A car is driving on a bridge having the shape of a parabola. The bridge height $h = 5m$, horizontal length $l = 60m$. Find the ratio of the force of pressure of the car on the road at the top of the bridge to its weight on a level road, if the car is traveling on the bridge at a constant speed $v = 54 km/h$.
Decision:
A car of mass m at the top of a bridge is subjected to a downward force $m \bar{g}$ of gravity and an upward force $\bar{N}$ of normal pressure from the side of the bridge. From the equation of motion of the car.
$\frac{mv^{2}}{R}=mg-N$,
we get the ratio we're looking for
$\frac{N}{mg}=1-\frac{v^{2}}{gR}$,
where the radius of curvature of the bridge $R$ at the top point is not yet known. Let us determine it from the following considerations. Let a small body be thrown from the top of the bridge in the horizontal direction with such a velocity $v_{0}$ that it flies along a parabola coinciding with the bridge. It is not difficult to obtain that
$v_{0}=l \sqrt{\frac{g}{8h}}=30 \frac{m}{s}$.
But in this case, the acceleration of the body at the top point of the bridge is equal to the acceleration of free fall and is the centripetal acceleration $\frac{v^{2}_{0}}{R}=g$. Hence we obtain
$R=\frac{v^{2}_{0}}{g}=\frac{l^{2}}{8h}$
and the ratio is equal to:
$\frac{N}{mg}=1-\frac{8hv^{2}}{g l^{2}}=0,75$
Decision:
A car of mass m at the top of a bridge is subjected to a downward force $m \bar{g}$ of gravity and an upward force $\bar{N}$ of normal pressure from the side of the bridge. From the equation of motion of the car.
$\frac{mv^{2}}{R}=mg-N$,
we get the ratio we're looking for
$\frac{N}{mg}=1-\frac{v^{2}}{gR}$,
where the radius of curvature of the bridge $R$ at the top point is not yet known. Let us determine it from the following considerations. Let a small body be thrown from the top of the bridge in the horizontal direction with such a velocity $v_{0}$ that it flies along a parabola coinciding with the bridge. It is not difficult to obtain that
$v_{0}=l \sqrt{\frac{g}{8h}}=30 \frac{m}{s}$.
But in this case, the acceleration of the body at the top point of the bridge is equal to the acceleration of free fall and is the centripetal acceleration $\frac{v^{2}_{0}}{R}=g$. Hence we obtain
$R=\frac{v^{2}_{0}}{g}=\frac{l^{2}}{8h}$
and the ratio is equal to:
$\frac{N}{mg}=1-\frac{8hv^{2}}{g l^{2}}=0,75$