# If cable AB is subjected to a tension

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

1. Find the position vector for each rope
2. Find the magnitude of each position vector
3. Find the unit vector of each position vector
4. Multiply the unit vector by the magnitude of the force in each cable Locations of each point:

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

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

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

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

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

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

We will now find the magnitude of each position vector:

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

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

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.

Next, we will find the unit vectors.

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

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

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

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 write each force in Cartesian vector form.

$F_{AB}\,=\,700\left(\dfrac{2}{7}i+\dfrac{3}{7}j-\dfrac{6}{7}k\right)$

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

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

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

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.

x-components:

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

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:

$F_{AC}\,=\,130$ N

$F_{AD}\,=\,510$ N

$F\,=\,1060$ N