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Free Body Diagram Finding Moments

by steve2510
Tags: body, diagram, free, moments
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steve2510
#1
Mar6-12, 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 –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??
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tiny-tim
#2
Mar7-12, 12:34 PM
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hi steve2510! welcome to pf!
Quote Quote by steve2510 View Post
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??
You also have to sum the forces left and right (to equal zero).
steve2510
#3
Mar7-12, 01:20 PM
P: 36
Okay So this is what I've got
ƩFx = Fcb + Fghcos26.57 + Fgbcos-36.87 = 0

ƩFy = 21.25 + Fghsin26.57 - Fgbsin-36.87 -5 - 10 = 0

ƩFy = Fghsin26.57 - Fgbsin-36.87 =- 6.25

ƩMb = (Fghcos26.57 x1.5 )+ (Fgbcos-36.87 x 1) +(2x5) -(2x21.25) = 0

ƩMb = (1.5 x Fghcos26.57 ) + (Fgbcos36.87) = 32.25

ƩMb = (Fghcos26.57 ) + (Fgbcos36.87) = 21.5 (divided by 1.5)

ƩFx = Fcb + Fghcos26.57 + Fgbcos-36.87 = 0
-((Fghcos26.57 ) + (Fgbcos36.87)) = 21.5
= Fcb = -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
#4
Mar7-12, 06:01 PM
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Free Body Diagram Finding Moments

Quote Quote by steve2510 View Post
ƩMb = (Fghcos26.57 x1.5 )+ (Fgbcos-36.87 x 1) +(2x5) -(2x21.25) = 0
shouldn't the moment of FGB about B be 0 ?

and the multiplier of FGH needs to be the distance from B to GH
steve2510
#5
Mar7-12, 06:25 PM
P: 36
Quote Quote by tiny-tim View Post
shouldn't the moment of FGB about B be 0 ?

and the multiplier of FGH needs to be the distance from B to GH
Am i right in saying Fgb 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
#6
Mar8-12, 04:20 AM
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hi steve2510!

(just got up )
Quote Quote by steve2510 View Post
Am i right in saying Fgb 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
#7
Mar8-12, 05:12 AM
P: 36
Cos is the horizontal component for both forces , am I right ?
tiny-tim
#8
Mar8-12, 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
steve2510
#9
Mar8-12, 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
tiny-tim
#10
Mar9-12, 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.
steve2510
#11
Mar9-12, 05:37 AM
P: 36
okay think its back to the library then, thank you for your help!


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