Determine the moment produced by force 4

Determine the moment produced by force $F_B$ about point O. Express the result as a Cartesian vector.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

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To find the moment, we will have to do the following:

1. Express force $F_B$ in Cartesian vector form
2. Write the position vector from O to A (you can also use a position vector from O to B, both will yield the same answer)
3. Take the cross product of $r_{OA}$ and $\vec F_B$

To express force $F_B$ in Cartesian vector form, we first need to express a position vector from A to B.

$r_{AB}\,=\,\left\{(0-0)i+(2.5-0)j+(0-6)k\right\}$

$r_{AB}=\left\{0i+2.5j-6k\right\}$

A position vector, denoted $\mathbf{r}$ is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was $(x_A,y_A,z_A)$ and the coordinates of point B was$(x_B,y_B,z_B)$, then $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

The magnitude of this position vector is:

magnitude of $r_{AB}\,=\,\sqrt{(0)^2+(2.5)^2+(-6)^2}=6.5$

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was $r\,=\,ai+bj+ck$, then the magnitude would be, $r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}$. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

The unit vector of this position vector is:

$u_{AB}\,=\,\left(0i+\dfrac{2.5}{6.5}j-\dfrac{6}{6.5}k\right)$

The unit vector is each corresponding unit of the position vector divided by the magnitude of the position vector. If the position vector was $r\,=\,ai+bj+ck$, then unit vector, $u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}$

We can now express the force in Cartesian form by multiplying the magnitude of the force by the unit vector.

$F_{AB}=780\left(0i+\dfrac{2.5}{6.5}j-\dfrac{6}{6.5}k\right)$

$F_{AB}=\left\{0i+300j-720k\right\}$ N

The next step is to write a position vector from O to A.

$r_{OA}=\left\{0i+0j+6k\right\}$

The last step is to take the cross product of the position vector and the force.

$M_A=\begin{bmatrix}\bold i&\bold j&\bold k\\0&0&6\\0&300&-720\end{bmatrix}$

$M_A=\left\{-1800i\right\}N\cdot m$

$M_A=\left\{-1800i\right\}N\cdot m$