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Chapter 3: Problem 121

A small card is placed on top of a spool as shown in Fig. \(P 3.121 .\) It is not possible to blow the card off the spool by blowing air through the hole in the center of the spool. The harder one blows, the harder the card "sticks" to the spool. In fact by blowing hard enough it is possible to keep the card against the spool with the spool turned upside down. Give this experiment a try. (Note: It may be necessary to use a thumb tack to prevent the card from sliding from the spool.) Explain this phenomenon.

Short Answer

Expert verified
The phenomenon of the card sticking to the spool when air is blown through it can be explained by Bernoulli's principle. The moving air inside the spool creates a lower pressure zone compared to the still air outside. The pressure difference between the two sides of the card creates a net force that 'sticks' the card to the spool.

Step by step solution

01

Describe the Observation

When observing the experiment, it's evident that air flowing through the hole in the center of the spool produces an effect that 'sticks' the card to the spool. The harder the air is blown, the stronger the card sticks to the spool.
02

Identify Relevant Principles

This phenomenon is primarily dictated by the principles of fluid dynamics, specifically Bernoulli's Principle. Bernoulli's Principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in the fluid's pressure or potential energy.
03

Apply Bernoulli's Principle

Applying this principle to the given situation, when air (a fluid) is blown through the spool, it speeds up compared to the surrounding still air. According to Bernoulli's Principle, the pressure of the moving air in the spool decreases. This decreased pressure creates a pressure difference between the top of the card (exposed to the lower pressure moving air) and the bottom of the card (exposed to the higher pressure still air). This pressure difference acts to hold the card against the spool.
04

Explain The phenomenon

The pressure difference acting on either side of the card creates a net upward force. This force counteracts the gravitational force acting on the card. As a result, the card 'sticks' to the spool, and with a large enough pressure difference, it can even be held upside down.

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Most popular questions from this chapter

Carbon dioxide flows at a rate of \(1.5 \mathrm{ft}^{3} / \mathrm{s}\) from a 3 -in. pipe in which the pressure and temperature are 20 psi (gage) and \(120 \%\) into a 1.5 -in. pipe. If viscous effects are neglected and incompressible conditions are assumed, determine the pressure in the smaller pipe.

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An incompressible fluid flows steadily past a circular cylinder as shown in Fig. P3.8. The fluid velocity along the dividing streamline \((-\infty \leq x \leq-a)\) is found to be \(V=V_{0}\left(1-a^{2} / x^{2}\right),\) where \(a\) is the radius of the cylinder and \(V_{0}\) is the upstream velocity. (a) Determine the pressure gradient along this streamline. (b) If the upstream pressure is \(p_{0},\) integrate the pressure gradient to obtain the pressure \(p(x)\) for \(-\infty \leq x \leq-a,\) (c) Show from the result of part (b) that the pressure at the stagnation point \((x=-a)\) is \(p_{0}+\rho V_{0}^{2} / 2,\) as expected from the Bernoulli equation.
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