Determine the moment produced by force F about the diagonal AF of the rectangular block. Express the result as a Cartesian vector.

#### Solution:

Show me the final answer↓

We will use the unit vector AF and the position vector AB to find the moment produced by F about the diagonal AF. Note that you can also use a position vector from F to B (try it yourself).

Our position vector, r_{AB} is:

r_{AB}=\left\{(0-0)i+(3-0)j+(1.5-1.5)k\right\}

r_{AB}=\left\{0i+3j+0k\right\} m

r_{AB}=\left\{0i+3j+0k\right\} m

Let us now calculate the unit vector AF. To do so, we need to find a position vector from A to F.

r_{AF}=\left\{(3-0)i+(3-0)j+(0-1.5)k\right\}

r_{AF}=\left\{3i+3j-1.5k\right\}

r_{AF}=\left\{3i+3j-1.5k\right\}

The magnitude of this position vector is:

magnitude of r_{AF}\,=\,\sqrt{(3)^2+(3)^2+(-1.5)^2}

magnitude of r_{AF}\,=\,4.5

magnitude of r_{AF}\,=\,4.5

We can now calculate the unit vector :(don’t remember?)

u_{AF}\,=\,\left(\dfrac{3}{4.5}i+\dfrac{3}{4.5}j-\dfrac{1.5}{4.5}k\right)

Now that we have both a position vector and a unit vector, we can find the moment about the diagonal AF. This can be found by:

M_{AF}=u_{AF}\cdot r_{AB}\times F

M_{AF}=\begin{bmatrix}\dfrac{3}{4.5}&\dfrac{3}{4.5}&\dfrac{1.5}{4.5}\\0&3&0\\-6&3&10\end{bmatrix}

(Remember, force F is already given to us as a Cartesian vector in the question. Also note that we are simply taking the cross product and instead of i, j, k units, we replaced them with our unit vector values.)

M_{AF}=\begin{bmatrix}\dfrac{3}{4.5}&\dfrac{3}{4.5}&\dfrac{1.5}{4.5}\\0&3&0\\-6&3&10\end{bmatrix}

(Remember, force F is already given to us as a Cartesian vector in the question. Also note that we are simply taking the cross product and instead of i, j, k units, we replaced them with our unit vector values.)

M_{AF}=14 N•M

#### Final Answer:

M_{AF}=14 N•M