How Do You Calculate the Total Moments for Various Points on a Beam?

[lemon]

1. Calculate the total moment of the forces on the thin beam shown in the attached file, about each of the points A, B, C, D, E.
Each of the distances AB, BC, CD, DE is 1m.



2.Clockwise moments = Anit-clockwise moments


3.(Axd)+(Exd)+[50sin(45)]=(20xd)+25
20+20+42.5=20+25
82.5=45
=37.5N in the upwards direction
There is also a force of 50cos(45)+5N=26.3+5=31.3N to the right

I have a sinking feeling about my method here
Please advise?

 

[lemon]

I am uploading a new image with this problem reconsidered:
I didn't take into account that the force needs to be multiplied by the "perpendicular" distance from the point.
So If I have calculated this correctly:
Drawing my line of force I get a (45°, 45°, 90°) triangle.

sinx=hypotenuse/1
=√2/2

F_D√2/2x50N=35.4N (2s.f.)

I now have to work out how to take the total moment of the forces

 

[lemon]

To calculate the total moments of the forces on the beam, do I just add and subtract to see what the resultant force is?

10N-20N-25N+35.4N+10N=10.4N

There is also a 5N east. Not sure what to do with this

 

[lemon]

So if I have a resultant force of 10.4N in the upwards direction, and I also have a force in the due east direction, do I just need to find the resultant of these forces?

c²=a²+b²
c²=10.4²+5²
=11.5 (3s.f.), North of east

 

[PhanthomJay]

Forget resultant forces, keep it simple. And you are confusing Force (in Newtons) with moments (in Newton-meters) . Break up the force applied at an angle (the 50 N force)into its x and y components. Then the moment of each force about the given point is the y component of that force times the perpendicular distance of that force to the point. Add each one up to get the total moment. Note that forces or force components in the x direction do not produce any moments about the points, because there is no perpendicular moment 'arm' distance. Also, when you reply to your own posts before someone has had a chance to respond, you stand a good chance of your question being ignored, since helpers think that someone else is responding and helping. Good thing I was watching..

 

[lemon]

Also, when you reply to your own posts before someone has had a chance to respond, you stand a good chance of your question being ignored, since helpers think that someone else is responding and helping. Good thing I was watching..

Yeah it's difficult. Thanks. I have so much work to get through so I'm kinda working my way through a lot of questions at the same time here, as the response is not immediate, of course. I'm new to physics too, so I'm really on uneasy ground with what I'm doing and finding it hard and confusing. When looking further into a question and realizing I have done things the wrong way I immediately want to change here as not to waste anyones time.