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Question: Prove that a motor car moving over a convex bridge is lighter than the same car resting on the same ...

Prove that a motor car moving over a convex bridge is lighter than the same car resting on the same bridge.

Explanation

Solution

In this question we have been asked to prove that moving cars on a convex surface is lighter than the stationary car on the same surface. Therefore, to solve this we will evaluate both the given instance and calculate the forces by the stationary and moving car. After calculating the forces for both instances, we can prove the given statement. To solve this, we will be using the equation of motion.

Complete answer:
Let us first consider the part where the car is resting stationary on the convex surface as shown in the diagram below.

         ![](https://www.vedantu.com/question-sets/0af45e10-408b-46fc-aea2-fc05a1d9999b7535554044565086112.png)

Now, from the diagram the we know that only force on stationary car will be the force due to gravity and to balance this will be the normal force
Therefore, the equation of motion can be given as
N = mg …………… (1)
Now, let the car be moving on the convex surface with velocity v as shown in the figure below

             ![](https://www.vedantu.com/question-sets/1c7d6b6b-7f45-4d96-a1d7-2d4c2359c68d4806700553519794504.png)

Now, the forces on the moving car will be the force due to gravity and the force due to the centripetal acceleration caused by the velocity v.
Therefore, the equation of motion can be given as,
Na=mg−mv2r{{N}_{a}}=mg-\dfrac{m{{v}^{2}}}{r} …………….. (2)
Therefore from (1) and (2)
N>NaN > {{N}_{a}}
Therefore, we can say that the moving car on the convex surface is lighter than the stationary car on the same surface.

Note:
The normal force is the support force that is exerted on every object having any mass and resting on any surface. The normal force is opposite in direction and equal in magnitude to the weight of the body that is resting stationary on a surface. For example, if a block is resting on the floor the force exerted by the block on the floor will be its weight and the balancing Normal force would be equal in magnitude and in upward direction.