Statics/Moments - Force perpendicular to slope?

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To determine the resultant moment at support points A and B for the truss, it is essential to calculate the X and Y components of forces acting at angles. For the 10kN force, it should be noted that while its X-component is neglected in calculations for point B, the Y-component must be included. The angle used for these calculations is 45 degrees, leading to the X-component being 10cos(45) and the Y-component being 10sin(45). Understanding the relationship between forces and their components is crucial for accurate moment calculations. Properly applying these principles will aid in resolving the moments at both support points.
AndrewZzz
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Determine the resultant moment at support points A and B due to the forces acting on the truss as shown.




Fx = F(cos∅), Fy = F(sin∅), M = F(distance)



Point A -88.98kN*m
Point B ?


I have no problem calculating most of the forces, but I'm lost when it comes to knowing what to do with the forces that are perpendicular to the sloped surfaces. Do I calculate X and Y components for them? Does it automatically turn into an X or Y component? When trying to calculate the moment on point B, is the 10kN multiplied by zero because it's on the same x-axis as the point, or is the 10kN a Y component? Some insight would be hugely appreciated.
 

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Yes, for any forces that are at an angle, generate the x and y components of the force. And good catch, the 10kN force will be multiplied by zero because it is on the same x-axis as point B (when calculating the reaction at B). However, only the x-component of that force is neglected in the calculations, be sure to include the y-component.
 
I figured it would still require the X and Y components! Which angle would I use to calculate those?
 
The angle should be 45degrees so the x-component is 10cos(45) and the y-component is 10sin(45). At least I hope it is haha.
 
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