Hi all,
Why was the thickness of the beam considered when moment was being calculated and why was the horizontal component of the 600 N force used in calculating the moment? That part is unclear to me. Please explain it. Am I looking for the logic behind using the horizontal comp. and the...
If you consider the horizontal component of Fhc, you have a vertical perpendicular distance of 1. This would be much easier for me to understand if you explained why we have to use a distance of 3 meters when Fhc is not 3 metres away from A, please explain this.
Since Fhc is at an angle, do we consider both its components when we are looking at the moment about point A? i.e. Fhc(1.5)sin(gamma) - Fhc(1)cos(gamma)? Gamma is the angle of Fhc
Yes, you take 2/sqrt(13) and multiply it by 3 to get said 1.66 m, I am asking why you are multiplying by 3 instead of 1 (the vertical perpendicular distance between A and Fhc)
Taking the moments about points A and C, why is a distance of 3 meters used in Fhc and Fhg? Isn’t it supposed to be 1.5 meters for both?
I acknowledge that Fhc and Fhg have x and y components, but were we considering the horizontal or vertical component (or both components) of either Fhc and Fhg...
Sum of forces in the y-direction = 0 and downwards is +ve
P + Fab,y = 0
P + Fab (4/5) = 0
Fab = -1.25P
ẟ = FL/AE -> ẟab = FabLab/AabE
ẟab = (-1.25P*.75)/(pi*(.01)^2*(200*10^3)) = -0.0149P
After this step, I am uncertain of how I can relate the vertical elongation with AB's elongation to find...
I cannot seem to start answering the question as a result of the path not being provided. How do I solve this when the path is not provided? See picture below
Induction motors:
pf = 0.8
triangle:
θ = arccos(0.8) = 36.86 degrees
Pa = 500/0.8 = 625 VA
Pr = sqrt(625^2-500^2) = 375 VAR
Synchronous motors:
pf = 0.707
triangle:
θ = arccos(0.707) = 45 degrees
Pa = 500/0.707 = 707 VA
Pr = sqrt(707^2-500^2) = 499.85 VAR
I am uncertain of how I can...