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

Answer:

net r=r2-r1=2i-4j-6k

f=4i-5j+3k

tork=f×r

=(2i-4j,-6k)(4i-5j+3k)

=8+20-18

=10Nm

hope it helps u...❤..

Answer 2

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

Explanation:

It is given that,

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

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

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

Displacement of the object,

$r=r_2-r_1=(3i-2j-3k)-(i+2j+3k)=(2i-4j-6k)\ m$

The cross product of force and the displacement of an object is called torque acting on it. It is given by :

$\tau=F\times r$

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

By calculating cross product of vectors, the torque acting on the object is equal to :

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

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

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

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