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 ​^ ​​ )

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?

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Maris 5 months 2021-09-01T03:10:49+00:00 1 Answers 0 views 0

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    2021-09-01T03:12:12+00:00

    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)

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