

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