91Ó°ÊÓ

$$\begin{gathered} \stackrel{â‡\P Ä}{r}=\left(0m\right)\hat{i}+\left(0.50m\right)\hat{j}-\left(2.0m\right)\hat{k} \\ \stackrel{â‡\P Ä}{F}=2N\hat{i}+0\hat{j}-3N\hat{k} \\ \stackrel{â‡\P Ä}{r}=\left(2m\right)\hat{i}+\left(0m\right)\hat{j}-\left(3.0m\right)\hat{k} \end{gathered}$$

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. Hence, through comparison of the torque given and calculated format, the force value can be given.

Formulae:

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

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

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

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

Using the formula for the torque, we have,

$\begin{array}{rcl}\widehat{\mathrm{Ï„}}& =& \left|\begin{array}{ccc}i& j& k\\ 0& 0.5& -2\\ 2& 0& -3\end{array}\right|\\ & =& \left\{\left(-1.5\right)\hat{i}+\left(-4\right)\hat{j}+\left(-1\right)\hat{k}\right\}Nm\end{array}$

Now,

$\begin{array}{rcl}\stackrel{â‡\P Ä}{r}\text{'}& =& \stackrel{â‡\P Ä}{r}-{\stackrel{â‡\P Ä}{r}}_0\\ & =& \left(-2.0\hat{i}+0.5\hat{j}+1\hat{k}\right)m\\ \stackrel{â‡\P Ä}{F}& =& 2N\hat{i}+0\hat{j}-3N\hat{k}\\ \stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \left|\begin{array}{ccc}i& j& k\\ -2& 0.5& 1\\ 2& 0& -3\end{array}\right|\\ & =& \left(-1.5\right)Nm\hat{i}+\left(-6+2\right)Nm\hat{j}+\left(-2×0-2×0.5\right)Nm\hat{k}\\ & =& \left\{\left(-1.5\right)\hat{i}+\left(-4\right)\hat{j}+\left(-1\right)\hat{k}\right\}Nm\end{array}$