The coordinates of a particle with respect to origin in a given reference frame is $(1,1,1)$ meters. If a force of $\overrightarrow{\mathrm{F}}=\hat{i}-\hat{j}+\hat{k}$ acts on the particle, then the magnitude of torque (with respect to origin) in z-direction is ______.

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Aicharya · Accepted Answer

The torque $\vec{\tau}$ acting on the particle with respect to the origin can be calculated using the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$ : $\vec{\tau}=\vec{r} \times \vec{F}$ Given the position vector $\vec{r}=(1,1,1) \mathrm{m}$ and the force vector $\vec{F}=\hat{i}-\hat{j}+\hat{k}$, we need to calculate the cross product: $\vec{\tau}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & -1 & 1 \end{array}\right|$ Calculating the determinant, we have: $\vec{\tau}=\hat{i}(1 \cdot 1-1 \cdot(-1))-\hat{j}(1 \cdot 1-1 \cdot 1)+\hat{k}(1 \cdot(-1)-1 \cdot 1)$ This simplifies to: $\begin{aligned} \vec{\tau} & =\hat{i}(1+1)-\hat{j}(1-1)+\hat{k}(-1-1) \ \vec{\tau} & =2 \hat{i}-0 \hat{j}-2 \hat{k} \end{aligned}$ The torque vector is $\vec{\tau}=2 \hat{i}-2 \hat{k}$. To find the magnitude of the torque in the $z$-direction, we look at the $\hat{k}$ component: $\tau_z=-2$ The magnitude of torque in the $z$-direction is: $\left|\tau_z\right|=2 \mathrm{Nm}$ Thus, the magnitude of the torque in the $z$-direction is 2 Newton-meters.

The coordinates of a particle with respect to origin in a given reference frame is \((1,1,1)\) meters. If a…

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