# Free Body Diagram Finding Moments

• steve2510
In summary, the homework statement says that the sum of the forces on a certain object must equal zero. The forces are in the up and down directions, and the moment of one of the forces is F*(d- perpendicular distance from the point to the line of force). The perpendicular distance between the point and the line of force is zero, so the sum of the moments is also zero.
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)

## 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??

Last edited by a moderator:
welcome to pf!

hi steve2510! welcome to pf!
steve2510 said:
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).

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

Last edited:
steve2510 said:
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

tiny-tim said:
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

hi steve2510!

(just got up :zzz:)
steve2510 said:
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

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

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

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

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.

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

## 1. What is a free body diagram (FBD)?

A free body diagram is a simplified representation of a physical system or object, showing all the external forces acting on it. It is used to analyze the forces and moments acting on an object in order to determine its motion or equilibrium.

## 2. How do you draw a free body diagram?

To draw a free body diagram, start by identifying the object or system you want to analyze. Then, draw a dot or a box to represent the object and label it with the appropriate symbol. Next, draw arrows to represent all the forces acting on the object, making sure to label each force with its magnitude and direction. Lastly, draw a coordinate system if needed and include any known dimensions or angles.

## 3. What is a moment in a free body diagram?

A moment, also known as torque, is a turning or twisting force that causes an object to rotate about a fixed point. In a free body diagram, moments are represented by curved arrows and are labeled with their magnitude and direction. Moments can be either clockwise or counterclockwise, and their value is calculated by multiplying the force by the distance from the point of rotation.

## 4. How do you find the moments in a free body diagram?

To find the moments in a free body diagram, first identify the point about which the object is rotating. Then, for each force acting on the object, determine its distance from the point of rotation and its perpendicular component to that distance. Multiply the force by its perpendicular distance to the point of rotation to calculate the moment. Finally, add up all the moments to determine the net moment acting on the object.

## 5. What is the importance of finding moments in a free body diagram?

Finding moments in a free body diagram is important because it helps us understand the rotational motion of an object and its equilibrium conditions. By analyzing the moments, we can determine if an object will rotate or remain in equilibrium, and we can also calculate the required forces to maintain or change its rotational motion.

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