Question

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

Answer 1

Hint: In this question, we need to find the value of \[x\]. As we know that, the torque is the cross product of radius and force. So, we will find the radius vector first using the points \[\left( {{\mathbf{4i}}{\text{ }} + {\text{ }}{\mathbf{3j}}{\text{ }}-{\text{ }}{\mathbf{k}}} \right)\] and \[\left( {{\mathbf{i}}{\text{ }} + {\text{ }}{\mathbf{2j}}{\text{ }} + {\text{ }}{\mathbf{k}}} \right)\]. After, we will take the cross product and will find the magnitude of the resultant vector.

Formula used: The formula for torque in the form of vector notation is given below.
\[\tau = r \times F\]
Here, \[\tau \] is the torque, \[r\] is the radius and \[F\] is the force.
Now, the cross of product of any two vectors such as \[A\left( {{a_1},{a_2},{a_3}} \right)\] and \[B\left( {{b_1},{b_2},{b_3}} \right)\]
So, \[A \times B = \left| {\begin{array}{*{20}{c}} i&j&k \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right| = i\left( {{a_2}{b_3} - {a_3}{b_2}} \right) - j\left( {{a_1}{b_3} - {a_3}{b_1}} \right) + k\left( {{a_1}{b_2} - {a_2}{b_1}} \right)\]
Also, the magnitude of vector \[\left( {m{\mathbf{i}}{\text{ }} + {\text{ n}}{\mathbf{j}}{\text{ }} + {\text{ q}}{\mathbf{k}}} \right)\] is given below.
\[\left| {\left( {m{\mathbf{i}}{\text{ }} + {\text{ n}}{\mathbf{j}}{\text{ }} + {\text{ q}}{\mathbf{k}}} \right)} \right| = \sqrt {{m^2} + {n^2} + {q^2}} \]

Complete step by step solution:
We know that \[\bar \tau = \bar r \times \bar F\]
Let us find \[\bar r\].
So, \[\bar r = \left( {{\mathbf{4i}}{\text{ }} + {\text{ }}{\mathbf{3j}}{\text{ }}-{\text{ }}{\mathbf{k}}} \right) - \left( {{\mathbf{i}}{\text{ }} + {\text{ }}{\mathbf{2j}}{\text{ }} + {\text{ }}{\mathbf{k}}} \right)\]
By simplifying, we get
\[\bar r = \left( {{\mathbf{4i}}{\text{ }} + {\text{ }}{\mathbf{3j}}{\text{ }}-{\text{ }}{\mathbf{k}}} \right) - i - 2j - k\]
\[\bar r = \left( {{\mathbf{4i}}{\text{ - i}} + {\text{ }}{\mathbf{3j}} - 2j{\text{ }}-{\text{ }}{\mathbf{k}} - k} \right)\]
\[\bar r = \left( {3{\mathbf{i}}{\text{ }} + {\text{ }}j{\text{ - 2}}{\mathbf{k}}} \right)\]
Also, we know that \[{\mathbf{F}}{\text{ }} = {\text{ }}\left( {{\mathbf{i}}{\text{ }} + {\text{ }}{\mathbf{2j}}{\text{ }} + {\text{ }}{\mathbf{3k}}} \right)\]
Now, we will find the torque vector.
\[\tau = r \times F\]
\[r \times F = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  3&1&{ - 2} \\
  1&2&3
\end{array}} \right| = i\left( {1\left( 3 \right) - \left( { - 2} \right)2} \right) - j\left( {3\left( 3 \right) - \left( { - 2} \right)1} \right) + k\left( {3\left( 2 \right) - 1\left( 1 \right)} \right)\]
\[r \times F = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  3&1&{ - 2} \\
  1&2&3
\end{array}} \right| = i\left( {3 + 4} \right) - j\left( {9 + 2} \right) + k\left( {6 - 1} \right)\]
By simplifying further, we get
\[r \times F = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  3&1&{ - 2} \\
  1&2&3
\end{array}} \right| = i\left( 7 \right) - j\left( {11} \right) + k\left( 5 \right)\]
The magnitude of the torque vector is \[\left( 7 \right)i - \left( {11} \right)j + \left( 5 \right)k\].

Now, we will find its magnitude.Thus, we get
\[\left| \tau \right| = \left| {\sqrt {{{\left( 7 \right)}^2} + 11{{\left( 1 \right)}^{^2}} + {{\left( 5 \right)}^2}} } \right|\]
Thus, the magnitude of a vector is \[\left| \tau \right| = \left| {\sqrt {49 + 121 + 25} } \right|\]
\[\left| \tau \right| = \left| {195} \right|\]
That is \[\left| \tau \right| = 195\]
Hence, the magnitude of torque is 195.

Therefore, the value of X is 195.

Note: Here, students generally make mistakes in calculating the radius vector using vectors \[\left( {{\mathbf{4i}}{\text{ }} + {\text{ }}{\mathbf{3j}}{\text{ }}-{\text{ }}{\mathbf{k}}} \right)\] and \[\left( {{\mathbf{i}}{\text{ }} + {\text{ }}{\mathbf{2j}}{\text{ }} + {\text{ }}{\mathbf{k}}} \right)\]. Also, they may confuse with signs while calculating the cross product of the vectors r and F. So, ultimately the end result depends on the cross product of the vectors r and F. If it is wrong then we will get the false magnitude of the resultant vector.