Q. 254.0( 15 Votes )

# 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

Answer :

Given:

Mass of person = 65 kg

Co-efficient of static friction, � = 0.2

Acceleration of the belt, a = 1 ms^{-2}

The net force F, acting on the man is given by Newton’s second law of motion as,

F_{net} = ma …(1)

Where,

m = mass of body

a = acceleration of body

By putting the values in equation (1) , we get

F_{net} = 65 Kg× 1 ms^{-2}

F_{net} = 65 N

The man on the conveyer will remain stationary with respect to conveyer until the force exerted on him is less than the frictional force between him and conveyer,

We can write,

F_{net} = f_{s}

m× a_{max} = �× m× g

→ a_{max} = �× g …(2)

Where,

� = coefficient of static friction

g = acceleration due to gravity

By putting the values in equation (2) we get,

→ a_{max} = 0.2× 10ms^{-2}

→ a_{max} = 2 ms^{-2}

As long as the acceleration of conveyer doesn’t exceed 2ms^{-2}, the man will stand stationary.

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