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

1. Mar 10, 2006

### VinnyCee

Note: This attempt is incorrect, scroll down for the most recent attempt, which is still incorrect!

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.

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

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

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

$$\overrightarrow{W} = (-W)\,\hat{k}$$

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

$$\sum\,F_z\,=\,\left(333.3\,+\,428.6\,-\,276.9\,+\,W\right)$$

$$W\,=\,485N$$

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

The weight hangs from point A.

Last edited: Mar 11, 2006
2. Mar 10, 2006

### lightgrav

Equilibrium requires that all 3 components of Force sum to zero.
To do that, F_AB cannot be 500N, because the x-component is too large.
(I suspect that all your Force components are the wrong sign;
if cable pulls from A toward B, F_AB points in direction ( - , + , -), etc).

Even if F_0A is 300N, F_AB(x-component) must be less than 236N,
or else it cannot be cancelled.
Worse, the y-components of the Tensions must be cancelled by the meager
y-component of the post... so F_AC can't be 500N either.

3. Mar 11, 2006

### VinnyCee

Like this?

$$\overrightarrow{F_{AB}} = F_{AB} \left(\frac{6\hat{i} - 3\hat{j} + 6\hat{k}}{\sqrt{6^2 + (-3)^2 + 6^2}}\right) = F_{AB} \left(\frac{2}{3}\,\hat{i} - \frac{1}{3}\,\hat{j} + \frac{2}{3}\,\hat{k}\right) = \frac{2F_{AB}}{3}\,\hat{i}\,-\,\frac{F_{AB}}{3}\,\hat{j}\,+\,\frac{2F_{AB}}{3}\,\hat{k}$$

$$\overrightarrow{F_{AC}} = F_{AC} \left(\frac{2\hat{i} - 3\hat{j} + 6\hat{k}}{\sqrt{2^2 + (-3)^2 + 6^2}}\right) = F_{AC} \left(\frac{2}{6.5}\,\hat{i} - \frac{1.5}{6.5}\,\hat{j} + \frac{6}{6.5}\,\hat{k}\right) = \frac{2F_{AC}}{6.5}\,\hat{i}\,-\,\frac{1.5F_{AC}}{6.5}\,\hat{j}\,+\,\frac{6F_{AC}}{6.5}\,\hat{k}$$

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

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.

$$\overrightarrow{W} = (-W)\,\hat{k}$$

$$\sum\,F_x\,=\,\frac{2F_{AB}}{3}\,+\,\frac{2F_{AC}}{6.5}\,-\,92.31N\,=\,0$$

$$\sum\,F_y\,=\,\frac{F_{AB}}{3}\,+\,\frac{1.5F_{AC}}{6.5}\,+\,69.23N\,=\,0$$

$$\sum\,F_z\,=\,\frac{2F_{AB}}{3}\,+\,\frac{6F_{AC}}{6.5}\,-\,W\,-\,276.9N\,=\,0$$

Convert these into three equations:

$$\frac{2}{3}\,F_{AB}\,+\,\frac{2}{6.5}\,F_{AC}\,+\,0W\,=\,92.31$$

$$\frac{1}{3}\,F_{AB}\,+\,\frac{1.5}{6.5}\,F_{AC}\,+\,0W\,=\,-69.23$$

$$\frac{2}{3}\,F_{AB}\,+\,\frac{6}{6.5}\,F_{AC}\,-\,W\,=\,276.9$$

Put this into a matrix:

$$\left[ \begin{array}{cccc}\frac{2}{3} & \frac{2}{6.5} & 0 & 92.31 \\ \frac{1}{3} & \frac{1.5}{6.5} & 0 & -69.23 \\ \frac{2}{3} & \frac{6}{6.5} & -1 & 276.9 \end{array} \right]$$

But when I RREF it, I get -1107N as the value for W. The answer in the book is 138N! I also get 831 for $F_{AB}$ and -1500 for $F_{AC}$! Max is supposed to be 500N!

Any suggestions?

