COM BYU 39

Force Analysis

The belt operates under the influence of two primary effects: the driving force from gravity acting on the sand, and the retarding force required to impart momentum to the newly falling sand.

Let $\mu$ be the rate at which sand falls onto the belt ($\mu = \frac{dm}{dt}$).

Let $v$ be the steady speed of the belt.

1. Driving Force ($F_{gravity}$)

The total mass of the sand on the belt at any instant is determined by the flow rate and the time it takes to traverse length $l$.

$$M_{sand} = \lambda \cdot l$$

Since $\mu = \lambda v$, the linear density is $\lambda = \frac{\mu}{v}$.

$$M_{sand} = \frac{\mu l}{v}$$

The component of gravity acting along the incline drives the system:

$$F_{drive} = M_{sand} g \sin \theta = \left(\frac{\mu l}{v}\right) g \sin \theta$$

2. Retarding Force ($F_{inertia}$)

The sand falls with negligible speed and must be accelerated to the belt’s speed $v$. The force required to change the momentum of the incoming sand is:

$$F_{retarding} = \frac{dP}{dt} = v \frac{dm}{dt} = \mu v$$

Steady State Condition

Since the resistive forces of the mechanism and inertia of the belt itself are negligible, the steady speed is reached when the driving force equals the retarding force:

$$F_{drive} = F_{retarding}$$ $$\left(\frac{\mu l}{v}\right) g \sin \theta = \mu v$$

Solving for $v$:

$$v^2 = g l \sin \theta$$