In US there are zones of maximum wind velocity and its pressure.

Majority of the zones assume (15 to 35) lbs/sqr ft pressure for heights of up to 60 feet, depending on combination of factors such as zone, exposure, utility of structure. Lets use just for illustration pick wind pressure of 20 #/sf.

We will have an overturning momentum of $$ m_{overturning} = (20\times 4 \times 10)_{total\space p} \times (10/2)_{arm} = 4000 \space lbsft \space moment$$

We need to counter this by adding enough width and weight to the base to resist this.

As a first guess say we pick the base to be 3 feet deep. then 4000/3= 1660lbs so you need a ballast of 2\times 1660 =3300 lbs at the center of base, which is not practical.

We add the depth to 5 feet and get the ballast= 4000/2.5 = 800.

I think there are reduction factores for temporary uses such as yours but this is just to get a feeling of the magnitude and order of wind pressure.

In US there are zones of maximum wind velocity and its pressure.

Majority of the zones assume (15 to 35) lbs/sqr ft pressure for heights of up to 60 feet, depending on combination of factors such as zone, exposure, utility of structure. Lets use just for illustration pick wind pressure of 20 #/sf.

We will have an overturning momentum of $$ m_{overturning} = (20\times 4 \times 10)_{total\space p} \times (10/2)_{arm} = 4000 \space lbsft \space moment$$

We need to counter this by adding enough width and weight to the base to resist this.

As a first guess say we pick the base to be 3 feet deep. then 4000/3= 1660lbs so you need a ballast of 2\times 1660 =3300 lbs at the center of base, which is not practical.

We add the depth to 5 feet and get the $ballast= 4000/2.5 = 2 \times 800 lbs.$

I think there are reduction factores for temporary uses such as yours but this is just to get a feeling of the magnitude and order of wind pressure.