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.

#### Solution:

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

- Locate the points A and B in Cartesian form.
- Express a position vector from A to B.
- Find the magnitude of this position vector
- Express the unit vector for this position vector
- 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:(0.25i+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\{-0.25i-0.067j+0.5k\right\}

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

The unit vector is then:

Now, we can express the force in Cartesian form.

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,

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

The position vector is:

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_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}

#### Final Answer: