A conveyor belt is moving horizontally with a uniform velocity of 2 m/s. Material is dropped at one end at the rate of 5 kg /s and discharge at the other end. Neglecting friction, the force required to move the belt is [in Newtons]
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Answer. Answer: The force generated on the conveyor belt due to falling of the particles will be F = Vdm/dt where the velocity is 2 m/s and the mass per unit time is 5 kg/s so the force will be 2*5 = 10 N.
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