Finding the Moment of a Force Using Varignon's Theorem

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sevag00
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Hi. I'm really confused in how to find the moment done by force.

I want to find the moment done by the force in this figure.

Diagram of a bracket pinned at point O on a wall, with a 0.4 m horizontal arm and a 0.2 m vertical dimension to the force application point, and a 400 N force F acting at 30 degrees, labelled (a).


The book has resolved the force into its components using principle of moments.

Diagram of a bent bracket attached to a wall at point O, with x and y axes; a 400 N force is resolved into 400 sin 30° N horizontal and 400 cos 30° N vertical components, with dimensions 0.2 m and 0.4 m marked.


So, how do i find the moments of the components of the force? I know I'm going to use the right hand rule, but i don't know where i am going to put my thumb.
 

Discussion

sevag00 said:
Hi. I'm really confused in how to find the moment done by force.

So, how do i find the moments of the components of the force? I know I'm going to use the right hand rule, but i don't know where i am going to put my thumb.

Show your work please.
Anyway,moment =Force x Perpendicular distance
 
I know how to find the magnitude. I'm asking for the direction.
 
sevag00 said:
I know how to find the magnitude. I'm asking for the direction.
If you know it,then show your work.
PF requires one to show his/her work before getting any help(In the homework section):smile:
 
Yeah. But first i should find the direction of the moments of the components of the force so that i get the right result.
 
sevag00 said:
Yeah. But first i should find the direction of the moments of the components of the force so that i get the right result.
You don't need to calculate that.Moment is a turning effect.It does not have any directions.It has direction of rotation.You just have to use common sense and decide whether it's anticlockwise or clockwise
For example,look at this:
Two diagrams labelled (a) and (b), each showing a horizontal line with point P at its left end and a force arrow labelled F acting near the right end: upward in (a) and downward in (b).

Look at diagram a,Force F produces an Anticlockwise moment.
Look at diagram b,Force F now produces a Clockwise moment.

Think about your diagram now
 
Attachments
  • Two diagrams, (a) and (b), each showing a horizontal line with a point P at its left end and an arrow labelled F acting near the right end: in (a) the arrow points up, in (b) it points down.
    Untitled.gif
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Okay. But there is something called right hand rule that you indirectly used to find the direction of (a) and (b).
I want to know how the right hand rule is used.
 
I know what your saying. Your taking a direction vector from point O to the line of action of the force. And by using cross product, you're getting the magnitude and direction. But I'm not using that method. I'm using Varignon's theorem.
 

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