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Statics street sign, help please?
A uniform rectangular sign is supported by two horizontal beams that are fixed to the top and bottom corners on the right edge (points a&b). The sign has a mass of 5kg. It is 0.5m tall and 0.8m wide. Find the loads acting on point a and point b.
Sum forces x = Fax + Fbx = 0
Sum forces y = Fay + Fby - mg = 0
Sum moments about a:
0.4mg + 0.5Fbx = 0
Fbx = (0.4*5*9.81) / (-0.5)
Fbx = -39.24N
Therefore, Fax = 39.24N
The problem is I can't seem to find the reaction forces in the y direction because i don't know where to take the moment. Please help
1 Answer
- 1 decade agoFavorite Answer
Take the moment about the center of the sign. That way the contribution due to gravity can be ignored (since gravity occurs at the centroid, which for a rectangle, is the center).
Looking at your sketch (which you likely drew the same way I did, since our signs agree for Fax and Fbx) and using the right hand rule:
Fax applies a negative moment, with a moment arm of 0.25 m. Fay applies a positive moment with a moment arm 0.4 m. Fbx applies a positive moment with moment arm 0.25 m (as long as Fbx is negative like you calculated). Fby applies a positive moment with a moment arm 0.4 m.
So summing the moments about the center of the sign,
-0.25Fax + 0.4Fay + 0.25 Fbx + 0.4Fby = 0
And you can solve from there.
Source(s): Myself (3rd year Structural Engineering major)
