# Statics - Crane boom

1. Mar 6, 2016

### newbphysic

1. The problem statement, all variables and given/known data

W engine = 500 kg * 9.81 = 4905 N
sin t = 3/5

2. Relevant equations

3. The attempt at a solution
1. use section AE
ΣMa = 0

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

$F = 12262.5 N$

2. use section EB

ΣMe = 0

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

$F = 16350 N$

why 1 and 2 doesn't result same answer ?

2. Mar 6, 2016

### SteamKing

Staff Emeritus
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. Mar 6, 2016

### newbphysic

There is only 3 equations right ?
ΣMoment =0
$Fy * distance - weight * distance = 0$

Σshear = 0
Σnormal = 0

4. Mar 6, 2016

### SteamKing

Staff Emeritus
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.

5. Mar 6, 2016

### newbphysic

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

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

ΣM = 0
is my formula correct ?

6. Mar 6, 2016

### SteamKing

Staff Emeritus
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.