Question

A man is planning to go out with his friends in a car of mass $1500\text{kg}$ via roadway. The road contains a hump of radius of curvature $30\text{m}$. Each of the friends including the man weighs an average of $80\text{kg}$. A force more than $5000\text{N}$ will damage the car seriously. If he plans to go at a maximum speed of $15\text{m/s}$, what is the maximum number of people that can ride in the car?

• A. $7$
• B. $8$
• C. $6$
• D. $5$

Answer 1

The correct option is C $6$


FBD of car when it comes in contact with hump:


Hump it is convex upwards, so centre of circle lies below the wheel of car. Applying Newton's $2^{nd}$ law in radial direction.
$m_{T}g-N=\frac{m_T{v}^2}{R}$
Hence, $N=(m_c+nm)(g-\frac{v^2}{R})$

where
$N=$ Force exerted by hump on the car
$R=$ Radius of curvature of circle = radius of hump $=30\text{m}$
$m_c=1500\text{kg}=$ Mass of car
$m=80\text{kg}=$ Avg. mass of each person
$n=$ Number of people
$m_T=m_c+nm=$ Total mass of system ($\text{kg}$)

On the applying condition of maximum force experienced by car, for safety purpose:
$N\le 5000$
$\Rightarrow (m_c+nm)(g-\frac{v^2}{R})\le 5000$
$\Rightarrow (1500+80n)(10-\frac{{15}^2}{30})\le 5000$
$\Rightarrow (1500+80n)\left(2.5\right)\le 5000$
$\therefore n\le \frac{2000-1500}{80}$
or, $n\le 6.25$
Hence, maximum number of people that can ride in the car is 6.