Question

A man standing stationary with respect to a horizontal conveyer belt which is accelerating with $1{\text{m/s}}^2$ as shown in figure. If the coefficient of static friction between man's shoes and the belt is $0.2$, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man $=65\text{kg}$). Take $g=10{\text{m/s}}^2$.

• A. $1{\text{m/s}}^2$
• B. $2.5{\text{m/s}}^2$
• C. $3{\text{m/s}}^2$
• D. $2{\text{m/s}}^2$

Answer 1

The correct option is D $2{\text{m/s}}^2$

Mass of man, $m=65\text{kg}$
Normal reaction on man, $N=mg=65\times 10=650\text{N}$
Let $a$ be the maximum acceleration of the conveyor belt.

$\Rightarrow$In order to remain stationary w.r.t conveyor belt, friction will act at its maximum value $f={\mu }_{s}N$ to provide necessary force for the man to move with same acceleration $a$ as that of conveyor belt.
$\therefore$Applying newton's $2^{\text{nd}}$ law along direction of acceleration,
$f=ma$
$\Rightarrow {\mu }_{s}N=ma$
Or, $a=\frac{{\mu }_{s}N}{m}$
$\therefore a=\frac{0.2\times 650}{65}=2{\text{m/s}}^2$