Statics: Crane Boom - Why Different Results?

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The discussion focuses on analyzing the static equilibrium of a crane boom with two different sections, AE and EB, leading to different force calculations. The calculations yield forces of 12,262.5 N for section AE and 16,350 N for section EB, raising questions about the discrepancies. Participants emphasize the importance of using free body diagrams and correctly applying the equations of static equilibrium, specifically ΣF = 0 and ΣM = 0. It is noted that the same letter should not be used for different moment arms to avoid confusion. The calculations for the forces are confirmed as correct, but further analysis of force components is necessary to understand the internal forces at point E.
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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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