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A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at \(11.0 \mathrm{~m} / \mathrm{s}\) over the top of the plate only, it is possible to keep the plate in a horizontal position, as illustrated in part \(a\) of the drawing. To what value should the air speed be reduced so that the plate is kept at a \(30.0^{\circ}\) angle with respect to the vertical, as in part \(b\) of the drawing? (Hint: Apply Bernoulli's equation in the form of Equation \(11.12 .)\)

Short Answer

Expert verified
The air speed should be reduced to approximately 12.7 m/s.

Step by step solution

01

Establish the Relationship Using Bernoulli’s Equation

Apply Bernoulli's equation to the top and bottom sides of the plate. On the top, where air speed is 11.0 m/s initially, the pressure is lower. Use Bernoulli's principle, which states: \[ P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}, \] where \( P \) is pressure, \( \rho \) is the air density, \( v \) is velocity, and \( h \) is height. Assume height is constant, removing the \( \rho gh \) term from the equation.
02

Determine the Necessary Air Speeds for Different Angles

At 0 degrees (horizontal position), the speed is 11.0 m/s. When the plate is at a 30-degree angle, the effective velocity parallel to the plate becomes \( v' = v \cos(30^\circ) \). Since pressure difference holds the plate due to velocity change, the initial and new conditions can be related.
03

Solve for the New Air Speed

Using the relation: \[ 11^2 = (v' \cos(30^\circ))^2, \] \[ v' = \frac{11}{\cos(30^\circ)}. \] Calculate \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), and solve for \( v' \): \[ v' \cos(30^\circ) = 11 \Rightarrow v' = \frac{11}{\cos(30^\circ)} = \frac{11}{\sqrt{3}/2} = \frac{22}{\sqrt{3}} \approx 12.7 \text{ m/s}. \]
04

Adjust and Confirm the Calculation

Check the calculation by evaluating \( \cos(30^\circ) \): \( \cos(30^\circ) \approx 0.866 \). Recalculate to confirm: \[ v' = \frac{11}{0.866} \approx 12.7 \text{ m/s}. \] This confirms the earlier steps.

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is a fascinating study of how fluids like air and water move, and it is essential in understanding many real-world applications. In this specific exercise, we're focusing on how air flows over a hanging plate. - **Key Principle**: The moving air reduces pressure on the surface it flows over, a principle best explained by Bernoulli's Equation. - **Bernoulli's Equation**: This equation relates pressure, velocity, and height for flowing fluids. It indicates that an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or potential energy. Understanding fluid dynamics helps us comprehend how airplanes fly, the functioning of air conditioning systems, and even weather patterns. When you see an airplane lift off the ground, it is the low pressure on the upper surface of the wings, thanks to increased air speed, that helps lift the plane into the sky. In this exercise, we apply these principles to determine how the air speed must adjust to keep the plate at a specific angle.
Air Speed Adjustment
Adjusting air speed is crucial to maintain different positions of the plate. By modifying the air flow, it is possible to keep the plate horizontal or at specific angles like 30 degrees.- **Horizontal Position**: Initially, with an air speed of 11.0 m/s, the plate remains horizontal.- **30-Degree Angle**: For the plate to hang at a 30-degree angle, the air speed should be reduced. The velocity at this angle is calculated using trigonometry, especially the cosine function, to find the effective air speed parallel to the plate.Here’s a simplified way to look at it:- **Initial Speed**: 11.0 m/s allows the plate to stay level.- **Speed Adjustment**: To achieve a 30-degree angle, calculate the speed using \[ v' = \frac{11}{\cos(30^\circ)} \] which results in approximately 12.7 m/s.This calculation shows how adjusting air speed directly affects the tilting angle of the plate by compensating for the change in pressure difference on the surface of the plate.
Pressure Difference
Pressure difference is created by changing air speed, and this plays a vital role in the movement of the plate. This concept is closely linked to Bernoulli's principle. When air moves faster over the top of the plate, the pressure decreases, allowing the plate to be lifted or maintained at certain angles. - **Pressure and Velocity**: Remember, according to Bernoulli's Principle, pressure decreases as velocity increases. - **Angle Maintenance**: By adjusting pressure through velocity changes, we can control the angle of the plate. For this exercise: - Initially, the plate stays horizontal due to high air speed creating a significant pressure difference between the top and bottom. - As the speed reduces for a 30-degree angle, the pressure difference also changes, balancing the forces at the new configuration. - This interrelation ensures that the right airspeed results in the desired position or angle of the plate. In everyday life, you'll see this principle in action with wind affecting open doors or even in race cars using spoilers to manage pressure for better handling.

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Most popular questions from this chapter

Interactive Solution \(\underline{11.14}\) at presents a model for solving this problem. A solid concrete block weighs \(169 \mathrm{~N}\) and is resting on the ground. Its dimensions are \(0.400 \mathrm{~m} \times 0.200 \mathrm{~m} \times 0.100 \mathrm{~m}\). A number of identical blocks are stacked on top of this one. What is the smallest number of whole blocks (including the one on the ground) that can be stacked so that their weight creates a pressure of at least two atmospheres on the ground beneath the first block?

A person who weighs \(625 \mathrm{~N}\) is riding a 98-N mountain bike. Suppose the entire weight of the rider and bike is supported equally by the two tires. If the gauge pressure in each tire is \(7.60 \times 10^{5} \mathrm{~Pa}\), what is the area of contact between each tire and the ground?

A hot-air balloon is accelerating upward under the influence of two forces, its weight and the buoyant force. For simplicity, consider the weight to be only that of the hot air within the balloon, thus ignoring the balloon fabric and the basket. (a) How is the weight of the hot air determined from a knowledge of its density \(\rho\) hot air and the volume \(V\) of the balloon? (b) For a given volume \(V\) of the balloon, does the buoyant force depend on the density of the hot air inside the balloon, the density of the cool air outside the balloon, or both? Provide a reason for your answer. (c) Draw a free-body diagram for the balloon, showing the forces that act on it. How is the upward acceleration of the balloon related to these forces and to its mass?

A person can change the volume of his body by taking air into his lungs. The amount of change can be determined by weighing the person under water. Suppose that under water a person weighs \(20.0 \mathrm{~N}\) with partially full lungs and \(40.0 \mathrm{~N}\) with empty lungs. Find the change in body volume.

Poiseuille's law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds number Re is less than about 2000: \(\mathrm{Re}=2 \bar{v} \rho R / \eta .\) Here \(\bar{v}, \rho,\) and \(\eta\) are, respectively, the average speed, density, and viscosity of the fluid, and \(R\) is the radius of the pipe. Calculate the highest average speed that blood \(\left(\rho=1060 \mathrm{~kg} / \mathrm{m}^{3}, \eta=4.0 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) could have and still remain in laminar flow when it flows through the aorta \(\left(R=8.0 \times 10^{-3} \mathrm{~m}\right)\).

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