Question

A force $F=(i+2j+3k)N$ acts at a point $(4i+3j–k)m$. Then the magnitude of torque about the point $(i+2j+k)m$ will be$\sqrt{x}N–m$. The value of $x$ is

Answer 1

Step 1: Given Data

Force, $F=(i+2j+3k)N$

Torque, $\tau =\sqrt{x}N-m$

Step 2: Find the radius

$$\begin{gathered} r=(4i-i)+(3j-2j)+(-k-k) \\ r=(3i+j-2k)m \end{gathered}$$

Step 3: Find the magnitude of the Torque

$\begin{array}{rcl}\tau & =& r\times F\\ \tau & =& (3i+j–2k)\times (i+2j+3k)\\ \tau & =& \left|\begin{array}{ccc}i& j& k\\ 3& 1& -2\\ 1& 2& 3\end{array}\right|\\ & =& i(3+4)–j(9+2)+k(6–1)\\ & =& 7i–11j+5k\\ \left|\tau \right|& =& \sqrt{7^2+{11}^2+5^2}\\ & =& \sqrt{195}\end{array}$

Step 4: Find the value of $x$

$$\begin{gathered} \left|\tau \right|=\sqrt{x}N–m \\ \sqrt{x}=\sqrt{195} \\ \Rightarrow x=195 \end{gathered}$$

Hence, The value of $x$ is $195$.