Question

Find the moment of the force $5\hat i + 10\hat j + 16\hat k$ acting at the point $2\hat i - 7\hat j + 10\hat k$ about the point $5\hat i + 6\hat j - 10\hat k$.
A) $408\hat i + 12\hat j - 135\hat k$
B) $ - 408\hat i + 148\hat j + 35\hat k$
C) $ - 31\hat i - 191\hat j + 5\hat k$
D) $31\hat i + 191\hat j - 5\hat k$

Answer 1

Hint: The moment is also known as torque. Apply the vector formula of torque $\vec \tau = \vec r \times \vec F$, where $\vec r$ is the radius vector (from the point about which torque is calculated to the point of application of force) and $\vec F$ is the applied force. Put the values and perform the vector cross multiplication.

Complete step by step solution:
According to the question, a force $5\hat i + 10\hat j + 16\hat k$ is applied at point $2\hat i - 7\hat j + 10\hat k$. We have to determine the moment of this force about point $5\hat i + 6\hat j - 10\hat k$.
We know that the moment of force is represented by a special physical quantity called torque. It is a vector quantity and the vector representation of torque is shown as:
$ \Rightarrow \vec \tau = \vec r \times \vec F$
Thus torque vector is equal to the cross product of radius vector and force vector, where radius vector is from the axis of rotation to the point of application of force.
So for the above question, we basically have to calculate the torque acting about point $5\hat i + 6\hat j - 10\hat k$ (axis of rotation passes through this line) due to force $5\hat i + 10\hat j + 16\hat k$ applied at point $2\hat i - 7\hat j + 10\hat k$.


In the above figure, a force $\vec F$ is acting at a point with position vector ${\vec r_2}$ and torque is to be calculated about a point with position vector ${\vec r_1}$.
As per question’s condition, we have:
$
   \Rightarrow \vec F = 5\hat i + 10\hat j + 16\hat k \\
   \Rightarrow {{\vec r}_1} = 5\hat i + 6\hat j - 10\hat k \\
   \Rightarrow {{\vec r}_2} = 2\hat i - 7\hat j + 10\hat k \\
 $
Radius vector can be calculated as:
$
   \Rightarrow \vec r = {{\vec r}_2} - {{\vec r}_1} \\
   \Rightarrow \vec r = 2\hat i - 7\hat j + 10\hat k - \left( {5\hat i + 6\hat j - 10\hat k} \right) \\
   \Rightarrow \vec r = - 3\hat i - 13\hat j + 20\hat k \\
 $
Applying the formula for torque, we have:
$
   \Rightarrow \vec \tau = \vec r \times \vec F \\
   \Rightarrow \vec \tau = \left( { - 3\hat i - 13\hat j + 20\hat k} \right) \times \left( {5\hat i + 10\hat j + 16\hat k} \right) \\
 $
We know that the cross product of two vectors can be calculated using determinants. Doing so, we’ll get:
$ \Rightarrow \vec \tau = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  { - 3}&{ - 13}&{20} \\
  5&{10}&{16}
\end{array}} \right|$
By expanding this determinant, we’ll get:
$
   \Rightarrow \vec \tau = \hat i\left( { - 208 - 200} \right) - \hat j\left( { - 48 - 100} \right) + \hat k\left( { - 30 + 65} \right) \\
   \Rightarrow \vec \tau = - 408\hat i + 148\hat j + 35\hat k \\
 $
Therefore the required moment of force or torque is $ - 408\hat i + 148\hat j + 35\hat k$. (B) is the correct option.

Note: Moment of force or torque can be zero in three different cases:
1. $\vec F = 0$ : No force is applied at the point of application.
2. $\vec r = 0$ : Force is applied at the same point about which torque is to be calculated.
3. $\vec r$ is parallel to $\vec F$ : The direction of the radius vector is the same as that of applied force i.e. they are parallel.