Mass required to prevent sign falling over with a set wind load - activity stations for disabled children I'm currently working on my thesis and I'm stuck on a question. I'm designing activity stations for disabled children to be used for equine therapy. 
The stand is 9ft tall and I've calculated the the wind load at 0.94lbs, now I need to calculate the mass required to stop the station falling over (the concrete base is not a foundation sunk in the ground - it needs to remain portable) 
The frontal area of the stand is evenly spread. Any help would be appreciated.  
 A: Assuming a square base of width $w$ with mass $M$, and a horizontal wind load $F_w$ at a height $h$, then the condition for static equilibrium is
$$\sum\tau = 0$$
$$\tau_{wind} + \tau_{base} = 0$$
Since if the sign tips, it would rotate around the edge of the base, that's a convenient axis about which to compute the torques:
$$- F_w \frac{h}{2} + M g \frac{w}{2} = 0$$
Solving for M...
$$M = \frac{F_w h}{g w}$$
Plugging in your numbers (0.94 lbs-force = 4.18 N, 9 ft = 2.7 meters),
$$M = \frac{11.29 ~\rm Nm}{9.81 ~\rm{m/s^2} ~w} $$
$$M = \frac{1.15 ~\rm kg~m}{w} $$
If you prefer lbs and feet,
$$M = \frac{8.32 ~\rm lbs~ft}{w}$$
For example, if your square base has a width of 3 feet, then you need a minimum of $M = 2.77 ~\rm lbs$
Note 1: This analysis assumes the mass of the sign itself is negligible compared to the base. This is a conservative assumption, since extra mass on the sign will make it more stable for initial tipping.
Note 2: I'm very skeptical that a sign of any significant size would only experience a wind load of 0.94 lbs in any significant wind. I would double-check that figure.
EDIT: I revised my answer now that the OP made it clear that the sign is a rectangle that extends from the ground up to 9 feet.
A: 
The weight of the base needed to prevent toppling of the stand also depends on it's width $x$.
In order to prevent toppling about the point $P$, there must not be a net moment about that point.
This means mathematically that:
$$F\frac{H}{2}=mg\frac{x}{2}$$
Where $H=9\:\mathrm{ft}$ and $F=0.04\:\mathrm{lbs}$. I'm assuming the load you assigned acts on the centre of gravity of the vertical panel and that the mass of the panel is negligible.
So the mass $m$ required is:
$$m=\frac{FH}{gx}$$
Note that $F$ and $H$ need to be converted to S.I. units, if you want to use $g=9.81\:\mathrm{ms^{-2}}$ as the Earth's acceleration.
