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.

#### Solution:

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

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_{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)*

The magnitude of each position vector is:

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 unit vector for each position vector is:

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)

We can now write each force in Cartesian vector form:

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.

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

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

y-components:

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

z-components:

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:

-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_{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.