Question

The force of 20\hat{j}~\text{N}20 ​j ​^ ​​ N is applied at \vec{r} =(4.0\hat{i}-2.0\hat{j})~\text{m} ​r ​⃗ ​​ =(4.0 ​i ​^ ​​ −2.0 ​j ​^ ​​ ) m. What is the torque of this force about the origin?

Answer 1

Answer:

$-80\hat{k}\text{ N/m}$

Explanation:

Torque is given by

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

Note that this is a cross product of vectors. To perform a cross product, we need to express the vectors in three components.

$\vec{F} =0.0\hat{i} + 20\hat{j} + 0.0\hat{k}$

$\vec{r} =4.0\hat{i} - 2.0\hat{j} + 0.0\hat{k}$

The torque is then

$\tau = \left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 0.0 & 20 & 0\\ 4.0 & - 2.0 & 0.0\end{matrix}\right|$

$\tau = -80\hat{k}\text{ N/m}$

(I'm not sure if the matrix will appear well because I'm constrained with the device I'm using but what I'm showing is the determinant of a matrix with i, j, k in the first row, the coefficients of F in the second and the coefficients of r in the third)