# Base length and Mass required to stabilize a large heavy sign [closed]

I have an 8ft x 7.5ft sign that is about 4inches thick and weighs about 150lbs. We can assume the whole thing is made of the same density

I am building the stand for this and I cannot anchor it to the ground. It must be free standing. I want a formula that I can play with wind speed, base length, and weight on the back of the base.

I found this: Mass required to prevent sign falling over with a set wind load - activity stations for disabled children but I only want a stand on the back and not both sides. That changes things, and the sign is pretty heavy where that was not a factor in the other.

What I have figured out so far:

1. Everything needs to be in meters so it is 2.4384 wide X 2.286 high
2. To get the wind force I use air mass times wind speed squared: $F=Am * \rm{a^{2}}$

• $\rm{1.229kg/m^3}$ is sea level air density so I will use that
• surface area is $\rm{2.4384m * 2.286m = 5.5741824 m^2}$
• $\rm{F = 1.229 * 5.5741824 * a^{2}}$ which equals $\rm{F = 6.85 * a^{2}}$
• $m/s^2$ = mph * 0.44704
• 5mph wind would be $\rm{2.2352m/s^2}$
• So a 5mph wind would exert 34.23N -- I will use that for now
3. The sited question yields the following formula and says it is for tipping at the edge of the base. Does that work for this? $$M = \frac{F_w h}{g w}$$ Plugging in numbers ( $F_w =34.23N$, $h=2.286m$, $\rm{g=9.81m/s^2}$ )

$$M = \frac{78.25 Nm}{9.81m/s^2 w}$$

$$M = \frac{7.9765545362 kg~m}{w}$$

or

$$M = \frac{57.70 lbs~ft}{w}$$

so given a 3ft base then I would get 19lb.

Does this need to change to account for the weight of the sign?

Do I need a different equation than the one I am using given the sign is not in the middle of the base? Here is where I am at right now on the build. I'm ready to build the base. I would love to get some science behind the construction. ## closed as off-topic by probably_someone, John Rennie, sammy gerbil, Kyle Kanos, ZeroTheHeroAug 28 '18 at 19:08

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• The pivot point is (initially) vertically below the centre of gravity of the sign, so the weight of the sign does not exert any moment (torque) about the pivot point. It does not become relevant until the sign tips at an angle to the vertical. – sammy gerbil Aug 24 '18 at 7:56
• Thanks @sammy so it sounds like that formula works given I am not trying to hold the sign at an angle. – Eric Aug 26 '18 at 12:36