Question

A car moves at constant speed on a road as shown in the figure. The normal force by the road on the car is when it is at point A and B respectively. Information to decide the relation of

A. ${{N}_{A}}={{N}_{B}}$
B. ${{N}_{A}}>{{N}_{B}}$
C. ${{N}_{A}}<{{N}_{B}}$
D. insufficient

Answer 1

In this question we are being asked to find the relation between ${{N}_{A}}$ and ${{N}_{B}}$. From the figure, it is clear that radius of curve a is less than radius of curve b. ${{N}_{A}}$ and ${{N}_{B}}$ are the normal forces at point A and B. We know that, formula for calculating the normal force on a curved road, states the relation between velocity and the radius of the curve. The velocity is not given as constant at both points.

Formula used:
$N=mg-\dfrac{m{{v}^{2}}}{r}$
Where,
Mg is the weight of the car
V is the velocity
And r is the radius of the curve

Complete step-by-step answer:
From the figure, we can say that radius of curve a say ${{r}_{A}}$is less than radius of curve B say ${{r}_{B}}$. We know that, weight of the car will remain constant at all times. The velocity at point a and point b is given as constant
Now,
We know from formula,
$N=mg-\dfrac{m{{v}^{2}}}{r}$
We can say that the normal force N is directly proportional to square of velocity v and inversely proportional to the radius of curve r.
Now, the normal force at point a say ${{N}_{A}}$ can be given as
${{N}_{A}}=mg-\dfrac{m{{v}_{A}}^{2}}{{{r}_{A}}}$ ……………. (1)
Similarly, the normal force at point b say${{N}_{B}}$ can be given as
${{N}_{B}}=mg-\dfrac{m{{v}_{B}}^{2}}{{{r}_{B}}}$ ………………………….. (2)
Now let us assume that ${{N}_{A}}$ = ${{N}_{B}}$
Therefore, from (1) and (2)
We can say that,
$mg-\dfrac{m{{v}_{A}}^{2}}{{{r}_{A}}}=mg-\dfrac{m{{v}_{B}}^{2}}{{{r}_{B}}}$
On solving we get,
$\dfrac{{{v}_{A}}^{2}}{{{r}_{A}}}=\dfrac{{{v}_{B}}^{2}}{{{r}_{B}}}$
But we know that velocity is constant
Therefore,
$\dfrac{1}{{{r}_{A}}}=\dfrac{1}{{{r}_{B}}}$
But from the diagram given, we know that ${{r}_{A}}$ < ${{r}_{B}}$
Also, we know that the normal force N is inversely proportional to radius r
Therefore, we can say that our assumption is wrong.
Since ${{r}_{A}}$ < ${{r}_{B}}$, we can now say that
${{N}_{A}}>{{N}_{B}}$, since the normal force is inversely proportional to r

So, the correct answer is “Option B”.

Note: The normal force is said to be the support force. For a normal force to act on a body the body must be in contact or lying on the surface. The normal force for static objects on a horizontal plane is usually taken as the weight of the object i.e. mg. In such cases normal force is counteracting the weight of the object.