Sand from a stationary hopper falls on a moving conveyor belt at the rate of 4.2
ID: 2181359 • Letter: S
Question
Sand from a stationary hopper falls on a moving conveyor belt at the rate of 4.20 kg/s as shown in the figure below. The conveyor belt is supported by frictionless rollers and moves at a constant speed of v = 0.630 m/s under the action of a constant horizontal external force supplied by the motor that drives the belt. Find the sand's rate of change of momentum in the horizontal direction. Find the force of friction exerted by the belt on the sand. Find the external force Find the work done by in 1 s. Find the kinetic energy acquired by the falling sand each second due to the change in its horizontal motion. Why are the answers to parts (d) and (e) different?Explanation / Answer
a) The mass of sand falling on the belt in time dt is
1) dm = dm/dt * dt
where dm/dt is the given rate at which the sand fall of the conveyor belt.
With a good approximation, we can assume that the initial momentum ofsand fallen on the belt in time dt has a zero value. Only the sand fallingon the belt changes its momentum changes its momentum in that time.Since the change is in the along the belt (x-direction), we can limitconsideration to one component. The change in total momentum of thesand is therefore
2) dPx/dt = 0 dm = v x dm/ dt *dt
where vx is v x dm/dtthe speed of the conveyor belt. Hence the rate of change inthe total momentum of the sand is
3) dPx/dt = v x dm/dt = 4.20 * 0.630 = 2.646N
b) There are three forces exerted on the sand but only the frictionalforce is not balanced. The net force is therefore equal to the frictionalforce. Hence, from Newton’s second law of motion (version 3), thex-component of the (total) frictional force exerted on the sand is 2.65 N.
c) From Newton’s third law of motion, the sand exerts an opposite,frictional force with the x-component of -2.65 N. According to Newton’ssecond law of motion, the net force exerted on the conveyor belt is zero,which requires that the motor exert a force with the x-component of 2.65 N.
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