Question 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.)
Figure 5.18
Chapter Laws Of Motion Page 111
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Conveyer belt is accelerating with 1 m/s². A man is stationary with respect to horizontal conveyor belt
Hence, acceleration of man = acceleration of belt = 1m/s²
Hence ,
Net force on the man = ma
= 65× 1 = 65 N direction of this force is opposite to direction of conveyor belt .
Coefficient of friction between the man's shoes and belt = 0.2
Let A is the acceleration of the belt can the man continue to be stationary relative to the belt .
In equilibrium condition,
Friction force = mA
0.2 × m × g = mA
A = 0.2×g = 0.2 × 10 = 2m/s²
Hence, acceleration of man = acceleration of belt = 1m/s²
Hence ,
Net force on the man = ma
= 65× 1 = 65 N direction of this force is opposite to direction of conveyor belt .
Coefficient of friction between the man's shoes and belt = 0.2
Let A is the acceleration of the belt can the man continue to be stationary relative to the belt .
In equilibrium condition,
Friction force = mA
0.2 × m × g = mA
A = 0.2×g = 0.2 × 10 = 2m/s²
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