Understanding the Force Required on a Conveyor Belt

This question asks us to determine the force needed to keep a conveyor belt moving at a constant speed while sand is falling onto it at a steady rate. The key physics principle involved here is the concept of momentum and how force relates to the rate of change of momentum, especially in systems where mass is being added.

Physics Principle: Force and Momentum Change

According to Newton's second law, the net force applied to an object is equal to the rate of change of its momentum (\(\text{F} = \frac{\text{dp}}{\text{dt}}\)). In this scenario, the momentum of the system (the belt plus the sand on it) changes because the mass of the sand on the belt is continuously increasing. To maintain a constant velocity, a force must be applied to give the newly added sand the same velocity as the belt.

If 'm' is the mass of the sand that has accumulated on the belt at any time 't', and 'v' is the constant velocity of the belt, the momentum of this accumulated sand (relative to a stationary frame) is \(\text{p} = \text{mv}\).

The force required to maintain the constant velocity 'v' is the rate at which the momentum of the system is changing due to the added mass. Since 'v' is constant, the rate of change of momentum is given by:

\(\text{F} = \frac{\text{dp}}{\text{dt}} = \frac{\text{d}}{\text{dt}}(\text{mv})\)

Since velocity 'v' is constant, we can take it out of the derivative:

\(\text{F} = \text{v} \frac{\text{dm}}{\text{dt}}\)

Here, \(\frac{\text{dm}}{\text{dt}}\) is the rate at which mass (sand) is being added to the belt.

Applying the Given Values

We are given:

• Rate of sand falling onto the belt, \(\frac{\text{dm}}{\text{dt}} = 0.1 \text{ kg/s}\)
• Uniform speed of the conveyor belt, \(\text{v} = 2 \text{ m/s}\)

Now, we can calculate the required force using the derived formula:

\(\text{F} = \text{v} \frac{\text{dm}}{\text{dt}}\)

Substitute the given values:

\(\text{F} = (2 \text{ m/s}) \times (0.1 \text{ kg/s})\)

\(\text{F} = 0.2 \text{ N}\)

Therefore, a force of 0.2 N is required to keep the conveyor belt moving at a uniform speed of 2 m/s while sand is falling on it at a rate of 0.1 kg/s.

Summary of Calculation

Formula used: \(\text{F} = \text{v} \frac{\text{dm}}{\text{dt}}\)

Calculation: \(\text{F} = (2 \text{ m/s}) \times (0.1 \text{ kg/s}) = 0.2 \text{ N}\)

Revision Table: Key Concepts

Additional Information: Force and Changing Mass Systems

Problems involving changing mass systems, like the conveyor belt with falling sand or a rocket expelling fuel, are common in mechanics. Newton's second law (\(\text{F} = \frac{\text{dp}}{\text{dt}}\)) is the fundamental principle, but its application needs careful consideration of how momentum changes due to both velocity changes and mass changes.

For a system with mass 'm' and velocity 'v', the momentum is \(\text{p} = \text{mv}\). The force is the derivative with respect to time:

\(\text{F} = \frac{\text{d}}{\text{dt}}(\text{mv})\)

Using the product rule of differentiation, we get:

\(\text{F} = \text{m}\frac{\text{dv}}{\text{dt}} + \text{v}\frac{\text{dm}}{\text{dt}}\)

In our conveyor belt problem:

• \(\text{m}\) is the instantaneous mass on the belt (which increases over time).
• \(\frac{\text{dv}}{\text{dt}}\) is the acceleration of the belt. The problem states the belt moves at a uniform speed, so \(\frac{\text{dv}}{\text{dt}} = 0\).
• \(\text{v}\) is the constant velocity of the belt.
• \(\frac{\text{dm}}{\text{dt}}\) is the rate at which mass is added to the belt.

Substituting \(\frac{\text{dv}}{\text{dt}} = 0\) into the equation, we get:

\(\text{F} = \text{m}(0) + \text{v}\frac{\text{dm}}{\text{dt}}\)

\(\text{F} = \text{v}\frac{\text{dm}}{\text{dt}}\)

This confirms the formula used in the solution. The force required is specifically to give the incoming sand (which initially has zero horizontal velocity) the horizontal velocity 'v' of the belt. This force acts on the belt, and the belt in turn accelerates the sand horizontally. The vertical component of the sand's momentum changes due to gravity and the normal force from the belt, but the force required to maintain the *horizontal* velocity of the belt is determined by the horizontal momentum transfer.