
The moment of the force, vector \[F=4\widehat{i}+5\widehat{j}-6\widehat{k}\]at (2,0,-3) about the point (2,-2,-2) is given by
A-\[-8\widehat{i}-4\widehat{j}-7\widehat{k}\]
B-\[-4\widehat{i}-\widehat{j}-8\widehat{k}\]
C-\[-7\widehat{i}-8\widehat{j}-4\widehat{k}\]
D-\[-7\widehat{i}-4\widehat{j}-8\widehat{k}\]
Answer
578.7k+ views
Hint: Here we need to find the moment of force at a point. First of all, let us define what is a moment of force. The moment of force is a measure of the ability of the force acting on a particular body to rotate it around a given axis.
Complete step by step answer:
We are given the coordinates of two points, say A (2,0,-3) and point B (2,-2,-2). In order to find the moment of force about point B, we need to have the distance between the point of application of force and the perpendicular distance between this and the axis of rotation. So we need to find first the displacement vector, \[\overrightarrow{r}\]. After this we know the moment of force commonly called torque is given by the cross product of the force and the displacement vector. So, we make use of matrix algebra to find out the value of the cross product.
The force is \[F=4\widehat{i}+5\widehat{j}-6\widehat{k}\]and the distance vector is given by,
\[\begin{align}
& \overrightarrow{r}=(2-2)\widehat{i}+(0+2)\widehat{j}+(-3+2)\widehat{k} \\
& =2\widehat{j}-\overrightarrow{k} \\
\end{align}\]
So, torque is given by the formula, Torque = \[\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}\]
\[\begin{align}
& \overrightarrow{\tau }=(2\widehat{j}-\widehat{k})\times (4\widehat{i}+5\widehat{j}-6\widehat{k}) \\
& =\left( \begin{matrix}
\widehat{i} & \widehat{j} & \widehat{k} \\
0 & 2 & -1 \\
4 & 5 & -6 \\
\end{matrix} \right) \\
& =-7\widehat{i}-4\widehat{j}-8\widehat{k} \\
\end{align}\]
So, the correct option is (D)
Note:While calculating the torque we have to keep in our mind that it is a vector quantity and is given by the cross product of force acting on the body and the perpendicular distance from the axis of rotation. So, we have to consider the angle between the two vectors
Complete step by step answer:
We are given the coordinates of two points, say A (2,0,-3) and point B (2,-2,-2). In order to find the moment of force about point B, we need to have the distance between the point of application of force and the perpendicular distance between this and the axis of rotation. So we need to find first the displacement vector, \[\overrightarrow{r}\]. After this we know the moment of force commonly called torque is given by the cross product of the force and the displacement vector. So, we make use of matrix algebra to find out the value of the cross product.
The force is \[F=4\widehat{i}+5\widehat{j}-6\widehat{k}\]and the distance vector is given by,
\[\begin{align}
& \overrightarrow{r}=(2-2)\widehat{i}+(0+2)\widehat{j}+(-3+2)\widehat{k} \\
& =2\widehat{j}-\overrightarrow{k} \\
\end{align}\]
So, torque is given by the formula, Torque = \[\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}\]
\[\begin{align}
& \overrightarrow{\tau }=(2\widehat{j}-\widehat{k})\times (4\widehat{i}+5\widehat{j}-6\widehat{k}) \\
& =\left( \begin{matrix}
\widehat{i} & \widehat{j} & \widehat{k} \\
0 & 2 & -1 \\
4 & 5 & -6 \\
\end{matrix} \right) \\
& =-7\widehat{i}-4\widehat{j}-8\widehat{k} \\
\end{align}\]
So, the correct option is (D)
Note:While calculating the torque we have to keep in our mind that it is a vector quantity and is given by the cross product of force acting on the body and the perpendicular distance from the axis of rotation. So, we have to consider the angle between the two vectors
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

