Could you please solve this mechanics problem for me?

Question

Gravel is dropped into a conveyor belt @ 0.4 kg/s. Find the extra force required to keep the belt moving at a speed of 4 m/s.

My doubt is, since the gravel is being dropped vertically, it should have no affect on the horizontal motion of the belt, as vertical & horizontal components of any vector quantity do not affect each other. So, please explain the answer & rectify me if I'm wrong.

Answer 1

Let the conveyor belt, having a system mass (Ms) be moving with a constant speed (V) under the influence of force (F1) applied on it generated by the electrical energy supplied to it. The work done on assembly (electrical work) is converted into its rotational kinetic energy with constant speed of V. It is closed mass system i.e. not trading mass so far. Under steady state v = V

Now let gravel be dumped “vertically” onto the belt, at entry point, from a height H with a mass flow rate (m*). This will add to the mass of the system, which in-turn will become – An open mass-system. The rate of change of momentum (dp/dt) will depend on mass entered, mass left and the net force acting on the assembly.

It may be noted that as gravel falls vertically onto belt, gravel’s horizontal component of velocity (along belt arrow) would be zero. So for the belt-assembly this new mass would start from rest (v-initial = 0). As the new system has donned additional momentum-dynamics, the force (electrical input) on the system will have to be increased to F2 so that same constant speed (V) is maintained. This V is generally maintained in belts because the output of conveyor has to be regulated so that dump at exit does not become too voluminous to manage (in industrial works, one has to carve out proper dump areas and these may require prior approvals from environmental clearances). Let there be no friction between belt and gravel, and mass flow rate is m* at entry and exit (same).

Under new steady state dp/dt = 0
= Net force – [m* (speed)exit - m* (speed)entry]
0 = (F2-F1) – [m* (speed)exit - m* (speed)entry]
0 = (F2-F1) – [m* (V) - m* (0)]
So F2-F1 = extra force = m* (V) = 0.4 * 4
= 1.6 kg-meter/sec^2

Extra expenditure goes in increasing the speed of gravel at exit.

* Explanation in question: It may be noted that vertical speed of gravel in this case is not moving or contributing to belt (V). But, if gravel is dumped from an angle (thita < 90 degree from belt line), then horizontal component of gravel will oppose V of belt and thus more FORCE will be needed. Oppositely, if (thita > 90 degree, then gravel fall will assist V or less Force or expenditure will be there.

Answer 2

Yes, but consider the horizontal component.

The gravel will need to be accelerated from 0 m/s to 4 m/s
From Newton's second law, this will require a (horizontal) force.