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Chapter 15: Problem 135

A power lawn mower hovers very close over the ground. This is done by drawing air in at a speed of \(6 \mathrm{~m} / \mathrm{s}\) through an intake unit \(A\), which has a cross-sectional area of \(A_{A}=0.25 \mathrm{~m}^{2}\), and then discharging it at the ground, \(B\) where the cross-sectional area is \(A_{B}=0.35 \mathrm{~m}^{2} .\) If air at \(A\) is subjected only to atmospheric pressure, determine the air pressure which the lawn mower exerts on the ground when the weight of the mower is freely supported and no load is placed on the handle. The mower has a mass of \(15 \mathrm{~kg}\) with center of mass at \(G\). Assume that air has a constant density of \(\rho_{a}=1.22 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The application of Bernoulli's equation and the principle of conservation of mass help to determine air pressure exerted on the ground by a lawn mower when no load is placed on the handle.

Step by step solution

01

Formulate Bernoulli's Equation

Start with the Bernoulli's equation - an equation that can be derived by conserving energy along a streamline in the flow where the forces are conservative. Often it's expressed as: \(P_A + 0.5 * \rho_a * v_A^2 = P_B + 0.5 * \rho_a * v_B^2\). \(P\) represents the pressure, \(v\) represents the velocity, \(\rho_a\) represents the air density, subscripts \(A\) and \(B\) refer to the intake unit and the ground respectively.
02

Calculate velocity at B

Using the principle of conservation of mass, which states that the product of area and velocity at any point in a flow tube must remain constant, we can calculate the velocity at B, \(v_B\). \(A_A * v_A = A_B * v_B\). Now, solve the equation for \(v_B\).
03

Calculate total force exerted on the ground

Using the weight of the mower, we can calculate the total force exerted by the lawnmower on the ground. The weight of the mower, \(F\) is given by \(F = m * g\), where \(m\) is the mass of the mower and \(g\) is the gravitational acceleration. In our case, \(g\) is approximately equal to \(9.8 m/s^2\).
04

Determine the air pressure at B

Initially, we have assumed pressure at A to be atmospheric pressure, So, \(P_A\) is equal to \(0\). By substituting the values of \(P_A\), \(v_A\), \(v_B\) and \(\rho_a\) from steps 1, 2 and 3 into the Bernoulli's equation, we can solve for \(P_B\). This will give us the air pressure exerted by the lawn mower on the ground.

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Key Concepts

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

Fluid Mechanics
Fluid mechanics is the study of the behavior of fluids, which includes both liquids and gases. In our scenario, we focus on air, a gas, and how it interacts with objects, such as a hovering lawn mower. The core idea is to analyze how the lawn mower uses air to stay afloat by manipulating pressure and velocity.

Fluid mechanics involves understanding concepts such as pressure differences, flow velocities, and fluid density. For instance, the air flowing into the mower at a specific speed develops a particular pressure pattern that keeps the mower above the ground. The flow dynamics are influenced by Bernoulli's equation, which links velocity and pressure in moving fluids, assuming incompressible and steady flow.
  • Pressure: the force exerted by the fluid per unit area.
  • Velocity: the speed at which fluid particles move.
  • Density: mass per unit volume of a fluid, which influences its inertia.
Conservation of Mass
The principle of conservation of mass is a fundamental part of fluid mechanics. It states that for any steady flow in a closed system, the mass flowing into the system must equal the mass flowing out. This principle is often represented in fluid dynamics equations, such as the continuity equation, especially useful for incompressible flows like air under regular conditions.

