newton's second law of motion - 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.)

Sandra (last edited 13 days ago) (Student, UG1 Grade)

Given that

Mass of the stationary man, m = 65 kg

Acceleration of the conveyor belt a = 1 m/s^2

Coefficient of the static friction \mu = 0.2

Acceleration due to gravity a = 9.8 m/s^2

Normal force is balanced by the weight of the man(mg)

N = mg

There is only one force acting on the man, and it is caused by the belt's acceleration.

F_{net} = F

Step 1: Calculating the net force acting on the man

According to the newton's second law of motion

Force, F = ma

F_{net} = ma

F_{net} = 65 kg * 1 m/s^2 = 65 N

Net force acting on the man = 65 N

Step 2: Find the maximum belt acceleration at which a man can remain stationary F_s = \mu N = \mu mg

The man will remain motionless in respect to the conveyor belt until his net force is less than or equal to the belt's frictional force,( F_s ). F_s = \mu N = \mu mg

F_{net} = F_s

ma = \mu mg

a = \mu g

a = 0.2 * 9.8 m/s^2

a = 1.96 m/s^2

Hence, the maximum belt acceleration at which a man can remain stationary a = 1.96 m/s^2