If vec{F} = (5 hat{i} - 10 hat{j}) text{ N} a | vedclass

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

Question

(B) The torque $\vec{	au}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:$\vec{	au} = \vec{r} 	imes

Vedclass · Accepted Answer

$-25 \hat{k} 	ext{ N m}$

Vedclass · Answer

(B) The torque $\vec{	au}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:$\vec{	au} = \vec{r} 	imes \vec{F}$Given $\vec{r} = (4 \hat{i} - 3 \hat{j})$ and $\vec{F} = (5 \hat{i} - 10 \hat{j})$:$\vec{	au} = (4 \hat{i} - 3 \hat{j}) 	imes (5 \hat{i} - 10 \hat{j})$Using the distributive property of the cross product:$\vec{	au} = 4 \hat{i} 	imes (5 \hat{i}) - 4 \hat{i} 	imes (10 \hat{j}) - 3 \hat{j} 	imes (5 \hat{i}) + 3 \hat{j} 	imes (10 \hat{j})$Since $\hat{i} 	imes \hat{i} = 0$ and $\hat{j} 	imes \hat{j} = 0$,and knowing $\hat{i} 	imes \hat{j} = \hat{k}$ and $\hat{j} 	imes \hat{i} = -\hat{k}$:$\vec{	au} = 0 - 40(\hat{k}) - 15(-\hat{k}) + 0$$\vec{	au} = -40 \hat{k} + 15 \hat{k}$$\vec{	au} = -25 \hat{k} 	ext{ N m}$

If $\vec{F} = (5 \hat{i} - 10 \hat{j}) \text{ N}$ and $\vec{r} = (4 \hat{i} - 3 \hat{j}) \text{ m}$,then the torque $\vec{\tau}$ acting on the object will be:

Similar Questions

$A$ force $\vec{F} = 4\hat{i} - 5\hat{j} + 3\hat{k}$ is applied at a point with position vector $\vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$. The torque about the point with position vector $\vec{r_2} = 3\hat{i} - 2\hat{j} - 3\hat{k}$ is:

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

$A$ body of mass $10 \, kg$ is hanging at rest with the help of a rope wrapped around a cylinder of diameter $0.2 \, m$. The torque applied about the horizontal axis of the cylinder is ....... $N-m$.

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:

$A$ rod of length $l$ is acted upon by a couple as shown in the figure. The moment of the couple is $\tau \text{ Nm}$. If the force at each end of the rod is $F$,then the magnitude of each force is (given $\sin 30^{\circ} = \cos 60^{\circ} = 0.5$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module