91Ó°ÊÓ

Location of jar (3.0m, -2.0m, 4.0m)

$$\begin{gathered} {\stackrel{â‡\P Ä}{F}}_1=\left(3\hat{i}-4\hat{j}+5\hat{k}\right)N \\ {\stackrel{â‡\P Ä}{F}}_2=\left(3\hat{i}-4\hat{j}-5\hat{k}\right)N \end{gathered}$$

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,$\stackrel{â‡\P Ä}{r}=x\hat{i}+y\hat{j}+z\hat{k}$

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,

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

$\begin{array}{rcl}\stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \stackrel{â‡\P Ä}{r}×\stackrel{â‡\P Ä}{F}\\ \stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \left|\begin{array}{ccc}i& j& k\\ 3& -2& 4\\ 3& -4& 5\end{array}\right|\\ & =& \left(-10+16\right)Nm\hat{i}+\left(12-15\right)Nm\hat{j}+\left(-12+6\right)Nm\hat{k}\\ & =& \left\{\left(6\right)\hat{i}+\left(-3\right)\hat{j}+\left(-6\right)\right\}\hat{k}Nm\end{array}$

$\begin{array}{rcl}\stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \stackrel{â‡\P Ä}{r}×\stackrel{â‡\P Ä}{F}\\ \stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \left|\begin{array}{ccc}i& j& k\\ 3& -2& 4\\ -3& -4& -5\end{array}\right|\\ & =& \left(10+16\right)Nm\hat{i}+\left(-12+15\right)Nm\hat{j}+\left(-12-6\right)Nm\hat{k}\\ & =& \left\{\left(26\right)\hat{i}+\left(3\right)\hat{j}+\left(18\right)\hat{k}\right\}Nm\end{array}$

$\begin{array}{rcl}{\stackrel{â‡\P Ä}{F}}_1+{\stackrel{â‡\P Ä}{F}}_2& =& -8\hat{j}N\\ \stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \stackrel{â‡\P Ä}{r}×\left({\stackrel{â‡\P Ä}{F}}_1+{\stackrel{â‡\P Ä}{F}}_2\right)\\ & =& \left|\begin{array}{ccc}i& j& k\\ 3& -2& 4\\ 0& -8& 0\end{array}\right|\\ & =& \left(32\right)\hat{i}+\left(0\right)\hat{j}+\left(-24+0\right)\hat{k}\\ & =& \left\{\left(32\right)\hat{i}+\left(-24\right)\hat{k}\right\}Nm\end{array}$

So,$\stackrel{â‡\P Ä}{r}=\stackrel{â‡\P Ä}{r}-\stackrel{â‡\P Ä}{r_0}$

$\stackrel{â‡\P Ä}{r}\text{'}=0\hat{i}-4\hat{j}m+0\hat{k}$

${\stackrel{â‡\P Ä}{F}}_1+{\stackrel{â‡\P Ä}{F}}_2=-8\hat{j}N$

Torque,

$\begin{array}{rcl}\stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \stackrel{â‡\P Ä}{r}×\left({\stackrel{â‡\P Ä}{F}}_1+{\stackrel{â‡\P Ä}{F}}_2\right)\end{array}$

$\begin{array}{rcl}\stackrel{â‡\P Ä}{\mathrm{Ï„}}& =& \stackrel{â‡\P Ä}{r}×\left({\stackrel{â‡\P Ä}{F}}_1+{\stackrel{â‡\P Ä}{F}}_2\right)\\ & =& \left|\begin{array}{ccc}i& j& k\\ 0& -4& 0\\ 0& -8& 0\end{array}\right|\\ & =& \left(0\right)\hat{i}Nm+\left(0\right)\hat{j}Nm+\left(0\right)\hat{k}Nm\\ & =& 0Nm\end{array}$