Find the torque of a force vec{F} = -3hat{i} | 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 $\vec{F}$:$\vec{	au} = \vec{r} 	imes \vec{F}$Given $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$,we calculate the determinant:$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 7 & 3 & 1 \ -3 & 1 & 5 \end{vmatrix}$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}$

Find the torque of a force $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acting at the point $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$.

Similar Questions

The center of an equilateral triangle $ABC$ is $O$. Three forces $F_1$,$F_2$,and $F_3$ are applied along the sides $AB$,$BC$,and $AC$ respectively. If the total torque about $O$ is zero,what is the relationship between $F_1$,$F_2$,and $F_3$?

$A$ force of $-P \hat{k}$ acts on the origin of the coordinate system. The torque about the point $(2, -3)$ is $P(a \hat{i} + b \hat{j})$. If the ratio $\frac{a}{b} = \frac{x}{2}$,find the value of $x$.

Find the torque of a force $\vec F = -3\hat i + \hat j + 5\hat k$ acting at the point $\vec r = 7\hat i + 3\hat j + \hat k$ with respect to the origin.

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:

Which factors should be considered for the calculation of torque,and which should not?

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