Figure shown a man standing stationary with respect to a horizontal converyor belt that is accelerationg with `1m//s^(-2)` . What is the net force on the man?If the coefficient of ststic friction between the man's shoes and the belt is `0.2` upto what maximum acceleration of the belt can the man continue to be stationary relative to the belt? Mass of the man`= 65kg (g =9.8m//s^(2))`

Explanation

Here acceleration of conveyor belt a = `1 ms^(-2)` , `mu_s` = 0.02 and mass of man m = 65 kg.t As the man is in an accelerating frame, he experiences a pseudo force `F_s`= ma as shown
in fig (a) Hence to maintain his equilibrium , he excerts a force F = `F_s` = ma = `65 xx 1 `= 65 N in forward direction i.e direction of motion of belt .
`therefore` Net force acting on man = 65 N (forward)
As shown in fig(b) , the man can continue to be stationary with respect to belt ,if force of friction `mu_s`N = `mu_s`mg = `ma_max`
`a_max` = `mu_s` .g = 0.2 `xx` 10 = 2`ms^(-2)`