If force vec{F} = -3 hat{i} + hat{j} + 5 hat{ | vedclass

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Question

(A) 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

$(14 \hat{i} - 38 \hat{j} + 16 \hat{k})$

Vedclass · Answer

(A) 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}$$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 7 & 3 & 1 \ -3 & 1 & 5 \end{vmatrix}$Expanding the determinant:$\vec{	au} = \hat{i} (3 	imes 5 - 1 	imes 1) - \hat{j} (7 	imes 5 - 1 	imes (-3)) + \hat{k} (7 	imes 1 - 3 	imes (-3))$$\vec{	au} = \hat{i} (15 - 1) - \hat{j} (35 + 3) + \hat{k} (7 + 9)$$\vec{	au} = 14 \hat{i} - 38 \hat{j} + 16 \hat{k}$Thus,the correct option is $A$.

If force $\vec{F} = -3 \hat{i} + \hat{j} + 5 \hat{k}$ acts at a position vector $\vec{r} = 7 \hat{i} + 3 \hat{j} + \hat{k}$,then the torque $\vec{\tau}$ acting at that point is:

Similar Questions

Give the formula for the moment of a couple.

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

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:

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

The torque of a force $\vec{F} = 5 \hat{i} + 3 \hat{j} - 7 \hat{k}$ about the origin is $\vec{\tau}$. If the force acts on a particle whose position vector is $\vec{r} = 2 \hat{i} + 2 \hat{j} + \hat{k}$,then the value of $\vec{\tau}$ will be:

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