In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0, -4.0m, 3.0m) if that torque is due to

(a) Force ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}}_{\mathbf{1}}$ with components F_1x = 2.0 N, F_1y = F_1z = 0,

(b) Force ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}}_{\mathbf{2}}$with components F_2x = 0,F_2x = 2.0N,F_2x = 4.0N?

For case a,$x=0m,y=-4.0m,z=3.0m,F_x=2.0N,F_y=0N,F_z=0N$

For case b,$x=0m,y=-4.0m,z=3.0m,F_x=0N,F_y=2.0N,F_z=4.0N$

Using the concept of torque, the unknown torque value is calculated. As per the concept, the torque acting on a body is due to the tangential force acting on a body along a radial path of the object in a circular motion. Thus, the cross-vector of the force and radial vector of the object will give the torque value.

Formulae:

The position vector in a 3-D diagram,$\stackrel{\rightharpoonup }{r}=x\hat{i}+y\hat{j}+z\hat{k}$

The force vector in 3-D,$\stackrel{\rightharpoonup }{F}=F_x\hat{i}+F_y\hat{j}+F_z\hat{k}$

The torque acting on the body due to the tangential force,

$\begin{array}{rcl}\overrightarrow{\tau }& =& \overrightarrow{r}\times \overrightarrow{F}\\ & =& \left|\begin{array}{ccc}i& j& k\\ x& y& z\\ F_x& F_y& F_z\end{array}\right|\\ & =& \left(y{F}_Z-z{F}_y\right)\hat{i}+\left(z{F}_x-x{F}_Z\right)\hat{j}+\left(x{F}_y-y{F}_x\right)\hat{k}\\ & & \end{array}$

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\tau }& =& \left|\begin{array}{ccc}i& j& k\\ 0& -4& 3\\ 2& 0& 0\end{array}\right|\\ & =& \left(0\right)\hat{i}Nm+\left(6-0\right)\hat{j}Nm+\left(8\right)\hat{k}Nm\\ & =& \left(6\hat{j}+8\hat{k}\right)Nm\end{array}$

Its magnitude is given by:

$\begin{array}{rcl}\tau & =& \sqrt{{\left(6Nm\right)}^2+{\left(8Nm\right)}^2}\\ & =& 10Nm\end{array}$

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\tau }& =& \left|\begin{array}{ccc}i& j& k\\ 0& -4& 3\\ 0& 2& 4\end{array}\right|\\ & =& \left(-16-6\right)\hat{i}Nm+\left(0\right)\hat{j}Nm=\left(0\right)\hat{k}Nm\\ & =& \left(-22\hat{i}\right)Nm\end{array}$