Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1\; m s^{-2}$. What is the net force on the man? If the coefficient of static friction between the 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 $= 65 \;kg.)$
Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1\; m s^{-2}$. What is the net force on the man? If the coefficient of static friction between the 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 $= 65 \;kg.)$
Mass of the man, $m=65\, kg$
Acceleration of the belt, $a=1 \,m / s ^{2}$
Coefficient of static friction, $\mu=0.2$
The net force $F$, acting on the man is given by Newton's second law of motion as:
$F_{m}=m a=65 \times 1=65\, N$
The man will continue to be stationary with respect to the conveyor belt until the net force on the man is less than or equal to the frictional force $f_{s}$, exerted by the belt, i.e., $F_{ na }^{\prime}=f_{s}$
$m a^{\prime}=\mu m g$
$\therefore a^{\prime}=0.2 \times 10=2\, m / s ^{2}$
Therefore, the maximum acceleration of the belt up to which the man can stand stationary is $2 \,m / s ^{2}$
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- [JEE MAIN 2023]