Determine the moment about a given point

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To determine the moments about points A, B, and D, the leftmost force at point A produces no moment. The discussion emphasizes breaking down the 52 kN force into its vertical and horizontal components to calculate the moments accurately. The moment of the 93 kN force about point A is calculated using its vertical line of action. Participants are advised to find the angles and distances needed for the calculations, particularly for the components of the 52 kN force. Additionally, a warning was issued regarding a potentially dangerous link in the thread.
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Homework Statement


I'm trying to find the moments about points A,B and D respectively.

wafhA.jpg


Homework Equations


Mx= F * d


The Attempt at a Solution


Correct me if I'm wrong, but on point A the leftmost force produces no moment. After that I'm a little confused about what to use for the distances of the other two forces.
 
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wannawin said:

Homework Statement


I'm trying to find the moments about points A,B and D respectively.

wafhA.jpg


Homework Equations


Mx= F * d


The Attempt at a Solution


Correct me if I'm wrong, but on point A the leftmost force produces no moment.
That is correct.
After that I'm a little confused about what to use for the distances of the other two forces.
Breakup the 52 kN force into its vertical and horizontal vector components, applied where the arrow of the 52 kN force is is touching the line. Then remember the golden rule: The moment of a force, or component of that force, is equal to the force, or component of that force, times the perpendicular distance from the line of action of the force, or component of that force, to the point about which you are summing moments.
For example, looking now at the 93 kN force, its line of action is a vertical line. The moment of that force about A is then 93*(___?__), clockwise. Now calculate the moments for each of the force components of the 52 kN force, and add them all up.
 
you may also need to find the angle in the triangle made after a square.
 
sorry rectangle
 
so I can use 1.8m as the vertical distance, but by the way it looks I guess I have to find out the length of x and add it to 7 in order to find the horizontal distance.

zVEWA.jpg
 
For 52 sin 28 its 1.8 m and for 52 cos 28 you have to find the angle to find the distance which should be added with 7 m
 
When I click on the picture posted in post 1, it says, "This web page has been identified as dangerous. For your own safety, please close this window now, and never return to this web site." Just wanted to warn you guys. It might be dangerous.
 

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