Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor
Question.
Figure $5.18$ shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1 \mathrm{~m} \mathrm{~s}^{-2}$. 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.)

solution:

Mass of the man, m = 65 kg

Acceleration of the belt, $a=1 \mathrm{~m} / \mathrm{s}^{2}$

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:

$F_{\text {nel }}=m a=65 \times 1=65 \mathrm{~N}$

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.,

$F_{\mathrm{net}}^{\prime}=f_{s}$

$m a^{\prime}=\mu m g$

$\therefore a^{\prime}=0.2 \times 10=2 \mathrm{~m} / \mathrm{s}^{2}$

Therefore, the maximum acceleration of the belt up to which the man can stand stationary is $2 \mathrm{~m} / \mathrm{s}^{2}$. [/question]
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