In the case of the lawn mower, air is drawn in and expelled, but the mass of air entering and exiting must remain constant. Thus, we can write: \[ A_A \cdot v_A = A_B \cdot v_B \]where \(A\) is the cross-sectional area and \(v\) is the velocity of the air. Using this equation, you can calculate the exit velocity at point B, given the incoming velocity at A and their respective areas. This approach ensures that the mower does not accumulate or lose more air than it processes, keeping it hovering steadily.
Airflow Dynamics
Airflow dynamics focus on how the air moves and behaves as it flows through different areas of cross-sections. In our exercise, airflow is manipulated by the lawnmower to generate lift, needed to hover. The lawns mower adjusts the velocity from the intake area A to discharge area B to maintain balance and lift.

Understanding airflow dynamics involves the application of Bernoulli's principle, which ties together pressure and velocity changes in a moving fluid. As the mower pulls in air at one velocity and discharges it at another larger cross-section, the velocities change proportionally with the areas due to the continuity equation. This results in a pressure difference which is crucial for maintaining the hover state.
  • Increased discharge area results in decreased velocity, leading to a change in pressure exerted on the ground.
  • Balance between velocity and pressure determines if an object can hover.
  • The lift is a result of dynamically adjusting airflow dynamics to counteract gravitational forces.
This means in simpler terms, that the smart use of airflow dynamics keeps the mower off the ground by carefully balancing pressure and air velocities.

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Most popular questions from this chapter

The water flow enters below the hydrant at \(C\) at the rate of \(0.75 \mathrm{~m}^{3} / \mathrm{s}\). It is then divided equally between the two outlets at \(A\) and \(B\). If the gauge pressure at \(C\) is \(300 \mathrm{kPa}\), determine the horizontal and vertical force reactions and the moment reaction on the fixed support at \(C\). The diameter of the two outlets at \(A\) and \(B\) is \(75 \mathrm{~mm}\), and the diameter of the inlet pipe at \(C\) is \(150 \mathrm{~mm}\). The density of water is \(\rho_{w}=1000 \mathrm{~kg} / \mathrm{m}^{3} .\) Neglect the mass of the contained water and the hydrant.

The truck has a mass of \(50 \mathrm{Mg}\) when empty. When it is unloading \(5 \mathrm{~m}^{3}\) of sand at a constant rate of \(0.8 \mathrm{~m}^{3} / \mathrm{s}\), the sand flows out the back at a speed of \(7 \mathrm{~m} / \mathrm{s}\), measured relative to the truck, in the direction shown. If the truck is free to roll, determine its initial acceleration just as the load begins to empty. Neglect the mass of the wheels and any frictional resistance to motion. The density of sand is \(\rho_{s}=1520 \mathrm{~kg} / \mathrm{m}^{3}\)

Sand is deposited from a chute onto a conveyor belt which is moving at \(0.5 \mathrm{~m} / \mathrm{s}\). If the sand is assumed to fall vertically onto the belt at \(A\) at the rate of \(4 \mathrm{~kg} / \mathrm{s}\), determine the belt tension \(F_{B}\) to the right of \(A\). The belt is free to move over the conveyor rollers and its tension to the left of \(A\) is \(F_{C}=400 \mathrm{~N}\)

Two smooth billiard balls \(A\) and \(B\) each have a mass of \(200 \mathrm{~g}\). If \(A\) strikes \(B\) with a velocity \(\left(v_{A}\right)_{1}=1.5 \mathrm{~m} / \mathrm{s}\) as shown, determine their final velocities just after collision. Ball \(B\) is originally at rest and the coefficient of restitution is \(e=0.85 .\) Neglect the size of each ball.

A 4-lb ball \(B\) is traveling around in a circle of radius \(r_{1}=3 \mathrm{ft}\) with a speed \(\left(v_{B}\right)_{1}=6 \mathrm{ft} / \mathrm{s}\). If the attached cord is pulled down through the hole with a constant speed \(v_{r}=2 \mathrm{ft} / \mathrm{s}\), determine how much time is required for the ball to reach a speed of \(12 \mathrm{ft} / \mathrm{s}\). How far \(r_{2}\) is the ball from the hole when this occurs? Neglect friction and the size of the ball.
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