Sol. (i) At the highest point of the bridge the equation of motion of the car is Mg – N = m ,

where N is the normal force acting on the car (and the negative of the required answer),
v = 20 ms –1 and ρ is the radius of curvature of the bridge there. The most difficult part of the problem is to find this radius of curvature.

If we could find a motion with this trajectory for which the normal acceleration is well known, the radius of curvature could be easily calculated. For a parabolic trajectory the flight of a projectile offers the required analogue. Let the projectile have an initial velocity of v 0 making an angle α with the horizontal.

The range (d = 100m) and height (h = 5m) of the projectile can be expressed using the initial data.

d = and h = .

The quotient h/d gives tan α = 4h/d (so α ≈ 11.3º), and the horizontal component of the initial velocity is

v x = v 0 cos α = d = 50 ms –1 .

Now the radius of curvature at the highest point can be calculated as ρ = v x 2 /g = 250m.

So the normal force at the highest point is

N = m = 8.40 kN.

(ii) The force exerted on any other part of the bridge can be calculated in the same way, i.e. using the radius of curvature. At a point three-

Quarters of the way across the bridge, the radius of curvature is approximately 254 m and the normal force about 8.37 kN. Away from the centre of there is also tangential (frictional) force; here is value is 995 the net force acting on the bridge is approximately 8.43 kN.