Question

A car is moving with constant speed on a road as shown in figure. The normal reaction by the road on the car is $N_A,N_B$ and $N_C$ when it is at the points A, B and C respectively. Then:

• A. $N_A=N_B$
• B. $N_A>N_B$
• C. $N_A<N_B$
• D. $N_A=N_C$

Answer 1

The correct option is B $N_A>N_B$

Let us suppose constant speed of car is $u$ and mass is $m$.


Car is in vertical equilibrium in each case. Hence $\sum F_y=0$
Case - $1$) when car is at point A
$mg-N_A=\frac{m{u}^2}{R_A}$
$\Rightarrow N_A=mg-\frac{m{u}^2}{R_A}----\left(1\right)$

Case - $2$) when car is at point $B$
$N_B=mg-\frac{m{u}^2}{R_B}---\left(2\right)$

Case - $3$) when car is at point $C$
$N_C=mg-\frac{m{u}^2}{R_C}---\left(3\right)$

From diagram, we can see that
$R_C>R_A>R_B$
$\Rightarrow \frac{m{u}^2}{R_C}<\frac{m{u}^2}{R_A}<\frac{m{u}^2}{R_B}$
$\Rightarrow mg-\frac{m{u}^2}{R_C}>mg-\frac{m{u}^2}{R_A}>mg-\frac{m{u}^2}{R_B}$
$\Rightarrow N_C>N_A>N_B$
[from equation 1, 2 & 3]
Normal Reaction at point A is greater than at point B.