Statics - Crane boom

  • Thread starter newbphysic
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  • #1
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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

[itex] 3/5 * Fcd *2 - 4905 * 3 = 0 [/itex]

[itex] F = 12262.5 N [/itex]

2. use section EB

ΣMe = 0

[itex] 3/5 * Fcd *1 - 4905 * 2 = 0 [/itex]

[itex] F = 16350 N [/itex]

why 1 and 2 doesn't result same answer ?
 

Answers and Replies

  • #2
SteamKing
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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

[itex] 3/5 * Fcd *2 - 4905 * 3 = 0 [/itex]

[itex] F = 12262.5 N [/itex]

2. use section EB

ΣMe = 0

[itex] 3/5 * Fcd *1 - 4905 * 2 = 0 [/itex]

[itex] F = 16350 N [/itex]

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.
 
  • #3
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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
[itex] Fy * distance - weight * distance = 0 [/itex]

Σshear = 0
Σnormal = 0
 
  • #4
SteamKing
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There is only 3 equations right ?
qDWn1Vk.png

ΣMoment =0
[itex] Fy * distance - weight * distance = 0 [/itex]

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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  • #5
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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
[itex] sin( t) * Fcd - W *d [/itex]
[itex](3/5 * Fcd *2) - (4905 * 3) = 0 [/itex]

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
ΣMe = 0

[itex] 3/5 * Fcd *1 - 4905 * 2 = 0 [/itex]

F=16350N
is my formula correct ?
 
  • #6
SteamKing
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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
[itex] sin( t) * Fcd - W *d [/itex]
[itex](3/5 * Fcd *2) - (4905 * 3) = 0 [/itex]

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 = 0


is my formula correct ?
 

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