Question

A car moves at a constant speed on a road as shown in figure. The normal force by the road on the car is N_A and N_B when it is at the points A and B respectively. Then?

• A. N_A = N_B
• B. N_A> N_B
• C. N_A< N_B
• D. Insufficient information to decide the relation of N_A and N_B.

Answer 1

The correct option is C

N_A< N_B

$\therefore N_A=mg-\frac{m{v}^2}{R_A}$
$N_B=mg-\frac{m{v}^2}{R_R}$
We see from the diagram that, the car traverses a much smaller curvature at A than at B.
$\therefore R_A<R_B$
$R_A=\frac{m{v}^2}{mg-N_A},R_B=\frac{m{v}^2}{mg-N_R}$
$\frac{m{v}^2}{mg-N_A}<\frac{m{v}^2}{mg-N_B}$
$\Rightarrow mg-N_B<mg-N_A$
$\Rightarrow -N_B<N_A$
$\Rightarrow N_B>N_A$