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

#### Solution:

This question involves expressing forces in Cartesian vector form. If you are unsure on how to do this, read the detailed guide on expressing forces in Cartesian notation.

Show me the final answer↓

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

- Express force F_B in Cartesian vector form
- 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)* - 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. (Don’t remember?)

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

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

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

#### Final Answer:

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

why we use Roa now ao ?

You can also use a position vector from O to B, both will yield the same answer. Please watch: https://www.youtube.com/watch?v=QNNnPZ68STI last example.

and why the answer of MA is negative? when i calculate it gives me positive 1800

You’re doing the calculations wrong. Please watch https://www.youtube.com/watch?v=F8IHrg3pc7g to see how it’s done 🙂