Q.How does the normal force on a concave suspension bridge change when a vehicle is travelling on it with constant speed?

A concave (sagging) bridge is the BOTTOM of a vertical circle — the mirror case of the convex crest. …

On a concave suspension bridge the roadway sags, so a vehicle at its lowest region is at the bottom of a vertical circular arc: the centre of curvature is ABOVE it. The net UPWARD force must supply the centripetal force, so $N - mg = mv^2/r$ and $N = m\left(g+\frac{v^2}{r}\right)$ — greater than the static weight, and increasing quadratically with speed even though the speed is constant in magnitude (the DIRECTION of the velocity is changing, which is what demands the extra force). This is the exact mirror of the convex over-bridge (§1.4.3), where $N = m(g - v^2/r)$ falls with speed. Consequences: the …

Feel it as apparent weight: at the dip your apparent weight is $(1 + v^2/rg)\,mg$ — for v = 20 m/s and r = 200 m that is a 20 % gain; comp …

Writing $mg - N = mv^2/r$ out of habit from the convex case — the sign flips because the centre of curvature is now ABOVE the vehicle; forgetting that 'consta …