Question

a force f=3i^ +2j^ -4k^ acts at the point (1,-1,2) find its torque about point (2,-1,3)

please help me out please ​

Answer 1

Answer: The torque is $\tau = -2\hat{i}+7\hat{j}+2\hat{k}$

Explanation:

Given that,

Force $F= 3\hat {i}+2\hat{j}-4\hat{k}$

First point $r_{1}= 1\hat{i}-1\hat{j}+2\hat{k}$

Second point $r_{2}= 2\hat{i}-1\hat{j}+3\hat{k}$

We know that,

The torque is the cross product of the position vector and the force.

$\tau = \vec{r}\times\vec{F}$

$\tau = (\vec{r}_{2}-\vec{r}_{1})\times F$

$\tau = (2\hat{i}-1\hat{j}+3\hat{k}-(1\hat{i}-1\hat{j}+2\hat{k}))\times3\hat {i}+2\hat{j}-4\hat{k}$

$\tau = (1\hat{i}+0\hat{j}+1\hat{k})\times 3\hat {i}+2\hat{j}-4\hat{k}$

$\tau = -2\hat{i}+7\hat{j}+2\hat{k}$

Hence, The torque is $\tau = -2\hat{i}+7\hat{j}+2\hat{k}$.

Answer 2

Answer:

The torque is (2i-7j-2k)

Explanation:

Given that,

Force $F= 3\hat{i}+2\hat{j}-4\hat{k}$

At point $r_{1}=1,-1,2$

About point $r_{2}=2,-1,3$

We need to calculate the displacement vector between the  points

Using formula of displacement

$r=r_{2}-r_{1}$

Put the value into the formula

$r=(2-1)i+(-1+1)j+(3-2)k$

$r=i+k$

We need to calculate the torque

Using formula of torque

$\tau=F\times r$

Put the value into the formula

$\tau=3\hat{i}+2\hat{j}-4\hat{k}\timesi+0j+k$

$\tau=2i-7j-2k$

Hence, The torque is (2i-7j-2k)