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  • 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

Q5.25 Figure 5.18 shows a man standing stationary for a horizontal conveyor belt that is accelerating 1ms^{-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.)

Answers (1)

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The net force on the man is given by : 

                                                                     F\ =\ ma

 or                                                                       =\ 65\times 1\ =\ 65\ N

The maximum force that can be exerted is given by :

                                            F_r\ =\ \mu mg

or                                                  =\ 0.20(650)\ =\ 130\ N

So the maximum acceleration is :

                                                         ma_{max}\ =\ \mu mg

or                                                           a_{max}\ =\ \mu g

or                                                           a_{max}\ =\ 0.2\times 10\ =\ 2\ m/s^2

Posted by

Devendra Khairwa

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