A 1 t car drives at 36 km/h over a convex bridge of radius 40 m. Find the force on the bridge and the take-off speed — Physics problems | PhysLabo
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A car on a convex bridge: force on the bridge and take-off speed

Medium 10 grade Mechanics / Circular Motion
Problem statement

A car of mass $m = 1$ t moves at a speed $v = 36$ km/h over a convex bridge of radius of curvature $R = 40$ m (see the figure). With what force does the car press on the bridge at its highest point? At what speed will the car stop pressing on the bridge at the top (lose contact with it)? Take $g = 10$ m/s².

A car at the top of a convex bridge of radius R; gravity mg, normal force N and velocity v are shown
Given
  • $m = 1$ t $= 1000$ kg
  • $v = 36$ km/h $= 10$ m/s
  • $R = 40$ m
  • $g = 10$ m/s²
  • $N = ?$
  • $v_{max} = ?$
Hints
Hint 1 / 2
At the top the centripetal acceleration $v^2/R$ points downward, so $mg - N = mv^2/R$. Do not forget to convert km/h to m/s.
Hint 2 / 2
The car stops pressing on the bridge when the normal force $N$ becomes zero: then all of gravity goes into the centripetal acceleration.
Solution

At the top the car moves along an arc of radius $R$; its centripetal acceleration $a = v^2/R$ points downward, toward the center of curvature. The forces on the car are gravity $mg$ (downward) and the normal force $N$ from the bridge (upward).

1. Convert the speed: $v = 36$ km/h $= 10$ m/s. Newton's second law along the vertical (downward):

$$mg - N = m\frac{v^2}{R} \;\Rightarrow\; N = m\left(g - \frac{v^2}{R}\right) = 1000\left(10 - \frac{100}{40}\right) = 7500 \text{ N}$$

By Newton's third law the car presses on the bridge with the same force: $7.5$ kN — less than the weight $mg = 10$ kN.

2. Contact is lost when $N = 0$:

$$mg = m\frac{v_{max}^2}{R} \;\Rightarrow\; v_{max} = \sqrt{gR} = \sqrt{10 \cdot 40} = 20 \text{ m/s} = 72 \text{ km/h}$$
Answer

$N = 7.5$ kN; $v_{max} = 20$ m/s $= 72$ km/h

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