well, you have two moments: the wind acting on the top, and the weight acting on the leeward side of the base, 9/12 of a foot from the base center..
We need distances from the center of the sign to the base, call it height h.
13.2*h-40*(9/12)=net tipping moment
if the net tipping moment is zero, then h= 40*8/(13.2*12) = 2.02 feet.
So if the height of the sign from the base is less than 2 feet, it wont tip. Now if it is more than two feet, it tips.
What if more weight W is added to the base?
13.2*h=(40+W)*(9/12) so put in h, and solve for W
A sign (including its post and base) weighs 40 pounds and is supported by an 18 in by 18 in square base. If the wind applies a force of 13.2 lbs to the sign at the geometric center of the sign surface, will the sign tip over? If so, how much evenly distributed weight (sand bags) should be added to the base of the sign to keep it from tipping when the wind blows?
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