Statics: Crane Boom - Why Different Results?

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newbphysic
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Diagram of a crane boom: a horizontal beam pinned to a vertical wall at A, supported by a diagonal cable from D on the wall to C, with points A, E, C and B spaced 1 m apart, a 1.5 m height marked on the wall, and a hanging engine at B.

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 * F<sub>cd</sub> *2 - 4905 * 3 = 0[/itex]

[itex]F = 12262.5 N[/itex]

2. use section EB

ΣMe = 0

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

[itex]F = 16350 N[/itex]

why 1 and 2 doesn't result same answer ?
 
newbphysic said:
Diagram of a crane boom fixed to a wall at A, with a cable from D to C supporting it, a 1.5 m height marked on the wall, points A, E and B spaced 1 m apart, and a hanging engine at B.

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 * F<sub>cd</sub> *2 - 4905 * 3 = 0[/itex]

[itex]F = 12262.5 N[/itex]

2. use section EB

ΣMe = 0

[itex]3/5 * F<sub>cd</sub> *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.
 
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 ?
Free-body diagram of a horizontal beam with supports at point A and point B, labelled 1m segments, a 1.5 dimension on an inclined orange line, an upward force Fy and a downward load W.

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

Σshear = 0
Σnormal = 0
 
newbphysic said:
There is only 3 equations right ?
Free-body diagram of a horizontal beam from A to B with a support at A, a force Fy acting upward at a point with a sloped orange line of 1.5 m height, a downward load W in red at B, and segment lengths of 1 m and 1 m marked along the beam with point E at the midpoint.

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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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
[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
Free-body diagram of a horizontal beam from E to B with a force Fy acting upward at an angled cable attached near the middle, a downward load W at B, labelled dimensions 1 m and 1, and a coordinate axis with N, V and M at the left end E.

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

ΣM = 0
newbphysic said:
ΣMe = 0

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

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
[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
Free-body diagram of a horizontal beam with end E at left and end B at right, a vertical force Fy applied upward at 1 m from E, a downward weight W at B, and an inclined orange line at the top-left. Blue axes labelled v and N and a brown moment arrow labelled M sit at E, with a blue length bracket labelled 2 below.

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

ΣM = 0is my formula correct ?