Free Body Diagram Finding Moments

steve2510

1. The problem statement, all variables and given/known data
I have to find the forces FGB FCB AND FGH
http://desmond.imageshack.us/Himg696...gif&res=medium
P1 = 5 kN, P2 = 10 kN, JY = 21.25 kN


a1 = 2 m, a2 = 1 m, a3 = 0.5 m, a4 = 1 m, a5 = 2.5 m

Angle FGH makes with the horizontal is: 26.57 degrees


Angle FGB makes with the horizontal is –tan-1(1.5/2)

2. Relevant equations




3. The attempt at a solution

Sum the forces up and down = 5+10 -21.25 +FGHsin26.57 - FGB sin38.87 = 0


Then summing moments =
2x10 - 2 x 21.25 +(a2+a3+a4)FGHcos26.57 + FGB cos 38.87= 0

I don't get what i can do from here and im a bit stuck and i don't no if the force Fcb would have an effect??

tiny-tim

hi steve2510! welcome to pf!
You also have to sum the forces left and right (to equal zero).

steve2510

Okay So this is what I've got
ƩF_x = F_cb + F_ghcos26.57 + F_gbcos-36.87 = 0

ƩF_y = 21.25 + F_ghsin26.57 - F_gbsin-36.87 -5 - 10 = 0

ƩF_y = F_ghsin26.57 - F_gbsin-36.87 =- 6.25

ƩM_b = (F_ghcos26.57 x1.5 )+ (F_gbcos-36.87 x 1) +(2x5) -(2x21.25) = 0

ƩM_b = (1.5 x F_ghcos26.57 ) + (F_gbcos36.87) = 32.25

ƩM_b = (F_ghcos26.57 ) + (F_gbcos36.87) = 21.5 (divided by 1.5)

ƩF_x = F_cb + F_ghcos26.57 + F_gbcos-36.87 = 0
-((F_ghcos26.57 ) + (F_gbcos36.87)) = 21.5
= F_cb = -21.5

Am i doing this right at the moment ? I'm not sure if i took moments about the correct point and whether the sin and cos are correct

tiny-tim

shouldn't the moment of F_GB about B be 0 ?

and the multiplier of F_GH needs to be the distance from B to GH

steve2510

Am i right in saying F_gb is the force that g exerts on b ? So there would be a moment force in the x direction but not in the y direction as the perpendicular distance would be 0. And i thought 1.5 was the mutliplier as the horizontal force acts along the top of the a4 line

tiny-tim

hi steve2510!

(just got up )
but isn't the perpendicular distance zero in any direction?
i was referring to the cos

steve2510

Cos is the horizontal component for both forces , am I right ?

tiny-tim

you seem to be mixing up the methods for components of force (in a direction) and moments of force (about an axis or point)

cos is for components

this is moments

with moments, it's usually sin

steve2510

I'm confused as the sin of both forces don't produce moments around B ? Is there any chance you could write what the sum of the moments is meant to equal

tiny-tim

The moment of a force F about a point B is F*d,

where d is the perpendicular distance from B to the line of F.

If A and C are any points on that line, then Fd = F*AB*sinBAC.

If you're not familiar with this, you need to go back to your book and study and practise it.

steve2510

okay think its back to the library then, thank you for your help!