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Laws of Motion

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Question : 25 of 40
Marks: +1, -0
Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s−2^{-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)
Solution:  
Here, acceleration of conveyor belt, a=1 m s−2a = 1 \, \mathrm{m} \, \mathrm{s}^{-2}
As the man is standing stationary w.r.t. the belt, so acceleration of the man = acceleration of belt a=1 m s−2a = 1 \, \mathrm{m} \, \mathrm{s}^{-2}
As m=65 kgm = 65 \, \mathrm{kg}
∴ Net force on the man, F=ma=65×1=65 NF = ma = 65 \times 1 = 65 \, \mathrm{N}
Now μ=0.2\mu = 0.2
Force of limiting friction; f=μR=μmgf = \mu R = \mu mg
If the man remains stationary upto maximum acceleration a′ of the belt, then
f=ma′=μmg,f = m a' = \mu m g,
a′=mg=0.2×10=2 m s−2a' = mg = 0.2 \times 10 = 2 \, \mathrm{m} \, \mathrm{s}^{-2}
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