Question

A force F=3i+j-4k is applied at a point (3,4,-2).find its torque about the point (-1,2,4).

Answer 1

hence, torque about the point (-1,2,4) is -2i - 2j - 2k

A force F = 3i + j - 4k is applied at a point (3, 4, -2) and we have to find torque about the point (-1,2,4).

first find position vector, r = (3, 4, -2) - (-1, 2, 4) = (4, 2, -6) = 4i + 2j - 6k

now torque = r × F

= (4i + 2j - 6k) × (3i + j - 4k)

= i{2 × (-4)- (1)(-6)} - j{4(-4) - 3(-6)} + k{4(1)-3(2)}

= i(-8 + 6) - j(-16 + 18) + k(4 - 6)

= -2i - 2j - 2k

hence, torque about the point (-1,2,4) is -2i - 2j - 2k and magnitude of torque = 4√3 Nm

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Answer 2

$\huge\bold\purple{Answer:-}$

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}