Let vec{F} be a force acting on a particle ha | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The torque $\vec{	au}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{	au} = \ve

Vedclass · Accepted Answer

$\vec{r} \cdot \vec{	au} = 0$ and $\vec{F} \cdot \vec{	au} = 0$

Vedclass · Answer

(D) The torque $\vec{	au}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{	au} = \vec{r} 	imes \vec{F}$.By the definition of the cross product,the resulting vector $\vec{	au}$ is perpendicular to both the position vector $\vec{r}$ and the force vector $\vec{F}$.Since $\vec{	au}$ is perpendicular to $\vec{r}$,the dot product $\vec{r} \cdot \vec{	au} = |\vec{r}| |\vec{	au}| \cos(90^{\circ}) = 0$.Similarly,since $\vec{	au}$ is perpendicular to $\vec{F}$,the dot product $\vec{F} \cdot \vec{	au} = |\vec{F}| |\vec{	au}| \cos(90^{\circ}) = 0$.Therefore,both dot products are equal to zero.

Let $\vec{F}$ be a force acting on a particle having position vector $\vec{r}$,and $\vec{\tau}$ be the torque of this force about the origin. Then:

Similar Questions

$A$ force of $3\hat{i} + 2\hat{j} - \hat{k} \text{ N}$ acts on a particle with position vector $\hat{i} + \hat{j} - \hat{k} \text{ m}$. The magnitude of torque of the given force is

$A$ force $\vec{F} = 4\hat{i} - 3\hat{j} + 4\hat{k} \, N$ acts on a particle at a position $\vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k}$ from the origin. The torque acting on the particle is:

The resultant of the system in the figure is a force of $8 \ N$ parallel to the given forces,acting through point $R$. The value of $PR$ is equal to . . . . . .

Find the torque of a force $\vec{F} = 7 \hat{i} + 3 \hat{j} - 5 \hat{k}$ about the origin. The force acts on a particle whose position vector is $\vec{r} = \hat{i} - \hat{j} + \hat{k}$.

The figure shows three forces $\overrightarrow{F}_{1}, \overrightarrow{F}_{2}$ and $\overrightarrow{F}_{3}$ acting along the sides of an equilateral triangle. If the total torque acting at point $O$ (the center of the triangle) is zero,then the magnitude of $\overrightarrow{F}_{3}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module