Moment of a pole of unknown height

In summary, the tension in cable AB creates a moment about point O (MO) equal to -900i+Myj-315k lb-ft, where My is unknown.
  • #1
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Homework Statement



Utility Pole BC is guyed by cable AB as shown. The tension in cable AB (TAB) acting at point B creates a moment about point O (MO) equal to -900i+Myj-315k lb-ft, where My is unknown.
a) What is the length of the pole (LBC)?
b) What is the magnitude of the tension, TAB?

Diagram.png


Homework Equations



MOABxF
TAB=uAB|TAB|

The Attempt at a Solution


Vector BO <0,-(h+1),6>
Vector BA <4,-h,-6>
|BA|= [itex]\sqrt{h^{2}+52}[/itex]
TBA= TBA[itex]\frac{<4,-h,-6>}{|BA|}[/itex]

I cross multiplied and got TBA<[itex]\frac{12h+6}{|BA|}[/itex],[itex]\frac{24}{|BA|}[/itex],[itex]\frac{4h+4}{|BA|}[/itex]>

In order to solve for the height of the pole, I need TBA. Where i get stuck is here, and a lack of confidence that my vectors BA and BO are correct. In order to solve for the height of the pole, I need the tension, but that is the second part of the problem.

EDIT: tips on how to fix the itex tags would be appreciated,too.
 
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  • #2
PensNAS said:

The Attempt at a Solution


Vector BO <0,-(h+1),6>
Do you need vector BO or vector OB?

Vector BA <4,-h,-6>
Check the diagram carefully to see if the y-displacement from B to A is -h.

In order to solve for the height of the pole, I need the tension, but that is the second part of the problem.
You will have to solve 2 equations simultaneously for h and TAB.

EDIT: tips on how to fix the itex tags would be appreciated,too.
When writing latex, don't use the superscript and subscript buttons on the tool bar. Instead, use standard latex notation: h^2, for example.
 
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  • #3
Vector OB should be the one I need, which is <0,h+1,-6> Vector BA should be <4,0.5-h,-6>.

This gives a new cross product of T[itex]_{BA}[/itex]<[itex]\frac{-12h-3}{|BA|}[/itex],[itex]\frac{-24}{|BA|}[/itex],-[itex]\frac{4h-4}{|BA|}[/itex]>

Am I right in thinking I can then set the i and k components of M[itex]_{O}[/itex] equal to the i and k components of the cross product? If so then I get:

From the i component:
T[itex]^{2}_{BA}[/itex]:[itex]\frac{90000(h^{2}-h+52.25)}{16h^{2}+h+52.25}[/itex]

From the k component:
T[itex]^{2}_{BA}[/itex]:[itex]\frac{6202(h^{2}-h+52.25)}{h^{2}+2h+1}[/itex]

Do I set those equal to each other? If so, I get h=14.74 feet. Plugging that value for h into
-900=T[itex]_{BA}[/itex][itex]\frac{-12h-3}{|BA|}[/itex] or
-315=T[itex]_{BA}[/itex][itex]\frac{-4h-4}{|BA|}[/itex]
gives me T[itex]_{BA}[/itex]=84.6 lb
 
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  • #4
PensNAS said:
Vector OB should be the one I need, which is <0,h+1,-6> Vector BA should be <4,0.5-h,-6>.

This gives a new cross product of T[itex]_{BA}[/itex]<[itex]\frac{-12h-3}{|BA|}[/itex],[itex]\frac{-24}{|BA|}[/itex],-[itex]\frac{4h-4}{|BA|}[/itex]>

Are you sure about the negative sign in the numerator of the last term? Shouldn't it be 4h + 4 instead of 4h - 4?

Am I right in thinking I can then set the i and k components of M[itex]_{O}[/itex] equal to the i and k components of the cross product? If so then I get:

From the i component:
T[itex]^{2}_{BA}[/itex]:[itex]\frac{90000(h^{2}-h+52.25)}{16h^{2}+h+52.25}[/itex]

I don't agree with your denominator here.

From the k component:
T[itex]^{2}_{BA}[/itex]:[itex]\frac{6202(h^{2}-h+52.25)}{h^{2}+2h+1}[/itex]

Do I set those equal to each other? If so, I get h=14.74 feet.

I agree with your value of h ( I get 14.75 ft ).

Plugging that value for h into
-900=T[itex]_{BA}[/itex][itex]\frac{-12h-3}{|BA|}[/itex] or
-315=T[itex]_{BA}[/itex][itex]\frac{-4h-4}{|BA|}[/itex]
gives me T[itex]_{BA}[/itex]=84.6 lb

I get somewhat less than 84.6 lb. But your approach seems correct.
[EDIT: Note that if you take the ratio of your last two equations, you get a simple equation for finding h.]
 
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  • #5
4h+4 is correct, I dropped a negative.

The denominator was a typo. I have it on my paper as [itex]\frac{90000(h^{2}-h+52.25)}{16h^{2}+8h+1}[/itex].

I redid the calculation for the tension and got 79.5 lb.
It would seem I got sloppy, this was the 8th time working through the problem.Thanks for the help!
 
  • #6
OK. I got 79.9 lb. Good work.

Pensacola?
 
  • #7
I get that when I spend the extra 15 seconds to calculate |BA| further than 15.9.

TSny said:
Pensacola?

Yep, home to beaches and Naval Aviation.:biggrin:
 

1. What is the moment of a pole of unknown height?

The moment of a pole of unknown height is a physical quantity that measures the tendency of the pole to rotate around its base. It is a product of the force applied to the pole and the perpendicular distance from the force to the base of the pole.

2. How is the moment of a pole of unknown height calculated?

The moment of a pole of unknown height is calculated by multiplying the force applied to the pole by the perpendicular distance from the force to the base of the pole. The formula is: moment = force * distance.

3. What is the unit of measurement for the moment of a pole of unknown height?

The unit of measurement for the moment of a pole of unknown height is newton-meter (N*m) in the SI system.

4. How does the height of the pole affect its moment?

The height of the pole does not directly affect its moment. However, a taller pole may require a greater force to be applied in order to generate the same moment as a shorter pole.

5. What is the significance of the moment of a pole of unknown height?

The moment of a pole of unknown height is an important concept in physics and engineering. It is used to analyze the stability of structures and determine the amount of force required to cause rotation. It also has practical applications in fields such as architecture and construction.

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