### Apparent weight of moving car over a convex or concave bridge

##### July 31, 2019

Apparent weight of moving car over a convex or concave bridge is the amount of * Normal Reaction* the car recives while travelling over the bridge.

Here the system consists of the car, the bridge which is either convex or concave and the motion of the car along the curvature.

So to start off we can write,

*Apparent weight of car = N (normal reaction)*In these cases,

*is to be considered because the car is moving along a circular path, while travelling through the concave or convex curvature,*

**centripetal force**## Apparent weight of moving car over a convex bridge

Lets consider the motion of the motor car over a **convex bridge**, along the segment of a circle (i.e. the circular curvature). *Say, the car is just at the midpoint of that circular segment.*

As the path of the car at that point is a part of a circle, we can say that the car is having a circular motion along that circular curvature.

And we know that to continue along the circular path, without deviating tangentially, we need a force called centripatal force.

Centripetal force directs towards the center of the circular path.

For the **convex bridge**, the **centripetal force** is provided by the difference of *weight mg of the car* and the *normal reaction N* of the bridge.

Pls note that the center of the circle is below the curvature in this case.

∴ for *convex bridge*

mg – N = mv^{2}/ r

or **N = mg – mv**^{2}**/ r **

Clearly N < mg, i.e., the apparent weight of the moving car is less than the weight of the stationary car.

## Convex bridge – car is tilted

Now if the car is in a position like the figure below over a convex bridge, then how to find out the N?

Here,

mg cosθ – N = mv^{2}/ r

or **N = mg** cosθ ** – mv**^{2}**/ r **

## Apparent weight of moving car over a Concave bridge

Lets consider the motion of the motor car over a concave bridge, along the segment of a circle (i.e. the circular curvature) . *Say, the car is just at the midpoint of that circular segment.*

Like the above case, here also centripetal force acts on the car to retain it along the circular path. But unlike the above case, this time the center of the circular path is above the bridge.

For the **concave bridge**, the **centripetal force** is provided by the difference of the *normal reaction N* of the bridge and *weight mg of the car*.

So for *Concave bridge *

N – mg = mv^{2}/ r

**Apparent weight N = mg + mv**^{2}**/ r**

Clearly N > mg, i.e., the apparent weight of the moving car is more than the weight of the stationary car.

## Concave bridge – car is tilted

Now if the car is in a position like the figure below over a concave bridge, then how to find out the N?

Here, N – mg cosθ = mv^{2}/ r**Apparent weight N = mg **cosθ **+ mv**^{2}**/ r**