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Chapter 4: Problem 100

When only the air of a sand-blasting gun is turned on, the force of the air on a flat surface normal to the stream and close to the nozzle is \(20 \mathrm{N}\). With the nozzle in the same position, the force increases to \(30 \mathrm{N}\) when sand is admitted to the stream. If sand is being consumed at the rate of \(4.5 \mathrm{kg} / \mathrm{min}\), calculate the velocity \(v\) of the sand particles as they strike the surface.

Short Answer

Expert verified
The velocity of the sand particles is 133.33 m/s.

Step by step solution

01

Identify Relevant Forces

First, identify the relevant forces acting on the surface. The force due to air alone is given as 20 N, and the force when both air and sand are used is given as 30 N.
02

Calculate Force Due to Sand

The force due to the sand particles can be found by subtracting the force with only air from the force with both air and sand. This gives:\[F_{sand} = F_{total} - F_{air} = 30 \, \mathrm{N} - 20 \, \mathrm{N} = 10 \, \mathrm{N}\]
03

Apply Momentum Principle

Using the momentum principle: the force exerted by the sand is equal to the rate of change of momentum of the sand. This can be expressed as:\[F_{sand} = \dot{m} \times v\] where \(\dot{m}\) is the mass flow rate of sand and \(v\) is the velocity of the sand particles.
04

Convert Mass Flow Rate to SI Units

Convert the sand flow rate from \(\mathrm{kg} / \mathrm{min}\) to \(\mathrm{kg} / \mathrm{s}\):\[\dot{m} = \frac{4.5 \, \mathrm{kg/min}}{60 \, \mathrm{s/min}} = 0.075 \, \mathrm{kg/s}\]
05

Solve for Velocity

Substitute the known values into the momentum equation to solve for \(v\):\[10 \, \mathrm{N} = 0.075 \, \mathrm{kg/s} \times v \v = \frac{10 \, \mathrm{N}}{0.075 \, \mathrm{kg/s}} = 133.33 \, \mathrm{m/s}\]

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Key Concepts

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

Momentum Principle
The momentum principle is a key concept in physics that relates force to the change in momentum of an object. In simple terms, it states that the force exerted on an object is equal to the rate of change of its momentum. For sand particles being emitted from a sand-blasting gun, this principle helps determine how the exerted force translates to particle velocity. Momentum is given by the product of mass and velocity. Therefore, a change in these factors will result in a change of momentum.
  • Momentum formula: \( p = m \cdot v \)
  • Rate of change of momentum equates to force: \( F = \frac{\Delta p}{\Delta t} \)
This principle not only aids in understanding how forces operate but also supports calculations that predict and control various physical processes like sand-blasting dynamics.
Force Calculation
Forces involved in sand-blasting include those exerted by the air and the sand particles. Calculating these forces helps you understand the impact energy delivered onto a surface. In this exercise, understanding the incremental force due to sand alone is crucial. First, only the force due to air is identified. Then, the total force when both sand and air are combined is measured. By subtracting the air-only force from the total, the force from the sand can be isolated:
  • Force exerted by air: \( F_{air} = 20 \, \mathrm{N} \)
  • Total force: \( F_{total} = 30 \, \mathrm{N} \)
  • Force from sand: \( F_{sand} = F_{total} - F_{air} \)
  • Result: \( F_{sand} = 10 \, \mathrm{N} \)
Understanding these forces better equips students to dissect similar problems in the future.
Velocity Determination
The velocity of the sand particles is crucial in understanding their impact during the blasting process. Using the calculated sand force and the momentum principle, you can determine how fast these particles strike the surface. The equation unfolds from the realization that force equals the rate of change of momentum:\[ F_{sand} = \dot{m} \times v \]Where:
  • \( F_{sand} \) is the force due to sand, which is 10 N
  • \( \dot{m} \) represents the mass flow rate of the sand
  • \( v \) is the velocity of the sand particles
Plugging in the values available, the velocity is then solved as follows:\[ v = \frac{10 \, \mathrm{N}}{0.075 \, \mathrm{kg/s}} \approx 133.33 \, \mathrm{m/s} \]This shows how dynamics operate at high velocities in processes like sand-blasting.
Mass Flow Rate Conversion
Understanding mass flow rate conversion simplifies the translation between different units of measurement. In this scenario, we initially have the sand mass flow rate in kilograms per minute. To effectively apply the momentum principle in calculations, converting to kilograms per second is essential.The conversion process is a straightforward modification that ensures consistency in unit usage across calculations,requiring a division by 60, since there are 60 seconds in a minute:
  • Initial flow rate: \( 4.5 \, \mathrm{kg/min} \)
  • Convert to SI units: \( \frac{4.5 \, \mathrm{kg/min}}{60 \, \mathrm{s/min}} = 0.075 \, \mathrm{kg/s} \)
This conversion is critical in achieving accurate results when substituted back into the calculations using the momentum principle.

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Most popular questions from this chapter

The mass \(m\) of a raindrop increases as it picks up moisture during its vertical descent through still air. If the air resistance to motion of the drop is \(R\) and its downward velocity is \(v,\) write the equation of motion for the drop and show that the relation \(\Sigma F=d(m v) / d t\) is obeyed as a special case of the variable-mass equation.

Three monkeys \(A, B,\) and \(C\) weighing \(20,25,\) and 15 Ib, respectively, are climbing up and down the rope suspended from \(D\). At the instant represented, \(A\) is descending the rope with an acceleration of \(5 \mathrm{ft} / \mathrm{sec}^{2}\) and \(C\) is pulling himself up with an acceleration of \(3 \mathrm{ft} / \mathrm{sec}^{2} .\) Monkey \(B\) is climbing up with a constant speed of \(2 \mathrm{ft} / \mathrm{sec}\). Treat the rope and monkeys as a complete system and calculate the tension \(T\) in the rope at \(D\).

The rocket shown is designed to test the operation of a new guidance system. When it has reached a certain altitude beyond the effective influence of the earth's atmosphere, its mass has decreased to \(2.80 \mathrm{Mg},\) and its trajectory is \(30^{\circ}\) from the vertical. Rocket fuel is being consumed at the rate of \(120 \mathrm{kg} / \mathrm{s}\) with an exhaust velocity of \(640 \mathrm{m} / \mathrm{s}\) relative to the nozzle. Gravitational acceleration is \(9.34 \mathrm{m} / \mathrm{s}^{2}\) at its altitude. Calculate the \(n\) - and \(t\) -components of the acceleration of the rocket.

The 50,000 -lb flatear supports a 15,000 -lb vehicle on a \(5^{\circ}\) ramp built on the flatear. If the vehicle is released from rest with the flatear also at rest, determine the velocity \(v\) of the flatcar when the vehicle has rolled \(s=40 \mathrm{ft}\) down the ramp just before hitting the stop at \(B\). Neglect all friction and treat the vehicle and the flatcar as particles.

The experimental ground-effect machine has a total weight of 4200 lb. It hovers 1 or \(2 \mathrm{ft}\) off the ground by pumping air at atmospheric pressure through the circular intake duct at \(B\) and discharging it horizontally under the periphery of the skirt \(C\). For an intake velocity \(v\) of \(150 \mathrm{ft} / \mathrm{sec},\) calculate the aver age air pressure \(p\) under the 18 -ft-diameter machine at ground level. The specific weight of the air is \(0.076 \mathrm{lb} / \mathrm{ft}^{3}\).
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