Free Body Diagram Finding Moments

[steve2510]

Homework Statement

I have to find the forces FGB FCB AND FGH
http://desmond.imageshack.us/Himg696/scaled.php?server=696&filename=img59wa.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)

Homework Equations

The Attempt at a Solution

Sum the forces up and down = 5+10 -21.25 +FGHsin26.57 - FGB sin38.87 = 0Then 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 I am a bit stuck and i don't no if the force Fcb would have an effect??

 

[tiny-tim]

welcome to pf!

hi steve2510! welcome to pf!

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 I am a bit stuck and i don't no if the force Fcb would have an effect??

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]

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

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]

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

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 :zzz:)

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.

but isn't the perpendicular distance zero in any direction?

And i thought 1.5 was the mutliplier as the horizontal force acts along the top of the a4 line

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!