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Chapter 13: Problem 1

An 80 -kg metal cylinder, \(2.0 \mathrm{~m}\) long and with each end of area 25 \(\mathrm{cm}^{2}\), stands vertically on one end. What pressure does the cylinder exert on the floor? $$ P=\frac{\text { Normal force }}{\text { Area }}=\frac{(80 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}{25 \times 10^{-4} \mathrm{~m}^{2}}=3.1 \times 10^{5} \mathrm{~Pa} $$

Short Answer

Expert verified
The pressure exerted on the floor by the cylinder is approximately \( 3.1 \times 10^{5} \) Pa.

Step by step solution

01

Understand the Problem

We need to find the pressure exerted by a metal cylinder on the ground. Pressure is defined as the force per unit area.
02

Calculate the Normal Force

The normal force is simply the weight of the cylinder. Calculate the weight using the formula: \( \text{Weight} = \text{mass} \times \text{gravity} \). Here, mass \( = 80 \text{ kg} \) and gravity \( = 9.81 \text{ m/s}^2 \), so the normal force is \( 80 \times 9.81 = 784.8 \text{ N} \).
03

Determine the Area

The area the cylinder acts on is one of its full end faces. Given that the area is \( 25 \text{ cm}^2 \), we need to convert this to \( \text{m}^2 \). Thus, \( 25 \text{ cm}^2 = 25 \times 10^{-4} \text{ m}^2 \).
04

Apply the Pressure Formula

Pressure is calculated using \( P = \frac{\text{Normal force}}{\text{Area}} \). Substitute the values: \( P = \frac{784.8}{25 \times 10^{-4}} = 3.1392 \times 10^{5} \text{ Pa} \).
05

Verify the Answer

The calculated pressure is approximately \( 3.1 \times 10^{5} \text{ Pa} \), which matches the given answer. Thus, identify any rounding used to present the final answer.

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Key Concepts

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

Normal Force
Let's start by understanding what the normal force is. In the context of this exercise, the normal force is the force that acts perpendicular to the surface of contact, which happens to be the ground in this case. For the metal cylinder, the normal force equates to its weight. Weight is the force due to gravity acting on the mass. To calculate it, you use the formula:
  • Weight = mass 脳 gravity
In our case, the metal cylinder has a mass of 80 kg and the standard value of gravitational acceleration is 9.81 m/s虏. Thus, the normal force is:\[ \text{Normal force} = 80 \times 9.81 = 784.8 \text{ N} \]This means the cylinder exerts a force of 784.8 Newtons on the floor due to gravity.
Area Conversion
Understanding the area conversion is crucial for accurately calculating pressure. The area that the cylinder's weight acts upon is the area of its circular base. It's given as 25 cm虏. However, to use it in the standard pressure formula, it needs to be converted into square meters because most SI units are in metric. To convert from cm虏 to m虏, remember that there are 100 cm in a meter, so one square meter is 10,000 cm虏 (which is 100 x 100). Therefore:
  • 25 cm虏 = 25 脳 10鈦烩伌 m虏
Now correctly converted, it becomes 0.0025 m虏. This allows us to move forward with calculations in SI units.
Pressure Formula
Knowing how the pressure formula works is key to solving this exercise. Pressure is defined as the force applied perpendicularly over a unit area. The formula used is:\[ P = \frac{\text{Normal force}}{\text{Area}} \]From previous steps, you have determined both the normal force (784.8 N) and the converted area (0.0025 m虏). You substitute these values into the formula to find the pressure exerted by the cylinder:\[ P = \frac{784.8}{0.0025} = 3.1392 \times 10^{5} \text{ Pa} \]Thus, the pressure exerted by the cylinder on the floor is approximately 3.1 脳 10鈦� Pascals, after considering significant figures for the final answer.
Cylinder Weight
Now let鈥檚 focus on the concept of cylinder weight. The weight of any object is the force with which it is pulled towards the Earth due to the planet鈥檚 gravitational field. Calculating the weight helps us understand how much force the object exerts on any surface it rests on. For the cylinder, its weight is calculated using the simple formula:
  • Weight = mass 脳 gravitational acceleration
With a mass of 80 kg and gravitational acceleration of 9.81 m/s虏, we compute the weight as 784.8 N. This weight, effectively, is what we interpret as the normal force acting downwards on the cylinder鈥檚 base.
SI Units
Let's delve into the importance of SI units in physics problems like this one. The International System of Units (SI) is a globally accepted metric system that standardizes measurements for better consistency and understanding in scientific computation. In this exercise:
  • Mass is measured in kilograms (kg)
  • Acceleration due to gravity is in meters per second squared (m/s虏)
  • Force/weight is in Newtons (N), where 1 N = 1 kg路m/s虏
  • Area is converted and measured in square meters (m虏)
  • Pressure results in Pascals (Pa), where 1 Pa = 1 N/m虏
Using SI units ensures that calculations and their outcomes are precise and understandable in worldwide contexts.

