The 50-kg pot is supported from A by the three cables. Determine the force acting in each cable for equilibrium. Take d = 2.5 m.

#### Solution:

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Let us first express each force in each rope in Cartesian vector form. To do so, the first step is to figure out the locations of points A, B, C, and D in cartesian vector notation.

From the diagram, the locations of the points are:

B:(6i+0j+2.5k) (Remember d=2.5m)

C:(-6i-2j+3k)

D:(-6i+2j+3k)

We can now write our position vectors:

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

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

The magnitude of each position vector is:

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

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

The unit vectors for each position vector is:

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

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

We can now express the force in each cable in Cartesian vector form:

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

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

(simplify by expanding brackets and writing fractions are decimals)

F_{AB}\,=\,0.923F_{AB}i+0j+0.385F_{AB}k

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

F_{AD}\,=\,-0.857F_{AD}i+0.286F_{AD}j+0.429F_{AD}k

F\,=\,\left\{0i+0j-490.5\right\}

(Force F is the weight of the pot, which has a mass of 50 kg, \therefore the weight is (50)(9.81) = 490.5 N)

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

F_{AB}+F_{AC}+F_{AD}+F\,=\,0

Furthermore, since each force added together must equal zero, then each component (x, y, z-component) added together must also equal zero.

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

y-components:

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

z-components:

Solving the three equations simultaneously gives us:

F_{AC}\,=\,312 N

F_{AD}\,=\,312 N