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Chapter 11: Problem 106

Pressure plays an important role in the behavior of fluids. As we have seen in this chapter, pressure is the magnitude of the force acting perpendicular to a surface divided by the area of the surface. Pressure should not be confused, however, with the force itself. Problem 106 serves to emphasize that pressure and force are different concepts. Problem 107 focuses on the essence of Archimedes' principle and its application to the situation of a buoyant force acting on a submerged object. The figure shows a rear view of a loaded two-wheeled wheelbarrow on a horizontal surface. It has balloon tires and a weight \(W=\) \(684 \mathrm{N},\) which is uniformly distributed. The left tire has a contact area with the ground of \(A_{\mathrm{L}}=6.6 \times 10^{-4} \mathrm{m}^{2},\) whereas the right tire is underinflated and has a contact area of \(A_{R}=9.9 \times 10^{-4} \mathrm{m}^{2} .\) Concepts: (i) Force is a vector. Therefore both a direction and a magnitude are needed to specify it. Are both a direction and magnitude needed to specify a pressure? (ii) How is the force each tire applies to the ground related to the force the ground applies to each tire? (iii) Do the left and right tires apply the same force to the ground? Explain. (iv) Do the left and right tires apply the same pressure to the ground? Calculations: Find the force and pressure that each tire applies to the ground.

Short Answer

Expert verified
Each tire applies 342 N force. Left tire: \(5.18 \times 10^{5}\, \mathrm{Pa}\); Right tire: \(3.45 \times 10^{5}\, \mathrm{Pa}\).

Step by step solution

01

Understanding Pressure and Force

Pressure signifies the amount of force applied per unit area. It is calculated as \( P = \frac{F}{A} \), where \( P \) is pressure, \( F \) is the force, and \( A \) is the area over which the force is distributed. Force, being a vector, needs direction and magnitude for description, while pressure doesn't have direction inherently and is therefore only a scalar quantity.
02

Distributing Weight Evenly

The total force applied by the wheelbarrow, which is its weight \( W = 684 \mathrm{N} \), is distributed evenly across both wheels, based on symmetry or equal weight distribution assumptions since no other information is given about balancing. Thus, each tire must support \( \frac{W}{2} = \frac{684}{2} = 342 \mathrm{N} \).
03

Calculating Force for Each Tire

Each tire applies the same force to the ground due to the even weight distribution: \( F_{L} = F_{R} = 342 \mathrm{N} \), where \( F_{L} \) and \( F_{R} \) are the forces applied by the left and right tires, respectively.
04

Calculating Pressure for Each Tire

For the left tire: \( P_{L} = \frac{F_{L}}{A_{L}} = \frac{342}{6.6 \times 10^{-4}} \approx 5.18 \times 10^{5} \mathrm{Pa} \). For the right tire: \( P_{R} = \frac{F_{R}}{A_{R}} = \frac{342}{9.9 \times 10^{-4}} \approx 3.45 \times 10^{5} \mathrm{Pa} \). This shows the left tire has higher pressure as it has a smaller contact area.
05

Comparison and Conclusion

While both tires apply the same force to the ground (342 N each), due to differing contact areas, they exert different pressures. The left tire has a smaller contact area and therefore exerts a greater pressure. This illustrates how the same force can result in different pressures depending on the area over which it is applied.

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Key Concepts

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

Archimedes' Principle
Archimedes' Principle is a fundamental concept in fluid dynamics that helps us understand buoyancy. It states that a fluid exerts an upward buoyant force on an object submerged in it. This force is equal to the weight of the fluid displaced by the object. In simple terms, when you submerge an object in a liquid, part of the upward force comes from the liquid trying to reclaim its space.

This principle is especially useful when you need to determine if an object will float or sink. If the buoyant force equals the object's weight, it will float. Otherwise, it sinks. This happens because the liquid provides resistance by pushing up against gravity.
  • Buoyant force = weight of displaced liquid
  • Helps predict floating or sinking
Archimedes' Principle is crucial for understanding fluid behaviors in everyday life, such as ship buoyancy and the operation of submarines. Remember, this principle only applies to objects fully or partially submerged in a fluid.
Buoyant Force
The buoyant force is the pushing force a fluid exerts upward on an object placed in it. Imagine when you push something heavy into a pool; the water pushes back up. That pushing up is the buoyant force. This force acts against gravity, effectively making the object seem lighter when submerged.

