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One of the concrete pillars that support a house is \(2.2 \mathrm{m}\) tall and has a radius of \(0.50 \mathrm{m}\). The density of concrete is about \(2.2 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\). Find the weight of this pillar in pounds \((1 \mathrm{N}=0.2248 \mathrm{lb})\).

Short Answer

Expert verified
The weight of the pillar is approximately 8393 lb.

Step by step solution

01

Calculate Volume of the Cylinder

The volume of a cylindrical pillar can be calculated using the formula for the volume of a cylinder: \( V = \pi r^2 h \). Here, \( r = 0.50 \) m and \( h = 2.2 \) m. Substituting these values, the volume \( V = \pi \times (0.50)^2 \times 2.2 \approx 1.73 \) m³.
02

Calculate Mass of the Pillar

To find the mass of the pillar, multiply its volume by the density of the concrete: \( \text{mass} = \text{density} \times \text{volume} = 2.2 \times 10^{3} \times 1.73 \approx 3806 \) kg.
03

Calculate Weight in Newtons

The weight of the pillar is its mass times the acceleration due to gravity \( g \approx 9.81 \) m/s². Therefore, \( \text{weight (N)} = 3806 \times 9.81 \approx 37335 \) N.
04

Convert Newtons to Pounds

To convert the weight from newtons to pounds, use the conversion factor \( 1 \text{ N} = 0.2248 \text{ lb} \). Thus, \( \text{weight (lb)} = 37335 \times 0.2248 \approx 8393 \) lb.

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Key Concepts

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

Cylinder Volume Calculation
Understanding how to calculate the volume of a cylinder is essential for this exercise. A cylinder's volume is determined by the formula: \( V = \pi r^2 h \). Let's break down what each part of this formula represents:
  • \(\pi\) (Pi) is a constant approximately equal to 3.14159. It's crucial in calculations involving circles.
  • \(r\) is the radius of the base of the cylinder. In this problem, the radius is 0.50 meters.
  • \(h\) is the height of the cylinder. Here, it's given as 2.2 meters.
To find the volume, simply plug these values into the formula. First, square the radius (0.50 m): \(0.50^2 = 0.25\). Then multiply this by \(\pi\) and the height (2.2 m):
\[V = \pi \times 0.25 \times 2.2 \approx 1.73 \text{ m}^3\]
This result tells us the amount of space the pillar occupies. Calculating volume is an initial step in many physics problems, preparing us to find mass next.
Mass and Weight Conversion
To solve physics problems involving mass and weight, understanding the conversion between the two is vital. Mass is a measure of how much matter an object contains, usually given in kilograms. To find the mass from the volume, multiply the volume by the material's density:
  • Density of concrete: \(2.2 \times 10^{3} \text{ kg/m}^3\)
  • Volume of the pillar: \(1.73 \text{ m}^3\)
Multiply these values to get the mass:
\[\text{mass} = 2.2 \times 10^{3} \times 1.73 \approx 3806 \text{ kg}\]
At this point, you have the mass and need to convert it to weight, which is the force exerted by gravity on that mass. Weight (in Newtons) is calculated by multiplying mass by the gravitational acceleration \(g\), approximately \(9.81 \text{ m/s}^2\).
\[\text{weight (N)} = 3806 \times 9.81 \approx 37335 \text{ N}\]
Finally, convert the weight from newtons to pounds using the given conversion factor: 1 Newton = 0.2248 pounds.
\[\text{weight (lb)} = 37335 \times 0.2248 \approx 8393 \text{ lb}\]
This conversion step is crucial in translating metrics into units used in everyday contexts.
Density and Weight
Density is a key concept when working out weight from volume. It represents how much mass is contained in a unit volume of a substance, typically expressed in kilograms per cubic meter (kg/m\(^3\)). It's important because it allows us to transition from volume—which was calculated from geometric properties—into mass.
  • Higher density means more mass in a given volume. Thus, a dense material like concrete will have significant weight because it concentrates a lot of mass into its structure.
  • The density of concrete, \(2.2 \times 10^{3} \text{ kg/m}^3\), indicates that every cubic meter of concrete weighs 2200 kg.
By understanding this, one can see that finding the weight isn't merely about size but also about what's contained within that size. Once density gives you the mass, weight comes from incorporating gravity—simply a force acting on that mass, pulling it toward the earth. These principles underpin much of physics when working with real-world applications, helping to bridge theoretical models and practical weight measurements.

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Most popular questions from this chapter

In the human body, blood vessels can dilate, or increase their radii, in response to various stimuli, so that the volume flow rate of the blood increases. Assume that the pressure at either end of a blood vessel, the length of the vessel, and the viscosity of the blood remain the same, and determine the factor \(R_{\text {dilated }} / R_{\text {normal }}\) by which the radius of a vessel must change in order to double the volume flow rate of the blood through the vessel.

The drawing shows a hydraulic system used with disc brakes. The force \(\overrightarrow{\mathbf{F}}\) is applied perpendicularly to the brake pedal. The pedal rotates about the axis shown in the drawing and causes a force to be applied perpendicularly to the input piston (radius \(\left.=9.50 \times 10^{-3} \mathrm{m}\right)\) in the master cylinder. The resulting pressure is transmitted by the brake fluid to the output plungers (radii \(\left.=1.90 \times 10^{-2} \mathrm{m}\right)\), which are covered with the brake linings. The linings are pressed against both sides of a disc attached to the rotating wheel. Suppose that the magnitude of \(\overrightarrow{\mathbf{F}}\) is \(9.00\) \(\mathrm{N}\). Assume that the input piston and the output plungers are at the same vertical level, and find the force applied to each side of the rotating disc.

(a) The mass and the radius of the sun are, respectively, \(1.99 \times 10^{30} \mathrm{kg}\) and \(6.96 \times 10^{8} \mathrm{m} .\) What is its density? (b) If a solid object is made from a material that has the same density as the sun, would it sink or float in water? Why? (c) Would a solid object sink or float in water if it were made from a material whose density was the same as that of the planet Saturn (mass \(=5.7 \times 10^{26} \mathrm{kg},\) radius \(\left.=6.0 \times 10^{7} \mathrm{m}\right) ?\) Provide a reason for your answer.

An airtight box has a removable lid of area \(1.3 \times 10^{-2} \mathrm{m}^{2}\) and negligible weight. The box is taken up a mountain where the air pressure outside the box is \(0.85 \times 10^{5}\) Pa. The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box?

An 81 -kg person puts on a life jacket, jumps into the water, and floats. The jacket has a volume of \(3.1 \times 10^{-2} \mathrm{m}^{3}\) and is completely submerged under the water. The volume of the person's body that is under water is \(6.2 \times 10^{-2} \mathrm{m}^{3} .\) What is the density of the life jacket?

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