Question

Given that $ \vec r = 2\hat i + 3\hat j $ and $ \vec F = 2\hat i + 6\hat k $ . Find the magnitude of the torque.

Answer 1

Hint :In order to solve this question, we need to use the formula for the torque experienced by a body. As the value for the force applied and the radial distance from the center to the direction of the force is given, we can directly compute the torque by finding the cross product and thereby calculating the magnitude of the torque vector.
The torque experienced by a body, when a force $ \vec F $ is applied and the radial distance $ \vec r $ is given by
 $ \tau = \vec r \times \vec F $
The magnitude of vector $ x\hat i + y\hat j + z\hat k $ is
 $ \sqrt {{x^2} + {y^2} + {z^2}} $

Complete Step By Step Answer:
The torque experienced by a body is equal to the cross product of the radial distance vector of the body from the center of the force and the force applied on it. Mathematically, it can be written in the form:
 $ \tau = \vec r \times \vec F $
Here the force and the radial distance vector are given as
 $ \vec F = 2\hat i + 6\hat k \\
  \vec r = 2\hat i + 3\hat j \\ $
Therefore, the torque produced can be calculated from the cross product as
 $ \tau = \left( {2\hat i + 3\hat j} \right) \times \left( {2\hat i + 6\hat k} \right) $
Calculating the vector product of the two vectors, we get
 $ \left( {2\hat i + 3\hat j} \right) \times \left( {2\hat i + 6\hat k} \right) = \hat i\left( {3 \times 6 - 0 \times 0} \right) - \hat j\left( {2 \times 6 - 2 \times 0} \right) + \hat k\left( {2 \times 0 - 2 \times 3} \right) \\
   \Rightarrow \left( {2\hat i + 3\hat j} \right) \times \left( {2\hat i + 6\hat k} \right) = 18\hat i - 12\hat j - 6\hat k \\ $
Therefore, the torque experienced by the body is
 $ \vec \tau = 18\hat i - 12\hat j - 6\hat k $
Now the magnitude of the torque is
 $ \left| {\vec \tau } \right| = \sqrt {{{\left( {18} \right)}^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 6} \right)}^2}} \\
  \left| {\vec \tau } \right| = \sqrt {504} \\ $
Thus, the torque is equal to $ \sqrt {504} Nm $

Note :
It is important to note that the torque of a body is equal to the cross product of the radial distance of the body from the center with force applied on the body and not the other way round. If the cross product $ \vec F \times \vec r $ is taken instead of $ \vec r \times \vec F $ then, the whole answer is changed and not even the same magnitude of the torque is obtained.