5.25 Figure 5.18 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.)
Fig. 5.18
Given,
Mass of the lady, m = 60 kg
Acceleration, a = 1 m/s2
Coefficient of friction, µ = 0.2
The total force acting on the boy, F = ma = 65 x 1 =65 N
The boy will continue to be stationary relative to the treadmill as long as the total force on the lady is equal to or less than the frictional force FS, provided by the treadmill
F’ = FS
ma’ = µmg
a’ = 0.2 x 10 = 2 m/s2 .
Thus, the lady can stand stationary relative to the treadmill up to an acceleration of 2 m/s2.
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