A car of mass m moves at constant speed v (a) on a…

A car of mass $m$ moves at constant speed $v$ (a) on a horizontal road, (b) over a convex bridge with radius of curvature $R$, (c) through a concave bridge with the same radius. Find the force the car exerts on the road (its weight) in each case, when the car is at the middle of the bridge.

Explanation of Method

A car on a flat road, a convex and a concave bridge.

Step 1: (a) Horizontal road

No acceleration, so the road's push balances gravity: $N = mg$.

Step 2: (b) Top of a convex bridge

The centripetal acceleration $v^2/R$ points down, toward the centre of curvature: $mg - N = \frac{mv^2}{R}$.

\[N = mg - \frac{mv^2}{R}\]

Step 3: (c) Bottom of a concave bridge

Now the centripetal acceleration points up: $N - mg = \frac{mv^2}{R}$.

\[N = mg + \frac{mv^2}{R}\]

$N$ $mg$ $N$ $mg$ $N$ $mg$ $N = mg$ $N = mg - \tfrac{mv^2}{R}$ $N = mg + \tfrac{mv^2}{R}$ $\text{(a) flat}$ $\text{(b) convex}$ $\text{(c) concave}$

At the middle of each bridge. The small orange arrow is the direction of the centripetal acceleration: down over a hump, up in a dip. The road's push $N$ (and so the car's weight on the road) is smaller over a hump and larger in a dip.

Result

\[\boxed{\,N = mg\,}\]

\[\boxed{\,N = mg - \frac{mv^2}{R}\,}\]

\[\boxed{\,N = mg + \frac{mv^2}{R}\,}\]

Remark

Note. The original results are correct. By Newton's third law the car presses on the road with the same force $N$, which is its weight. On the convex bridge, for $v \ge \sqrt{gR}$ the formula gives $N \le 0$: the car loses contact at the top.