In your second case, the maximum force exerted by the truck will be the
tire coefficient of friction times the weight portion of the truck on the driven wheels (hence, if it is a 4WD, the total weight of the truck).
For example, tires on gravel (CoF = 0.60) with a 2WD truck weight 4000 lb and 45% of its weight on the rear axle:
0.60 X 0.45 X 4000 lb = 1080 lb
Like AlephZero said, if the chain pulls the truck up (chain goes up from the truck to the post), the effective weight will be decrease; if it pulls the truck down (chain goes down from the truck to the post), the effective weight will be increase.
For the first case, in addition of the previous force calculated, you need to add the force produced by the energy of the truck (= ½ X truck mass X (truck velocity)²). You could find its speed after 15 ft of acceleration
here to determine the energy available. The force produced could be defined as the energy divided by the displacement. Unfortunately, that displacement is the key variable that is almost impossible to determine. Assuming your truck and the post are also perfectly rigid, you would need to know how much the chain would elongate with that much energy. Your chain is then looked at as a spring with some elasticity. The equations would be as follow:
[itex]K=\frac{2E}{x^2}[/itex]
and
[itex]F=Kx=\frac{2E}{x}[/itex]
Where [itex]E[/itex] is the energy stored in the moving truck, [itex]x[/itex] is the chain elongation, [itex]K[/itex] is the chain «spring stiffness» and [itex]F[/itex] is the added force produced by the moving truck.
If you knew the «spring stiffness» of your chain, then you could also determine the force with [itex]F=\sqrt{2EK}[/itex].
Unfortunately, your truck and post are not rigid and it will lower the equivalent «spring stiffness» of you set-up (or increase the total displacement, if you prefer). By how much? that's the «almost impossible» part of the equation.