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Here is the problem I am working on:

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

The points are as follows:

A(2, -1.5, 6)

B(-4, 1.5, 0)

C(0, 1.5, 0)

A lamp is supposed to hang off of point A. I tried the calculations with the answer being way off of the answer in the back of the book.

[tex]\overrightarrow{F_{AB}} = F_{AB} \left(\frac{6\hat{i} - 3\hat{j} + 6\hat{k}}{\sqrt{6^2 + (-3)^2 + 6^2}}\right) = 500N \left(\frac{2}{3}\,\hat{i} - \frac{1}{3}\,\hat{j} + \frac{2}{3}\,\hat{k}\right) = 333.3\,\hat{i}\,\,-\,\,166.7\,\hat{j}\,\,+\,\,333.3\,\hat{k}[/tex]

[tex]\overrightarrow{F_{AC}} = F_{AC} \left(\frac{2\hat{i} - 3\hat{j} + 6\hat{k}}{\sqrt{2^2 + (-3)^2 + 6^2}}\right) = 500N \left(-\frac{2}{7}\,\hat{i} + \frac{1.5}{7}\,\hat{j} - \frac{6}{7}\,\hat{k}\right) = 142.9\,\hat{i}\,\,-\,\,214.3\,\hat{j}\,\,+\,\,428.6\,\hat{k}[/tex]

[tex]\overrightarrow{F_{OA}} = F_{OA} \left(\frac{-2\hat{i} + 1.5\hat{j} - 6\hat{k}}{\sqrt{(-2)^2 + 1.5^2 + 6^2}}\right) = 300N \left(-\frac{2}{6.5}\,\hat{i} + \frac{1.5}{6.5}\,\hat{j} - \frac{6}{6.5}\,\hat{k}\right) = -92.31\,\hat{i}\,\,+\,\,69.23\,\hat{j}\,\,-\,\,276.9\,\hat{k}[/tex]

[tex]\overrightarrow{W} = (-W)\,\hat{k}[/tex]

Now, adding up the forces for equilibrium in z-direction and solving for W:

[tex]\sum\,F_z\,=\,\left(333.3\,+\,428.6\,-\,276.9\,+\,W\right)[/tex]

[tex]W\,=\,485N[/tex]

However, the answer in the back of the book is 138N, where am I screwing up?

The weight hangs from point A.

http://img62.imageshack.us/img62/6594/problem3516xa.jpg [Broken]

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# Homework Help: Equilibrium of a Particle (3D-space, 4 points, 3 vectors and one weight)

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