if radius vector r=2i+j-k m and force f=i+j+3k n find torque
Answers
Torque of a Force \vec{F}
F
acting on a point with position vector \vec{r}
r
is given by:
\boxed{\vec{\tau}=\vec{r}\times \vec{F}}
τ
=
r
×
F
So, we can find Torque by finding the cross product of \vec{r}
r
and \vec{F}
F
We have:
\begin{gathered}\vec{r} = 3 \hat{\imath} + 2\hat{\jmath} + 3\hat{k} \, \, m \\ \\ \vec{F} = 2 \hat{\imath} - 3\hat{\jmath} + 4\hat{k} \, \, N\end{gathered}
r
=3
^
+2
^
+3
k
^
m
F
=2
^
−3
^
+4
k
^
N
So, torque will be:
\begin{gathered}\vec{\tau} = |\begin{array}{ccc}\hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{array}| \\ \\ \\ \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}\end{gathered}
τ
=∣
^
3
2
^
2
−3
k
^
3
4
∣
⟹
τ
=
^
((2)(4)−(−3)(3))−
^
((3)(4)−(2)(3))+
k
^
((3)(−3)−(2)(2))
⟹
τ
=
^
(8+9)−
^
(12−6)+
k
^
(−9−4)
⟹
τ
=17
^
−6
^
−13
k
^
Nm
This is the torque of the force acting about Origin.