# Cables AB and AC can sustain a maximum tension

Cables AB and AC can sustain a maximum tension of 500 N, and the pole can support a maximum compression of 300 N. Determine the maximum weight of the lamp that can be supported in the position shown. The force in the pole acts along the axis of the pole.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

Using the diagram, the locations of the points A, B, C, and O are:

$A:(2i-1.5j+6k)$

$B:(-4i+1.5j+0k)$

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

$O:(0i+0j+0k)$

The position vectors for points from A to B, A to C, and A to O are:

$r_{AB}\,=\,\left\{(-4-2)i+(1.5-(-1.5))j+(0-6)k\right\}\,=\,\left\{-6i+3j-6k\right\}$

$r_{AC}\,=\,\left\{(0-2)i+(1.5-(-1.5))j+(0-6)k\right\}\,=\,\left\{-2i+3j-6k\right\}$

$r_{OA}\,=\,\left\{(2-0)i+(-1.5-0)j+(6-0)k\right\}\,=\,\left\{2i-1.5j+6k\right\}$
(Why did we write the position vector from O to A instead of A to O? Unlike ropes, which can only be in tension, the pole will actually be in compression. That means a force from the pole is heading upwards, where as in the cables, the force is heading away from the lamp)

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

The magnitude of each position vector is:

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

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

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

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.

The unit vector for each position vector is:

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

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

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

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

We can now write each force in Cartesian vector form:

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

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

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

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

$F_{AB}\,=\,\left\{-0.667F_{AB}i+0.333F_{AB}j-0.667F_{AB}k\right\}$

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

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

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

(Force F is the weight of the lamp, which only has a z-component)

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

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

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

Furthermore, as each force added together must equal zero, than each individual component (x, y, z-components) added together must also equal zero.

x-components:

$-0.667F_{AB}-0.286F_{AC}+0.308F_{OA}\,=\,0$

y-components:

$0.333F_{AB}+0.429F_{AC}-0.231F_{OA}\,=\,0$

z-components:

$-0.667F_{AB}-0.857F_{AC}+0.923F_{OA}-W\,=\,0$

The question states that the pole can only support 300 N in compression. Thus, we will assume force $F_{AO}$ is 300 N and substitute it into our equations. Doing so gives us the following equations:

x-components:

$-0.667F_{AB}-0.286F_{AC}+92.4\,=\,0$

y-components:

$0.333F_{AB}+0.429F_{AC}-69.3\,=\,0$

z-components:

$-0.667F_{AB}-0.857F_{AC}+276.9-W\,=\,0$

Solving the three equations simultaneously gives us:

$F_{AB}\,=\,103.8$ N

$F_{AC}\,=\,80.9$ N

$W\,=\,138.3$ N

Notice how our values for the cables are less then 500 N, thus our assumption was right. To double check, you can assume each cable is 500 N and solve the equations again, doing so will show that the forces in either the pole or the cable will exceed the maximum allowable limit.