Let's take the question seriously, even if as it is phrased it is totally devoid of physical meaning.
We can use the Principle of Equivalence as follows.
Put yourself in a reference frame in free vertical fall on Earth.
Furthermore, assume there is no bridge, and that the car driver relies on its speed to jump over the river.
The question now becomes: how much will the car fall during the jump?
In the free-falling frame, the car will not fall at all: it will move horizontally, as in this frame gravity is absent.
So, reverting to an earth-bound frame, we can safely assert that the car will have a vertical acceleration equal to $g$ (independent of its mass and speed).
Let $s$ be the bridge span: then the car will fall $h=(g s^2)/(2 v^2)$.
Just for fun, let's put some numbers. Speed $v$ is given; as to $s$, the longest suspension bridge on Earth does not reach 2000 m, and the longest one under construction barely exceeds this length. So we may take $s=2000\rm m$.
We find $h=2.7\times10^{-10}\rm m$: the size of an atom.
The driver would hardly care of such a drop.
Anyhow, to answer the question: if you want to cancel the drop, you will have to apply a vertical force causing an acceleration opposite to gravity. The SR formula relating force to transverse acceleration is $F=m\gamma a$, and the force the bridge must apply to the car to sustain it is $m\gamma g$, greater than the maximum load the bridge can bear.
Would it mean that the bridge would collapse? Not at all! The maximum load is a static one, whereas the stress due to the car would last $s/v=0.74\mu\rm s$. Surely the bridge would not even notice the passage of the car.
Someone could doubt about the above reasoning, arguing that we should get in the full GR machinery.
This I can't do here, but the result would be the same.
