If cable AB is subjected to a tension of 700 N, determine the tension in cables AC and AD and the magnitude of the vertical force F.

#### Solution:

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We will first express each force along the ropes in Cartesian form. To do so, we will follow the following steps:

- Find the position vector for each rope
- Find the magnitude of each position vector
- Find the unit vector of each position vector
- Multiply the unit vector by the magnitude of the force in each cable

Locations of each point:

B:(2i+3j+0k)

C:(-1.5i+2j+0k)

D:(-3i-6j+0k)

The position vectors are:

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

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

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

We will now find the magnitude of each position vector:

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

magnitude of r_{AD}\,=\,\sqrt{(-3)^2+(-6)^2+(-6)^2}\,=\,9 m

Next, we will find the unit vectors.

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

u_{AD}\,=\,\left(-\dfrac{3}{9}i-\dfrac{6}{9}j-\dfrac{6}{9}k\right)

We can now write each force in Cartesian vector form.

F_{AB}\,=\left\{200i+300j-600k\right\} N

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

F_{AC}\,=\left\{-0.23F_{AC}i+0.308F_{AC}j-0.923F_{AC}k\right\} N

F_{AD}\,=\,F_{AD}\left(-\dfrac{3}{9}i-\dfrac{6}{9}j-\dfrac{6}{9}k\right)

F_{AD}\,=\left\{-0.333F_{AD}i-0.667F_{AD}j-0.667F_{AD}k\right\} N

F\,=\,\left\{0i+0j+Fk\right\} N

(This is the force that is applied directly upwards. It only has a z-component)

We can now write our equation of equilibrium. All the forces added together must equal zero because the system is in equilibrium.

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

Since all forces added together must equal zero, that means each of the components ( x, y, z-components) added individually must add up to zero as well.

200-0.23F_{AC}-0.333F_{AD}\,=\,0

y-components:

300+0.308F_{AC}-0.667F_{AD}\,=\,0

z-components:

-600-0.923F_{AC}-0.667F_{AD}+F\,=\,0

Solving all three equations simultaneously gives us:

F_{AD}\,=\,510 N

F\,=\,1060 N