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Most popular questions from this chapter

The density of ice is \(917 \mathrm{~kg} / \mathrm{m}^{3}\). What fraction of the volume of a piece of ice will be above the liquid when floating in fresh water? The piece of ice will float in the water, since its density is less than \(1000 \mathrm{~kg} / \mathrm{m}^{3}\), the density of water. As it does, \(F_{B}=\) Weight of displaced water = Weight of piece of ice But the weight of the ice is \(\rho_{\mathrm{ice}} g V\), where \(V\) is the volume of the piece. In addition, the weight of the displaced water is \(\rho_{w} g V^{\prime}\) ' where \(V^{\prime}\) is the volume of the displaced water. Substituting into the above equation $$ \begin{aligned} \rho_{\text {ice }} g V &=\rho_{w} g V^{\prime} \\ V^{\prime} &=\frac{\rho_{\text {ice }}}{\rho_{w}} V=\frac{917}{1000} V=0.917 V \end{aligned} $$ The fraction of the volume that is above water is then $$ \frac{V-V^{\prime}}{V}=\frac{V-0.917 V}{V}=1-0.917=0.083 \text { or } 8.3 \% $$

A partly filled beaker of water sits on a scale, and its weight is \(2.30 \mathrm{~N}\). When a piece of metal suspended from a thread is totally immersed in the beaker (but not touching bottom), the scale reads \(2.75 \mathrm{~N}\). What is the volume of the metal? The water exerts an upward buoyant force on the metal. According to Newton's Third Law of action and reaction, the metal exerts an equal downward force on the water. It is this force that increases the scale reading from \(2.30 \mathrm{~N}\) to \(2.75 \mathrm{~N}\). Hence the buoyant force is \(2.75-2.30=0.45 \mathrm{~N}\). Then, because \(F_{B}=\) weight of displaced water \(=\rho_{w} g V=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(V)\) we have the volume of the displaced water, and of the piece of metal, namely, $$ V=\frac{0.45 \mathrm{~N}}{9810 \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}^{2}}=46 \times 10^{-6} \mathrm{~m}^{3}=46 \mathrm{~cm}^{3} $$

A 2.0-cm cube of metal is suspended by a fine thread attached to a scale. The cube appears to have a mass of \(47.3 \mathrm{~g}\) when measured submerged in water. What will its mass appear to be when submerged in glycerin, sp gr = 1.26? [Hint: Find \(\rho\) too.]

A barrel will rupture when the gauge pressure within it reaches 350 \(\mathrm{kPa}\). It is attached to the lower end of a vertical pipe, with the pipe and barrel filled with oil \(\left(\rho=890 \mathrm{~kg} / \mathrm{m}^{3}\right)\). How long can the pipe be if the barrel is not to rupture? From \(P=\rho g h\) we have $$ h=\frac{P}{\rho g}=\frac{350 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}}{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(890 \mathrm{~kg} / \mathrm{m}^{3}\right)}=40.1 \mathrm{~m} $$

The area of a piston of a force pump is \(8.0 \mathrm{~cm}^{2}\). What force must be applied to the piston to raise oil \(\left(\rho=0.78 \mathrm{~g} / \mathrm{cm}^{2}\right)\) to a height of \(6.0 \mathrm{~m}\) ? Assume the upper end of the oil is open to the atmosphere.
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