Last edited: Mar 11, 2006
4. Mar 11, 2006

### lightgrav

Summing x-component to zero gives F_ab in terms of F_ac;
Check whether that solution is consistent with y-component sum.

Sum of z-component (replacing F_ab with its formula) gives F_ac in terms of W. invert to find W in terms of F_ac.

This is over-constrained if you assume that Compression is 300N ; there might be NO solution.

5. Mar 11, 2006

### VinnyCee

There is a solution though

It says that the answer is 138N in the back of the book.

6. Mar 11, 2006

### VinnyCee

Anyone know why this problem cannot be done using these methods? I know that there is an answer, but I just cannot find it. Please help!

7. Mar 11, 2006

### topsquark

Take a closer look at the square root. $$\sqrt{2^2 + (-3)^2 + 6^2}$$ is 7 not 6.5. I don't know if that's a typo or not. If it isn't that'd certainly be enough to screw up the system of equations.

Also in your net force equations, where are the 92.31 N, 69.23 N, and 276.9 N coming from?

-Dan

Last edited: Mar 11, 2006
8. Mar 12, 2006

### lightgrav

you still have an overall negative sign error
and Fac y-multiplier is 3 (not 1.5)

9. Mar 14, 2006

### VinnyCee

Thanks, here is solution

$$\overrightarrow{F_{AB}} = F_{AB} \left(\frac{-6\hat{i} + 3\hat{j} + 6\hat{k}}{\sqrt{6^2 + (-3)^2 + 6^2}}\right) = F_{AB} \left(-\frac{2}{3}\,\hat{i} + \frac{1}{3}\,\hat{j} + \frac{2}{3}\,\hat{k}\right) = -\frac{2F_{AB}}{3}\,\hat{i}\,+\,\frac{F_{AB}}{3}\,\hat{j}\,+\,\frac{2F_{AB}}{3}\,\hat{k}$$

$$\overrightarrow{F_{AC}} = F_{AC} \left(\frac{2\hat{i} - 3\hat{j} + 6\hat{k}}{\sqrt{2^2 + (-3)^2 + 6^2}}\right) = F_{AC} \left(\frac{2}{7}\,\hat{i} - \frac{1.5}{7}\,\hat{j} + \frac{6}{7}\,\hat{k}\right) = \frac{2F_{AC}}{7}\,\hat{i}\,-\,\frac{1.5F_{AC}}{7}\,\hat{j}\,+\,\frac{6F_{AC}}{7}\,\hat{k}$$

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

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.

$$\overrightarrow{W} = (-W)\,\hat{k}$$

$$\sum\,F_x\,=\,-\frac{2F_{AB}}{3}\,-\,\frac{2F_{AC}}{7}\,+\,92.31N\,=\,0$$

$$\sum\,F_y\,=\,\frac{F_{AB}}{3}\,+\,\frac{3F_{AC}}{7}\,-\,69.23N\,=\,0$$

$$\sum\,F_z\,=\,-\frac{2F_{AB}}{3}\,-\,\frac{6F_{AC}}{7}\,+\,W\,=\,0$$

Convert these into three equations:

$$-\frac{2}{3}\,F_{AB}\,-\,\frac{2}{7}\,F_{AC}\,=\,-92.31$$

$$\frac{1}{3}\,F_{AB}\,+\,\frac{3}{7}\,F_{AC}\,=\,69.23$$

Put this into a matrix:

$$\left[ \begin{array}{ccc}-\frac{2}{3} & -\frac{2}{7} & -92.31 \\ \frac{1}{3} & \frac{3}{7} & 69.23 \end{array} \right]$$

RREF it and substitute into this equation from above:

$$\frac{2}{3}\,F_{AB}\,+\,\frac{6}{7}\,F_{AC}\,-\,W\,=\,0$$

$$\frac{2}{3}\,F_{AB}\,+\,\frac{6}{7}\,F_{AC}\,=\,W$$

$$\frac{2}{3}\,\left(103.8525N\right)\,+\,\frac{6}{7}\,\left(80.7625N\right)\,=\,W$$

$$W\,=\,138N$$

Last edited: Mar 14, 2006