91Ó°ÊÓ

For case a,

$x=-2.0m,y=0m,z=4.0m,F_x=6.0N,F_y=0N,F_z=0N$

For case b,

$x=-2.0m,y=0m,z=4.0m,F_x=-6.0N,F_y=0N,F_z=6.0N$

For case c,

$x=-2.0m,y=0m,z=4.0m,F_x=0N,F_y=0N,F_z=6.0N$

For case d,

$x=-2.0m,y=0m,z=4.0m,F_x=6.0N,F_y=0N,F_z=-6.0N$

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:

$\begin{array}{rcl}\stackrel{â‡\P Ä}{r}& =& x\hat{i}+y\hat{j}+z\hat{k}\&\stackrel{â‡\P Ä}{F}=F_x\hat{i}+F_y\hat{j}+F_z\hat{k}\\ \stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \stackrel{â‡\P Ä}{r}×\stackrel{â‡\P Ä}{F}\\ & =& \left|\begin{array}{ccc}i& j& k\\ x& y& z\\ F_x& F_y& F_z\end{array}\right|\\ & =& \left(yFz-z{F}_y\right)\hat{i}+\left(z{F}_x-x{F}_x\right)\hat{j}+\left(x{F}_y-y{F}_x\right)\hat{k}\end{array}$

$\begin{array}{rcl}\stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \left|\begin{array}{ccc}i& j& k\\ -2& 0& 4\\ 6& 0& 0\end{array}\right|\\ & =& \left(0\right)\hat{i}N.m+\left(24-0\right)\hat{j}N.m\\ & =& 24\hat{j}N.m\end{array}$

$\begin{array}{rcl}\stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \left|\begin{array}{ccc}i& j& k\\ -2& 0& 4\\ -6& 0& 0\end{array}\right|\\ & =& \left(0\right)\hat{i}Nm+\left(0-24\right)\hat{j}Nm+\left(0\right)\hat{k}Nm\\ & =& 24\hat{j}Nm\end{array}$

$\begin{array}{rcl}\stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \left|\begin{array}{ccc}i& j& k\\ -2& 0& 4\\ 0& 0& 6\end{array}\right|\\ & =& \left(0\right)\hat{i}Nm+\left(0-\left(-12\right)\right)\hat{j}Nm+\left(0\right)\hat{k}Nm\\ & =& 12\hat{j}Nm\end{array}$

$\begin{array}{rcl}\stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \left|\begin{array}{ccc}i& j& k\\ -2& 0& 4\\ 0& 0& -6\end{array}\right|\\ & =& \left(0\right)\hat{i}Nm+\left(0-12\right)\hat{j}Nm+\left(0\right)\hat{k}Nm\\ & =& -12\hat{j}Nm\end{array}$