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.)
Given: Acceleration of conveyer belt is 1 ms – 2 , coefficient of static friction between the man’s shoes and the belt is 0.2 and mass of the man is 65 kg .
The net force acting on the man is given as,
F = m a
Where, m is mass of man and a is acceleration of conveyor belt.
By substituting the values in the above equation, we get
F = 65 × 1 = 65 N
Thus, the net force on the man is 65 N .
The frictional force exerted by the belt on the man is given as,
F r = μ m g
Where, μ is the coefficient of static friction and g is acceleration due to gravity.
By substituting the given values in the above equation, we get
F r = 0.2 × 65 × 10 = 130 N
The man will remain stationary with respect to the conveyor belt until the net force on him is less than or equal to the frictional force exerted by the belt.
F ′ ≤ F r m a ′ = F r a ′ = F r m
By substituting the values in the above equation, we get
a ′ ≤ 130 65 a ′ ≤ 2 ms − 2
Thus, the maximum acceleration of the belt up to which the man can stand stationary is 2 ms − 2 .
Given: Acceleration of conveyer belt is
The net force acting on the man is given as,
Where,
By substituting the values in the above equation, we get
Thus, the net force on the man is
The frictional force exerted by the belt on the man is given as,
Where,
By substituting the given values in the above equation, we get
The man will remain stationary with respect to the conveyor belt until the net force on him is less than or equal to the frictional force exerted by the belt.
By substituting the values in the above equation, we get
Thus, the maximum acceleration of the belt up to which the man can stand stationary is
