Question

please answer this question?????​

Answer 1

To find:

• Moment of the force, $\sf{\vec{F}=4\hat{i}+5\hat{j}-6\hat{k}}$ at point $\sf{(2,0,-3)}$, about the point $\sf{(2,-2,-2)}$

Solution:

• In three dimensional motion the torque (moment of force) becomes a pseudovector, And is given by the cross product of displacement vector and force vector.

Here,

position vector at the point (2,0,-3), $\sf{\vec{r_1}=2\hat{i}+0\hat{j}-3\hat{k}}$

is about the position vector at point (2,-2,-2)

$\sf{\vec{r_2}=2\hat{i}-2\hat{j}-2\hat{k}}$

so,

change in position vector, $\sf{\vec{r}}$would be

$\implies\sf{\vec{r}=\vec{r_1}-\vec{r_2}}$

$\implies\sf{\vec{r}=2\hat{i}+0\hat{j}-3\hat{k}-(2\hat{i}-2\hat{j}-2\hat{k})}$

$\implies\sf{\vec{r}=2\hat{i}+0\hat{j}-3\hat{k}-2\hat{i}+2\hat{j}+2\hat{k}}$

$\implies\underline{\underline{\large{\sf{\vec{r}=2\hat{j}-\hat{k}}}}}$

Now,

force vector is given as

$\sf{\vec{F}=4\hat{i}=5\hat{j}-6\hat{k}}$

Now,

Since torque ($\sf{\tau}$) is given by the cross product of Force and displacement vector therefore,

$\implies\sf{\tau=\vec{F}\times \vec{r} = (4\hat{i}+5\hat{j}-6\hat{k})\times (2\hat{j}-\hat{k}) }$

$\implies\sf{\tau=\vec{F}\times \vec{r} = \Big[\begin{array}{ccc}\sf{\hat{i}}&\sf{\hat{j}}&\sf{\hat{k}}\\0&2&-1\\4&5&-6\end{array}\Big] }$

$\implies\sf{\tau=\vec{F}\times \vec{r} =}$ $\sf{\hat{i}[(2)(-6)-(-1)(5)]}$ + $\sf{\hat{j}[(-1)(4)-(-6)(0)]}$ + $\sf{\hat{k}[(0)(5)-(2)(4)]}$

$\implies\boxed{\boxed{\large{\sf{\tau=\vec{F}\times \vec{r} = -7\hat{i} -4\hat{j}-8\hat{k}}}}}$

Therefore,

• Moment of force in this case would be -7$\bf{\hat{i}}$ -4$\bf{\hat{j}}$ -8$\bf{\hat{k}}$

Hence,

OPTION (3) is correct.

Answer 2

Explanation:

given

• $\sf \: \vec F=4\hat i+5\hat j-6\hat k$

To Find : :

The force is acting at the point (2, 0,-3).

The point under consideration is (2,-2, -2). The moment of force is calculated around this point.

Hence, the position vector of (2,0,-3) with respect to (2,-2,-2) will be -

$\sf (2-2)\hat i+(0-[-2])\hat j+([-3]-[-2])\hat k$

$\sf \: =\ 0\hat i+2\hat j-1\hat k$

The moment of the force is calculated as -

$\sf \: \vec\tau\ =\ \vec r\times \vec F$

$\sf \: = (2\times [-6]-[-1]\times 5)\hat i-(0\times [-6]-[-1]\times 4)\hat j+(0\times 5-4\times 2)\hat k$

$\sf \: = \ -7\hat i-4\hat j-8\hat k$

Hence option 3 is correct

$\sf \: 7\hat i-4\hat j-8\hat k$