The magnitude of the buoyant force depends on three major factors:
  • The density of the fluid
  • The volume of the object submerged in the fluid
  • The gravitational force
To calculate the buoyant force experienced by an object, you can use the formula:
\[F_b = \rho \cdot V \cdot g\]
where \( F_b \) is the buoyant force, \( \rho \) is the fluid density, \( V \) is the volume of fluid displaced, and \( g \) is the acceleration due to gravity. This formula shows why objects like balloons filled with a lighter gas can float upwards. The buoyant force is stronger than the force of gravity pulling them down.
Force and Pressure Relationship
Force and pressure are related through the area upon which the force is applied. Force is the interaction that causes an object to be pushed, pulled, or change motion. Pressure is how this force is distributed over an area. It's calculated using the formula:
\[P = \frac{F}{A}\]
where \( P \) is pressure, \( F \) is force, and \( A \) is the area.

In everyday life, this relationship explains why sharp objects can cut more easily than blunt ones. A sharp blade concentrates force onto a smaller area, resulting in greater pressure, making it easier to penetrate materials.
  • Force: Acts on an area, vector (magnitude + direction)
  • Pressure: Scalar, no direction, just magnitude
In the context of the exercise, even though the tires apply the same force due to even weight distribution, their differing contact areas result in different pressure values. Recognizing this distinction is vital for understanding phenomena like why balloon tires distribute weight differently than regular tires.

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Most popular questions from this chapter

Planners of an experiment are evaluating the design of a sphere of radius \(R\) that is to be filled with helium \(\left(0^{\circ} \mathrm{C}, 1\right.\) atm pressure). Ultrathin silver foil of thickness \(T\) will be used to make the sphere, and the designers claim that the mass of helium in the sphere will equal the mass of silver used. Assuming that \(T\) is much less than \(R,\) calculate the ratio \(T / R\) for such a sphere.

One way to administer an inoculation is with a "gun" that shoots the vaccine through a narrow opening. No needle is necessary, for the vaccine emerges with sufficient speed to pass directly into the tissue beneath the skin. The speed is high, because the vaccine \(\left(\rho=1100 \mathrm{kg} / \mathrm{m}^{3}\right)\) is held in a reservoir where a high pressure pushes it out. The pressure on the surface of the vaccine in one gun is \(4.1 \times 10^{6}\) Pa above the atmospheric pressure outside the narrow opening. The dosage is small enough that the vaccine's surface in the reservoir is nearly stationary during an inoculation. The vertical height between the vaccine's surface in the reservoir and the opening can be ignored. Find the speed at which the vaccine emerges.

Spring Gun. A hydraulic press is used to compress a spring that will then be used to project a \(25.0-\mathrm{kg}\) steel ball. The system is similar, but smaller in scale, to the car jack illustrated in Figure \(11.14 .\) In this case, the smaller cylinder has a diameter of \(d_{1}=0.75 \mathrm{cm}\) and has a manually operated plunger. The larger cylinder has a diameter of \(d_{2}=10.0 \mathrm{cm},\) and its piston compresses the spring. The idea is that the gun operator pulls a lever that pushes the plunger on the small cylinder, which transmits a pressure to the larger piston that, in turn, exerts a force on the spring and compresses it. Once the spring is compressed, the steel ball is loaded and the spring is released, ejecting the ball. (a) If a force of \(950 \mathrm{N}\) is exerted on the primary (smaller) piston to compress the spring \(1.25 \mathrm{m}\) from its equilibrium (uncompressed) position, what is the spring constant \(k\) of the spring? (b) What is the velocity of the steel ball just after it is ejected? (c) Neglecting air resistance, what is the maximum range of this "spring gun"?

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Friction is absent. The spring constant of the spring is \(3600 \mathrm{N} / \mathrm{m}\). The piston has a negligible mass and a radius of \(0.024 \mathrm{m}\). (a) When the air beneath the piston is completely pumped out, how much does the atmospheric pressure cause the spring to compress? (b) How much work does the atmospheric pressure do in compressing the spring?

Crossing a River. You and your team come to a slow flowing river that you need to cross. The nearest bridge is 20 miles to the north, too dangerous and too far to trek with your group. You explore the area down river and discover an abandoned shed with a stash of 55 -gallon drums (all empty and weighing 35.0 lb each) and a stack of 10 -foot planks: You will build a raft. The six members of your team have a combined weight of 925 pounds (assuming everyone was truthful). You also have a four-wheeler (all-terrain vehicle), which weighs 450 lb, and other gear, which adds another 315 lb. Your simple raft design is as follows: a platform of planks with barrels strapped to the bottom. You estimate that the planks weigh about \(45.0 \mathrm{lb}\) each, and you will need 20 of them to make a platform that can accommodate everything for the one-way trip. You measure the dimensions of the cylindrical drums and find they have diameter \(D=22.5\) inches and height \(h=33.5\) inches (note: not exactly \(^{455}\) gallons"). (a) What is the minimum number of barrels that you will need so that the raft will float when fully loaded? (b) What is the minimum number of barrels you will need if you want the platform to be at least 6 inches above the water when fully loaded?
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