# : Researchin the forces in rasing and lowering a boom

1. May 3, 2007

URGENT: Researchin the forces in rasing and lowering a boom

1. The problem statement, all variables and given/known data
In an experiment a group of students used a boom, and some string to represent a tie as in a crane. The figure shows how the apparatus was set up.

Boom and other variables
Centre of mass boom: 0.5m
Distance to tie (d): 0.6m
Mass of boom: 0.35kg
Height of tie above pivot (h): 0.62cmAcceleration due to gravity: 9.8m/s/s
2. The attempt at a solution
Following are the questions/calculations asked for the experiment and the answers which i think are correct.

Graphs for reference to questions 2 and 3:

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1. What happens to the tension in the tie as the boom is raised? Give an explanation for this in terms of moment arm lengths.

As the boom is raised the tension in the wire decreases. As the boom is lowered the perpendicular force exerted by the end mass will increase. As moment is dependent upon the perpendicular force and the distance from the pivot point, and increase in the perpendicular force, and a constant distance from the pivot, the moment or torque will increase. The increase in torque will place the tie under greater stress, thus resulting in the tie having a greater tension as the boom is lowered, or the angle (measured from the vertical to the boom) increases.

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2. From the graphs find the angle of the boom that produces the maximum horizontal reaction at the pivot. Explain why the maximum will always be at this angle in a boom supported this way.

The maximum horizontal reaction at the pivot is at an angle of 90 degrees. The maximum will always be at this angle in a boom supported this way because the tie holding up the boom is at the furtherest possible from the pivot. therefore resulting in the greatest horizontal reaction force.

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3. From the graph of the pivot reaction determine the boom angle that produces a resultant reaction force horizontally. How will this angle change if a larger load is placed on the boom? Give an explanation for your answer

This question I am not sure on how to solve, or the theory behind it. Any help/hints greatly appreciated.

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4. Suggest a reason why the tension in the tie does not change uniformly with the angle of the boom.

This question I am not sure on how to answer it because I am not too sure how this system works in this situation. The vertical distance changes gradually at the beginning and the end of the system, while around the 80-100 degree mark the component changes at a more rapid rate, even though the increments of the angle remain the same, but I do not know why this happens

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5. For a boom angle of 90 degrees calculate the tension in the tie. Use values you obtained in the table above (results shown listed below the image of the apparatus).

$$\begin{array}{l} \sum {\tau _{anticlockwise} } = \sum {\tau _{clockwise} } \\ 0.5\left( {0.35} \right)\left( {9.8} \right) + 0.97\left( {0.45} \right)\left( {9.8} \right) = 0.6\left( F \right) \\ F = 9.9878 = T_{vertical} \\ \theta = \tan ^{ - 1} \frac{{0.6}}{{0.62}} = 44.06 \\ T = \frac{{T_{vertical} }}{{\cos \theta }} = 13.8989 \\ \end{array}$$

Therefore for a boom of angle 90 degrees and with the values obtained in the table above, the tension in the tie is 13.9N

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6. For the boom angle of 90 degrees calculate from the magnitude and direction of the reaction at the pivot. Use the values you obtained in the table above.

$$\begin{array}{l} \sum {F = 0} \\ = {\rm{Tesion in the tie + weight force + reaction force of the pivot}} \\ 13.8989 + 0.35\left( {9.8} \right) + 0.45\left( {9.8} \right) = R_{pivot} = 21.7389 \\ \end{array}$$

vertical components:
$$\begin{array}{l} \sum {F \uparrow = } \sum {F \downarrow } \\ 9.9878 + 0.35\left( {9.8} \right) + 0.45\left( {9.8} \right) = R_{pivot\left( {vertical} \right)} = 17.8278 \\ \end{array}$$

horizontal components:
$$\begin{array}{l} \sum {F \leftarrow = } \sum {F \to } \\ \sin \theta = \frac{O}{H} \\ O = T\sin \theta = R_{pivot\left( {horizontal} \right)} = 9.6656 \\ \end{array}$$

angle of reaction force:
$$\theta = \tan ^{ - 1} \frac{{17.8278}}{{9.6656}} = 61.535$$

therefore reaction force on the pivot (magnitude) = 21.7389N
therefore reaction force on the pivot (direction) = 61.535 degrees

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They are the question which were given to me and the answers to most of them. I am unsure about my logics which are behind my questions, so if i have done anything wrong, please point it out. Basically for all the questions which i have answered, I'm asking if I have done them correctly. Questions 3 and 4 I have no idea how to solve. All help, suggestions, hints given will be GREATLY appreciated.
Thank you for all the replies

Last edited: May 3, 2007
2. May 3, 2007

### cam_alp

"3. From the graph of the pivot reaction determine the boom angle that produces a resultant reaction force horizontally. How will this angle change if a larger load is placed on the boom? Give an explanation for your answer."

The boom angle that will produce a resultant reaction force horizontally is around 70 degrees, as it is the point on the first graph where the vertical reaction force decreases to 0. This angle will change as when a larger load is placed on the boom, the boom will have to be raised for it to be able to stay in the same position but to be able to withstand the extra weight foce, since the boom is moved up the angle will increase.

Sorry for my bad physics terminoligy, but i only just got the answer, with the help of clint.
Cam

Last edited: May 3, 2007
3. May 3, 2007

4. May 5, 2007

...........

5. May 6, 2007

6. May 7, 2007

In order for me to get some help on this once great forums, what must i do?? Can I please! get some help?

7. May 7, 2007

### Mindscrape

This is a very involved problem, and it is a lot of work for someone to go through and work out every one of your questions. If you specifically went into one of the questions you are having the most problems with, I think you might get some more responses.

8. May 7, 2007

### denverdoc

Also part 3a, was answered some time ago, just find the zero crossing for the vertical rxn force. You have a complete set of eqns with the exception of one linking the two angles.