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Chapter 12: Problem 13

Many cultures around the world still use a simple weapon called a blowgun, a tube with a dart that fits tightly inside. A sharp breath into the end of the tube launches the dart. When exhaling forcefully, a healthy person can supply air at a gauge pressure of \(6.0 \mathrm{kPa} .\) What force does this pressure exert on a dart in a 1.5 -cm-diameter tube?

Short Answer

Expert verified
The air pressure exerts a force of approximately \(1.06 N\) on the dart.

Step by step solution

01

Calculate the cross-sectional area of the tube

The cross-sectional area of the tube can be calculated using the formula for the area of a circle, which is \( \pi r^2 \), where \( r \) is the radius of the tube. Given that the diameter of the tube is 1.5 cm, the radius is half of the diameter, 0.75 cm. Since the radius needs to be in meters for SI units, this translates to 0.0075 m. Thus, the cross-sectional area of the tube is \( \pi \times (0.0075)^2 \), approximately equal to \(1.7671 \times 10^{-4} m^2\).
02

Calculate the exerted force

Next, use the formula for pressure, which rearranged to find force is Force = Pressure \times Area. Substitute the given pressure and the calculated area into the formula, \( Force = 6.0 \mathrm{kPa} \times 1.7671 \times 10^{-4} m^2\). Note that the pressure needs to be converted from \( \mathrm{kPa} \) to \( \mathrm{Pa} \) for the units to match. So, \( 6.0 \mathrm{kPa} = 6000 \mathrm{Pa} \). Thus, the force is \(6000 \mathrm{Pa} \times 1.7671 \times 10^{-4} m^2 \), equalling approximately \(1.06 N\).

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Key Concepts

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

Blowgun Mechanics
Blowguns, also known as blowpipes, are fascinating tools that have been used by various cultures for hunting and even in sports. The basic principle involves using your own breath to generate enough pressure to propel a dart out of a tube.
It is a simple yet effective weapon that capitalizes on the mechanics of pressure and force.
When you blow into the tube, the air speed increases due to the narrow passage, which helps in building pressure behind the dart.
With enough force, this pressure can launch the dart at a significant speed.
  • Using a blowgun requires skill to maintain a consistent and accurate breath that provides adequate force without sacrificing accuracy.
  • These mechanics are directly tied to the principles of fluid dynamics, where fast-moving air inside the tube creates a pocket of high pressure directly behind the projectile, effectively pushing it forward.

  • Another interesting aspect is the fit of the dart inside the tube. It needs to create an airtight seal, ensuring no air escapes past the dart, maximizing the force exerted.
Cross-Sectional Area
The concept of cross-sectional area is crucial in understanding how blowguns function. The cross-section of a blowgun is the circular opening through which the dart is propelled. To comprehend the pressure calculations, it is important to determine this area accurately.
To calculate the cross-sectional area, we use the formula for the area of a circle, \[A = \pi r^2 \], where \( r \) is the radius.
This seems straightforward, but remember:
  • The diameter must be halved to find the radius.
  • It鈥檚 vital to convert measurements into the appropriate units to maintain consistency in calculations, typically meters in physics problems.
  • In our exercise, the diameter is given as 1.5 cm, making the radius 0.75 cm, or 0.0075 m. Thus, the cross-sectional area is \( \pi \times (0.0075)^2 \), which is approximately \( 1.7671 \times 10^{-4} \ m^2 \).
This small area plays a big role in determining how much force can be exerted due to pressure.
Force and Pressure Conversion
Converting forces and pressures is a key step in tackling physics problems, particularly those involving objects where units must match. For pressure, we often deal with kPa (kilopascal) in problems similar to our blowgun example.
Pressure is defined as force per unit area, expressed as \( P = \frac{F}{A} \). To find the force exerted by pressure on a surface, this equation is rearranged to \( F = P \times A \).
In our exercise:
  • The pressure initially given is 6.0 kPa. In SI units, this needs conversion to pascals since 1 kPa is 1000 pascals. This conversion yields 6000 Pa.
  • When inserting these values into our equation, \( F = 6000 \ \text{Pa} \times 1.7671 \times 10^{-4} \ \text{m}^2 \), the force calculates to be approximately \( 1.06 \ N \).
  • This showcases the importance of consistent units in calculations, ensuring that all factors are compatible for accurate force determination.
Converting correctly is essential to not only solve the problem correctly but also to understand the physics behind the numbers.

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Most popular questions from this chapter

The lowest pressure ever obtained in a laboratory setting is \(4.0 \times 10^{-11}\) Pa. At this pressure, how many molecules of air would there be in a \(20^{\circ} \mathrm{C}\) experimental chamber with a volume of \(0.090 \mathrm{m}^{3} ?\)

A football is inflated in the locker room before the game. The air warms as it is pumped, so it enters the ball at a temperature of \(27^{\circ} \mathrm{C} .\) The ball is inflated to a gauge pressure of 13 psi. The ball is used for play at \(10^{\circ} \mathrm{C}\). Once the ball cools, what is the pressure in the ball? Assume that atmospheric pressure is 14.7 psi.

For a normal car riding on tires with relatively flexible sidewalls, the weight of the car is held up, in large measure, by the pressure of the air in the tires. If you look at one of your car's tires, you'll note that the tire is flattened slightly to make a rectangle where it touches the ground. The area of the resulting "contact patch" depends on the pressure in the tires. To a good approximation, the upward normal force of the ground (which we can assume is equal to \(1 / 4\) of the car's weight) on this patch of the tire is equal to the downward pressure force on the patch. a. Suppose you inflate your \(2000 \mathrm{kg}\) car's tires to the recommended pressure, as measured by a gauge. The resulting contact patch is \(18 \mathrm{cm}\) wide and \(12 \mathrm{cm}\) long. What does the gauge read? b. If you let a bit of air out of your tire, what happens to the area of the contact patch?

An ideal gas is at \(20^{\circ} \mathrm{C}\). If we double the average kinetic energy of the gas atoms, what is the new temperature in \({ }^{\circ} \mathrm{C} ?\)

How many atoms of hydrogen are in 100 g of hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right) ?\)
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