Figure below shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1ms–2. If the coefficient of static friction between the man’s shoes and the belt is 0.2, upto what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg).

As the man is standing stationary w.r.t. the horizontal conveyor belt, he is also accelerating at 1ms–2, the acceleration of the belt. Thus, Acceleration of the man, a =1ms–2 Mass of the man,M = 65 kgTherefore, net force on the man,ma = 65 x1= 65 NThe limiting friction between the shoes of the man and the belt is given byF=μMg=0.2×65×9.8NIf, the man can remain stationary upto an acceleration say ‘a’, thenMa′=For a′=FM=0.2×65×9.865=1.96ms–2