# If the mass of the flowerpot

If the mass of the flowerpot is 50 kg, determine the tension developed in each wire for equilibrium. Set x = 1.5 m and z = 2 m. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will first find the locations of the points A, C, and D. We do not need to find the location of point B as the force in cable AB would lie along the y-axis. Using the diagram, the locations of the points in Cartesian vector form is:

$A:(0i+6j+0k)$

$C:(2i+0j+3k)$

$D:(-1.5i+0j+2k)$

(remember x = 1.5 m, and z = 2 m)

We can now write position vectors for cables AC and AD.

$r_{AC}\,=\,\left\{(2-0)i+(0-6)j+(3-0)k\right\}\,=\,\left\{2i-6j+3k\right\}$

$r_{AD}\,=\,\left\{(-1.5-0)i+(0-6)j+(2-0)k\right\}\,=\,\left\{-1.5i-6j+2k\right\}$

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

Next, we will find the magnitude of each position vector.

magnitude of $r_{AC}\,=\,\sqrt{(2)^2+(-6)^2+(3)^2}\,=\,7$

magnitude of $r_{AD}\,=\,\sqrt{(-1.5)^2+(-6)^2+(2)^2}\,=\,6.5$

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.

Now, we can find the unit vectors for each position vector.

$u_{AC}\,=\,\left(\dfrac{2}{7}i-\dfrac{6}{7}j+\dfrac{3}{7}k\right)$

$u_{AD}\,=\,\left(-\dfrac{1.5}{6.5}i-\dfrac{6}{6.5}j+\dfrac{2}{6.5}k\right)$

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

Our next step is to express each force in the cables.

$F_{AC}\,=\,F_{AC}\left(\dfrac{2}{7}i-\dfrac{6}{7}j+\dfrac{3}{7}k\right)$

$F_{AD}\,=\,F_{AD}\left(-\dfrac{1.5}{6.5}i-\dfrac{6}{6.5}j+\dfrac{2}{6.5}k\right)$

(simplify by expanding the brackets and writing the fractions in decimal form)

$F_{AC}\,=\,\left\{0.286F_{AC}i-0.857F_{AC}j+0.429F_{AC}k\right\}$

$F_{AD}\,=\,\left\{-0.231F_{AD}i-0.923F_{AD}j+0.308F_{AD}k\right\}$

$F_{AB}\,=\,\left\{0i+F_{AB}+0k\right\}$

(Remember, $F_{AB}$ is the force that lies along the y-axis, it only has the y-component)

$W\,=\,\left\{0i+0j-490.5k\right\}$

(Force W is the weight of the pot, which has only a z-component. The weight is equal to (50)(9.81)=490.5 N)

As the system is in equilibrium, all forces added together must equal zero.

$\sum \text{F}\,=\,0$

$F_{AB}+F_{AC}+F_{AD}+W\,=\,0$

Since all the forces added together must equal zero, then the individual components added together must also equal zero.

x-components:

$0.286F_{AC}-0.231F_{AD}\,=\,0$

y-components:

$-0.857F_{AC}-0.923F_{AD}+F_{AB}\,=\,0$

z-components:

$0.429F_{AC}+0.308F_{AD}-490.5\,=\,0$

Solving the three equations gives us:

$F_{AB}\,=\,1210$ N

$F_{AC}\,=\,605$ N

$F_{AD}\,=\,749$ N