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.

#### Solution:

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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:

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

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

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

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

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

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

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.

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.

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

y-components:

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

z-components:

Solving the three equations gives us:

F_{AC}\,=\,605 N

F_{AD}\,=\,749 N