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

In summary: W\,=\,485N...W\,=\,485N...... I cannot figure out how to do this. The answer is 138N, but I cannot reach it like this.
  • #1
VinnyCee
489
0
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.

[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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  • #2
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 canceled by the meager
y-component of the post... so F_AC can't be 500N either.
 
  • #3
Like this?

[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) = 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}[/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) = 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}[/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]

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

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

[tex]\sum\,F_x\,=\,\frac{2F_{AB}}{3}\,+\,\frac{2F_{AC}}{6.5}\,-\,92.31N\,=\,0[/tex]

[tex]\sum\,F_y\,=\,\frac{F_{AB}}{3}\,+\,\frac{1.5F_{AC}}{6.5}\,+\,69.23N\,=\,0[/tex]

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

Convert these into three equations:

[tex]\frac{2}{3}\,F_{AB}\,+\,\frac{2}{6.5}\,F_{AC}\,+\,0W\,=\,92.31[/tex]

[tex]\frac{1}{3}\,F_{AB}\,+\,\frac{1.5}{6.5}\,F_{AC}\,+\,0W\,=\,-69.23[/tex]

[tex]\frac{2}{3}\,F_{AB}\,+\,\frac{6}{6.5}\,F_{AC}\,-\,W\,=\,276.9[/tex]

Put this into a matrix:

[tex]\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][/tex]

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 [itex]F_{AB}[/itex] and -1500 for [itex]F_{AC}[/itex]! Max is supposed to be 500N!

Any suggestions?
 
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  • #4
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
There is a solution though

It says that the answer is 138N in the back of the book.
 
  • #6
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
VinnyCee said:
[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) = 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}[/tex]

Take a closer look at the square root. [tex]\sqrt{2^2 + (-3)^2 + 6^2}[/tex] 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
 
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  • #8
you still have an overall negative sign error
and Fac y-multiplier is 3 (not 1.5)
 
  • #9
Thanks, here is solution

[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) = 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}[/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) = 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}[/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]

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

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

[tex]\sum\,F_x\,=\,-\frac{2F_{AB}}{3}\,-\,\frac{2F_{AC}}{7}\,+\,92.31N\,=\,0[/tex]

[tex]\sum\,F_y\,=\,\frac{F_{AB}}{3}\,+\,\frac{3F_{AC}}{7}\,-\,69.23N\,=\,0[/tex]

[tex]\sum\,F_z\,=\,-\frac{2F_{AB}}{3}\,-\,\frac{6F_{AC}}{7}\,+\,W\,=\,0[/tex]

Convert these into three equations:

[tex]-\frac{2}{3}\,F_{AB}\,-\,\frac{2}{7}\,F_{AC}\,=\,-92.31[/tex]

[tex]\frac{1}{3}\,F_{AB}\,+\,\frac{3}{7}\,F_{AC}\,=\,69.23[/tex]

Put this into a matrix:

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

RREF it and substitute into this equation from above:

[tex]\frac{2}{3}\,F_{AB}\,+\,\frac{6}{7}\,F_{AC}\,-\,W\,=\,0[/tex]

[tex]\frac{2}{3}\,F_{AB}\,+\,\frac{6}{7}\,F_{AC}\,=\,W[/tex]

[tex]\frac{2}{3}\,\left(103.8525N\right)\,+\,\frac{6}{7}\,\left(80.7625N\right)\,=\,W[/tex]

[tex]W\,=\,138N[/tex]
 
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1. What is the definition of equilibrium of a particle in 3D-space?

The equilibrium of a particle in 3D-space refers to a state in which the particle is at rest or moving with a constant velocity, with no net external forces acting on it.

2. How many points are required to determine the equilibrium of a particle in 3D-space?

Four points are required to determine the equilibrium of a particle in 3D-space. These points are the particle itself, two points through which the forces act, and a fourth point that serves as a reference point for calculating the forces.

3. What are the three vectors involved in the equilibrium of a particle in 3D-space?

The three vectors involved in the equilibrium of a particle in 3D-space are the position vector, force vector, and weight vector. The position vector represents the location of the particle, the force vector represents the magnitude and direction of the forces acting on the particle, and the weight vector represents the force of gravity acting on the particle.

4. How do you determine the equilibrium of a particle in 3D-space?

To determine the equilibrium of a particle in 3D-space, you must first draw a free body diagram of the particle, showing all the forces acting on it. Then, you can use the vector sum of forces to determine if the particle is in equilibrium. If the vector sum of forces is equal to zero, the particle is in equilibrium.

5. What is the significance of equilibrium of a particle in 3D-space in real-life situations?

The equilibrium of a particle in 3D-space is important in many real-life situations, such as in engineering and physics problems. It allows us to analyze and predict the behavior of objects and structures under different forces and conditions. Understanding equilibrium can also help in designing and optimizing structures for maximum stability and safety.

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