Question

A force F=-2i+j+k acts at point A(1,5,1) on a solid which is free to turn about a fixed point Q(-1,2,1). Find the torque exerted by the force on the body.

Answer 1

The torque exerted by the force on the body is $\vec \tau = 3\,\hat{i}-2\,\hat{j} + 8\,\hat{k}$.

Vectorially speaking, the torque exerted ($\vec \tau$) by the force on the body ($\vec F$) is resulted from this expression:

$\vec \tau = \vec r_{QA} \times \vec F$ (1)

Where:

• $\vec r_{QA}$ - Vector distance between points A and Q.
• $\vec F$ - Vector force exerted at point A.

If we know that $\vec r_{QA} = (2, 3, 0)$ and $\vec F = (-2, 1, 1)$, then the torque is:

$\vec \tau = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\2&3&0\\-2&1&1\end{array}\right|$

$\vec \tau = (3, -2, 8)$

The torque exerted by the force on the body is $\vec \tau = 3\,\hat{i}-2\,\hat{j} + 8\,\hat{k}$.

To learn more on torque, we kindly invite to check this verified question: brainly.com/question/3388202