Question

A truck of mass $3000\text{kg}$ is travelling on a road with a velocity of $90\text{km/h}$. The road has a pit which is concave upward. The approximate radius of curvature of the pit is $40\text{m}$. What is the force ($\text{kN}$) experienced by the truck while crossing through the pit? Take $g=10{\text{m/s}}^2$.

• A. $80.5\text{kN}$
• B. $97\text{kN}$
• C. $76.87\text{kN}$
• D. $67.57\text{kN}$

Answer 1

The correct option is C $76.87\text{kN}$


FBD of wheel when it enters pit:


Given $v=90\text{km/h}=25\text{m/s}$

Here pit is concave upwards, so centre of circle lies above the wheel. Applying Newton's $2^{nd}$ in radial direction:

$N-mg=\frac{m{v}^2}{R}$.........($1$)
where
$R=\text{radius of curvature for wheel}$
$\text{= radius of pit=40 m}$
$m=\text{mass of truck=3000 kg}$
$N=\text{Normal force exerted by ground on truck}$

Rearranging the Eq.($1$), we get:
$N=mg+m\frac{v^2}{R}$
$\therefore N=3000\times 10+\frac{3000\times {25}^2}{40}=76,875\text{N}=\frac{76875}{1000}=76.875\text{kN}$
Hence the truck experiences a force of $76.875\text{kN}$