# Strut AB of the 1-m-diameter hatch door

Strut AB of the 1-m-diameter hatch door exerts a force of 450 N on point B. Determine the moment of this force about point O.

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

#### Solution:

Let us first express the 450 N force in Cartesian vector form. To do so, we will follow the following steps.

1. Locate the points A and B in Cartesian form.
2. Express a position vector from A to B.
3. Find the magnitude of this position vector
4. Express the unit vector for this position vector
5. Multiply the magnitude of the force by the unit vector

These steps are almost always the same when expressing a force in Cartesian form. For further reading or if you are unfamiliar with this, take a look at expressing forces in Cartesian form.

Using the image, let us write the locations of points A and B.

$A:((0.5\sin30^0)i+(0.5\cos30^0+0.5)j+0k)$

$A:(0.25i+0.933j+0k)$

$B:(0i+(1\cos30^0)j+(1\sin30^0)k)$

$B:(0i+0.866j+0.5k)$

We can now write our position vector from A to B.

$r_{AB}=\left\{(0-0.25)i+(0.866-0.933)j+(0.5-0)k\right\}$

$r_{AB}=\left\{-0.25i-0.067j+0.5k\right\}$

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 $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

Let us now calculate the magnitude of this position vector. We will use the magnitude to express a unit vector in the next step.

magnitude of $r_{AB}\,=\,\sqrt{(-0.25)^2+(-0.067)^2+(0.5)^2}=0.563$

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 is then:

$u_{AB}\,=\,\left(-\dfrac{0.25}{0.563}i-\dfrac{0.067}{0.563}j+\dfrac{0.5}{0.563}k\right)$

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, $u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}$

Now, we can express the force in Cartesian form.

$F=450\left(-\dfrac{0.25}{0.563}i-\dfrac{0.067}{0.563}j+\dfrac{0.5}{0.563}k\right)$

$F=\left\{-200i-53.5j+400k\right\}$

To calculate the moment at O, we need to now express a position vector from O to A. O is the location where we are calculating the moment, and A is the location where the force is being applied from.

From earlier, we already know the location of point A. O is the origin, so it’s at $\left\{oi+0j+0k\right\}$. Thus,

$A:(0.25i+0.933j+0k)$

$O:(0i+0j+0k)$

The position vector is:

$r_{OA}=\left\{(0.25-0)i+(0.933-0)j+(0-0)k\right\}$

$r_{OA}=\left\{0.25i+0.933j+0k\right\}$

To calculate the moment, we must take the cross product between the position vector and the force. Thus, we have:

$M_O=r_{OA}\times F$

$M_A=\begin{bmatrix}\bold i&\bold j&\bold k\\0.25&0.933&0\\-200&-53.5&400\end{bmatrix}$

$M_A=\left\{373.2i-100j+173.2k\right\}\,\text{N}\cdot\text{m}$

$M_A=\left\{373.2i-100j+173.2k\right\}\,\text{N}\cdot\text{m}$