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Chapter 8: Problem 35

Sand drops at a rate of \(2000 \mathrm{~kg} / \mathrm{min}\) from the bottom of a stationary hopper onto a belt conveyer moving horizontally at \(250 \mathrm{~m} / \mathrm{min}\). Determine the force needed to drive the conveyer, neglecting friction. [Hint: How much momentum must be imparted to the sand each second?]

Short Answer

Expert verified
The force needed is approximately 138.89 N.

Step by step solution

01

Understand the problem

We are given that sand drops at a rate of 2000 kg/min onto a conveyor belt moving at 250 m/min. We need to calculate the force needed to keep the conveyor moving at this speed as sand is continuously added. The hint suggests considering the change in momentum of the sand.
02

Calculate Mass Flow Rate

First, we convert the mass flow rate into kg/s from kg/min. Given the sand drops at 2000 kg/min, converting this to kg/s involves dividing by 60, because there are 60 seconds in a minute.\[ \text{Mass flow rate} = \frac{2000 \, \text{kg}}{60 \, \text{s}} = \frac{200}{6} \, \text{kg/s} \approx 33.33 \, \text{kg/s} \]
03

Determine Velocity of Sand

Next, we identify the velocity at which the sand needs to move as it lands on the conveyor belt. This is given directly as 250 m/min. We will also convert this to m/s.\[ \text{Velocity} = \frac{250 \, \text{m/min}}{60 \, \text{s/min}} = \frac{25}{6} \, \text{m/s} \approx 4.167 \, \text{m/s} \]
04

Calculate Change in Momentum per Second

The change in momentum per second is given by the product of the mass flow rate and the velocity. This is essentially the force required.\[ \text{Force} = \text{Mass flow rate} \times \text{Velocity} = 33.33 \, \text{kg/s} \times 4.167 \, \text{m/s} \approx 138.89 \, \text{N} \]
05

Conclusion

Since momentum is mass times velocity, the force required to drive the conveyor is the change in momentum per unit time. From our calculations, we find that approximately 138.89 N of force is needed to move the sand continually alongside the conveyor belt.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Flow Rate
When discussing mass flow rate, we are referring to the amount of mass passing through a given point per unit time. It is a fundamental concept in fluid dynamics and is vital in understanding how materials like sand are transferred on conveyor systems.
In the given problem, sand drops at a rate of 2000 kg per minute onto a conveyor belt. To work with this system in terms of force and dynamics, it's easier to convert this mass flow rate into kilograms per second.
  • Given: 2000 kg/min
  • Convert minutes to seconds: There are 60 seconds in a minute
  • Mass flow rate in kg/s: \( \frac{2000 \, \text{kg}}{60 \, \text{s}} = 33.33 \, \text{kg/s} \)
This conversion allows us to use the rate directly in further calculations, especially when calculating force related to momentum changes.
Force Calculation
Force is a vector quantity, often calculated as the product of mass and acceleration. However, when dealing with moving objects and momentum changes, it can also be viewed as the rate of change of momentum per second. In our scenario, it's the force required to keep the conveyor belt moving at a constant speed while sand is continuously added.

The change in momentum, related to the force, is the product of the mass flow rate and the speed of the sand moving with the conveyor. By calculating this, we understand the force necessary:
  • Momentum change per second (also the force needed): \( \text{Force} = \text{Mass flow rate} \times \text{Velocity} \)
  • Here, \( \text{Force} = 33.33 \, \text{kg/s} \times 4.167 \, \text{m/s} \approx 138.89 \, \text{N} \)
Thus, approximately 138.89 Newtons of force is needed to maintain the conveyor's speed against the weight of the falling sand.
Velocity Conversion
Velocity is a measure of how fast an object is moving and in what direction. In this problem, we need to ensure all our quantities are in compatible units. The conveyor belt moves horizontally at 250 meters per minute, but since our mass flow rate is in kg/s, it's preferable to also express velocity in meters per second.

Thus, converting from m/min to m/s involves:
  • Given velocity: 250 m/min
  • Convert minutes to seconds: Again, with 60 seconds in a minute
  • Velocity in m/s: \( \frac{250 \, \text{m/min}}{60 \, \text{s/min}} = 4.167 \, \text{m/s} \)
This streamlined unit system lets us proceed with calculations more efficiently and helps directly apply these values in determining the required force.

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Most popular questions from this chapter

A 40000 -kg freight car is coasting at a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) along a straight level track when it strikes a 30000 -kg stationary freight car and couples to it. What will be their combined speed after impact?

A billiard ball moving at a speed \(v_{1 i}\) strikes, head-on, another billiard ball that is at rest. Assuming the collision is completely elastic, show that $$ v_{1 i}=v_{1 f}+v_{2 f} $$

A force of \(1000 \mathrm{~N}\) is applied to a small space satellite for a time of \(10.0\) minutes. If the craft has a mass of \(200 \mathrm{~kg}\), what will be its final speed? [Hint: Be careful with those exponents when using a calculator.]

A ball having a mass of \(0.500 \mathrm{~kg}\) is thrown at a speed of \(20 \mathrm{~m} / \mathrm{s}\). Determine the magnitude of its momentum.

Imagine that a \(1.20-\mathrm{kg}\) hard-rubber ball traveling at \(10.0 \mathrm{~m} / \mathrm{s}\) bounces off a brick wall in an essentially elastic collision. Determine the change in the momentum of the ball. [Hint: What change in momentum will just stop the ball?]
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