Statics: Crane Boom - Why Different Results?

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Discussion Overview

The discussion revolves around the calculations involved in determining the forces acting on a crane boom, specifically addressing why different results are obtained when analyzing two sections of the boom (AE and EB) using static equilibrium equations. The focus is on the application of static equilibrium principles in a mechanical context.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Participants calculate the weight of the engine as 4905 N and use it in their static equilibrium equations.
  • Some participants propose using section AE and section EB to derive forces, resulting in different values (12262.5 N for AE and 16350 N for EB), leading to confusion about the discrepancy.
  • One participant suggests writing the equations of static equilibrium for the crane and drawing a free body diagram to clarify the situation.
  • There is a discussion about the equations of static equilibrium, with participants noting that there are two main equations: ∑F = 0 and ∑M = 0, and that generally only one moment equation can be written.
  • Another participant emphasizes the need to calculate the components of the force to find the internal forces acting at point E and cautions against using the same letter for different moment arms, as it may cause confusion.
  • There is uncertainty regarding the correctness of the formulas used for calculating moments and forces, with participants seeking validation of their approaches.

Areas of Agreement / Disagreement

Participants express differing views on the application of static equilibrium equations and the resulting calculations, indicating that there is no consensus on the correct approach or resolution of the discrepancies in results.

Contextual Notes

Participants highlight the importance of distinguishing between different moment arms and the components of forces, but there are unresolved assumptions regarding the geometry and specific conditions of the crane setup.

newbphysic
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rbd6X51.jpg

1. Homework Statement
W engine = 500 kg * 9.81 = 4905 N
sin t = 3/5

Homework Equations

The Attempt at a Solution


1. use section AE
ΣMa = 0

3/5 * F<sub>cd</sub> *2 - 4905 * 3 = 0

F = 12262.5 N

2. use section EB

ΣMe = 0

3/5 * F<sub>cd</sub> *1 - 4905 * 2 = 0

F = 16350 N

why 1 and 2 doesn't result same answer ?
 
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newbphysic said:
rbd6X51.jpg

1. Homework Statement
W engine = 500 kg * 9.81 = 4905 N
sin t = 3/5

Homework Equations

The Attempt at a Solution


1. use section AE
ΣMa = 0

3/5 * F<sub>cd</sub> *2 - 4905 * 3 = 0

F = 12262.5 N

2. use section EB

ΣMe = 0

3/5 * F<sub>cd</sub> *1 - 4905 * 2 = 0

F = 16350 N

why 1 and 2 doesn't result same answer ?
Beats me.

You should write the equations of static equilibrium for this crane rather than try to jump into the middle of a calculation.

Try drawing a free body diagram for the beam AEB and putting that member in equilibrium.
 
SteamKing said:
Beats me.

You should write the equations of static equilibrium for this crane rather than try to jump into the middle of a calculation.

Try drawing a free body diagram for the beam AEB and putting that member in equilibrium.

There is only 3 equations right ?
qDWn1Vk.png

ΣMoment =0
Fy * distance - weight * distance = 0

Σshear = 0
Σnormal = 0
 
newbphysic said:
There is only 3 equations right ?
qDWn1Vk.png

ΣMoment =0
Fy * distance - weight * distance = 0

Σshear = 0
Σnormal = 0
There are two equations of static equilibrium: ∑F = 0 and ∑M = 0.

The force equation can be applied in the horizontal or the vertical direction, but generally only one moment equation can be written.
 
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SteamKing said:
There are two equations of static equilibrium: ∑F = 0 and ∑M = 0.

The force equation can be applied in the horizontal or the vertical direction, but generally only one moment equation can be written.
ok, so moment is caused by force in the vertical direction so formula for ΣM = Fy * d - W * d

For moment at point A
ΣMa = 0
Fy *d - W * d = 0
sin( t) * Fcd - W *d
(3/5 * Fcd *2) - (4905 * 3) = 0

F=12262.5N
is this correct ?

If i cut to segment EB
rxZcgoL.png

sin t will be the same since angle t is not change

ΣM = 0
newbphysic said:
ΣMe = 0

3/5 * Fcd *1 - 4905 * 2 = 0

F=16350N

is my formula correct ?
 
newbphysic said:
ok, so moment is caused by force in the vertical direction so formula for ΣM = Fy * d - W * d

For moment at point A
ΣMa = 0
Fy *d - W * d = 0
sin( t) * Fcd - W *d
(3/5 * Fcd *2) - (4905 * 3) = 0

F=12262.5N
is this correct ?
Your calculation for F is correct. However, you need to calculate the components of F in order to find the internal forces acting at point E.

I would not use the same letter "d" for both moment arms as this suggests each arm is the same distance.
If i cut to segment EB
rxZcgoL.png

sin t will be the same since angle t is not change

ΣM = 0is my formula correct ?
 

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