
#1
Mar612, 01:17 PM

P: 36

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 –tan1(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?? 



#2
Mar712, 12:34 PM

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hi steve2510! welcome to pf!




#3
Mar712, 01:20 PM

P: 36

Okay So this is what I've got
ƩF_{x} = F_{cb} + F_{gh}cos26.57 + F_{gb}cos36.87 = 0 ƩF_{y} = 21.25 + F_{gh}sin26.57  F_{gb}sin36.87 5  10 = 0 ƩF_{y} = F_{gh}sin26.57  F_{gb}sin36.87 = 6.25 ƩM_{b} = (F_{gh}cos26.57 x1.5 )+ (F_{gb}cos36.87 x 1) +(2x5) (2x21.25) = 0 ƩM_{b} = (1.5 x F_{gh}cos26.57 ) + (F_{gb}cos36.87) = 32.25 ƩM_{b} = (F_{gh}cos26.57 ) + (F_{gb}cos36.87) = 21.5 (divided by 1.5) ƩF_{x} = F_{cb} + F_{gh}cos26.57 + F_{gb}cos36.87 = 0 ((F_{gh}cos26.57 ) + (F_{gb}cos36.87)) = 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 



#4
Mar712, 06:01 PM

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P: 26,167

Free Body Diagram Finding Momentsand the multiplier of F_{GH} needs to be the distance from B to GH 



#5
Mar712, 06:25 PM

P: 36





#6
Mar812, 04:20 AM

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P: 26,167

hi steve2510!
(just got up ) 



#7
Mar812, 05:12 AM

P: 36

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




#8
Mar812, 05:54 AM

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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 



#9
Mar812, 06:13 AM

P: 36

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




#10
Mar912, 04:09 AM

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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. 



#11
Mar912, 05:37 AM

P: 36

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



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