# The curved rod lies in the x–y plane

The curved rod lies in the x–y plane and has a radius of 3 m. If a force F = 80 N of acts at its end as shown, 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 force being applied at the end of the rod in Cartesian vector form. To do so, we first need a position vector starting from the location of the force to the location where it is headed. The locations of points A and C in Cartesian form is:

$A:(3i+3j+0k)$

$C:(4i+0j-2k)$

Now, we can write our position vector.

$r_{AC}=\left\{(4-3)i+(0-3)j+(-2-0)k\right\}$

$r_{AC}=\left\{1i-3j-2k\right\}$

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 $r_{AC}=\sqrt{(1)^2+(-3)^2+(-2)^2}=3.74$

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:

$u_{AB}\,=\,\left(\dfrac{1}{3.74}i-\dfrac{3}{3.74}j-\dfrac{2}{3.74}k\right)$

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.

$F=80\left(\dfrac{1}{3.74}i-\dfrac{3}{3.74}j-\dfrac{2}{3.74}k\right)$

$F= \left\{21.4i-64.2j-42.8k\right\}$

The next step is to express another position vector from O to A. This is the location of where the moment is calculated to the location where the force is being applied. From the image, we see that points O and A are located at:

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

$A:(3i+3j+0k)$

The position vector from O to A is:

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

$r_{OA}=\left\{3i+3k+0k\right\}$

We can now calculate the moment by taking the cross product between the position vector and the force.

$M_O=r_{OA}\times F$

$M_A=\begin{bmatrix}\bold i&\bold j&\bold k\\3&3&0\\21.4&-64.2&-42.8\end{bmatrix}$

$M_A=\left\{-128i+128.4j-257k\right\}N\cdot m$

$M_A=\left\{-128i+128.4j-257k\right\}N\cdot m$