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Chapter 15: Problem 101

BIO A person weighs \(685 \mathrm{N}\) in air but only \(497 \mathrm{N}\) when standing in water up to the hips. Find (a) the volume of each of the person's legs and (b) the mass of each leg, assuming they have a density that is 1.05 times the density of water.

Short Answer

Expert verified
Each leg has a volume of approximately 0.0096 m鲁 and a mass of about 10.08 kg.

Step by step solution

01

Understand the Problem

We are given two weights, one in air and one in water. The difference in weight is due to the buoyant force in water, which can be used to find the volume displaced by the submerged part of the body (legs, in this case). We need to calculate the volume and mass of each leg.
02

Calculate Buoyant Force

The buoyant force is the difference between the weight in air and the weight in water. Calculate it using the formula: \[ F_b = W_{air} - W_{water} \] where \( W_{air} = 685 \text{ N} \) and \( W_{water} = 497 \text{ N} \).
03

Apply Archimedes' Principle

According to Archimedes' principle, the buoyant force is equal to the weight of the fluid displaced. In water, this is the weight of the water displaced:\[ F_b = \rho_{water} \cdot g \cdot V_{legs} \]where \( \rho_{water} = 1000 \text{ kg/m}^3 \) and \( g = 9.81 \text{ m/s}^2 \). Solve for the volume \( V_{legs} \).
04

Calculate Volume of Legs

Re-arrange the formula to calculate the volume:\[ V_{legs} = \frac{F_b}{\rho_{water} \cdot g} \]Substitute the values and compute \( V_{legs} \).
05

Calculate Volume of Each Leg

Since both legs are submerged, divide the total volume by 2 to find the volume of each leg:\[ V_{each leg} = \frac{V_{legs}}{2} \].
06

Calculate the Density of Legs

Given that the density of the legs is 1.05 times that of water, calculate the density using:\[ \rho_{legs} = 1.05 \cdot \rho_{water} \] where \( \rho_{water} = 1000 \text{ kg/m}^3 \).
07

Calculate Mass of Each Leg

With the density of the legs, use the formula for mass:\[ m_{leg} = \rho_{legs} \cdot V_{each leg} \] Substitute the calculated density and volume to find the mass of each leg.

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Key Concepts

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

Understanding Buoyant Force
When a person is submerged in water, they experience an upward force that makes them weigh less. This is called the buoyant force. The principle behind this is Archimedes' Principle, which states that any object submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object. In the scenario where a person weighs 685 N in air but only 497 N when submerged up to the hips in water, the difference in these two weights gives us the buoyant force. This difference is calculated as:
  • Buoyant force, \( F_b = W_{air} - W_{water} \)
  • \( F_b = 685 \, \text{N} - 497 \, \text{N} = 188 \, \text{N} \)
Understanding this force is crucial as it helps us determine how much volume of water is displaced, which in turn tells us the volume of the legs submerged.
Concept of Density
Density is a measure of how much mass is contained in a given volume. It is one of the key factors in understanding buoyancy. In this exercise, we are told that the density of each leg is 1.05 times the density of water. Standard water density is \( 1000 \, \text{kg/m}^3 \). Hence, the density of the leg can be calculated as:
  • \( \rho_{legs} = 1.05 \times \rho_{water} \)
  • \( \rho_{legs} = 1.05 \times 1000 \, \text{kg/m}^3 = 1050 \, \text{kg/m}^3 \)
Understanding the density of the legs helps us to calculate the mass when we know the volume. This is because mass is the product of density and volume.
Volume Calculation Method
Volume indicates the amount of three-dimensional space an object occupies. In the given task, we need to discover how much water the legs displace when submerged. This is determined using the buoyant force. According to Archimedes' principle, this buoyant force is the weight of the water displaced:
  • \( F_b = \rho_{water} \cdot g \cdot V_{legs} \)
Rearranging the formula to solve for the volume of the legs gives us:
  • \( V_{legs} = \frac{F_b}{\rho_{water} \cdot g} \)
  • Substituting known values: \( V_{legs} = \frac{188 \, \text{N}}{1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2} \)
  • \( V_{legs} = 0.01916 \, \text{m}^3 \)
Since both legs are submerged, we divide the total volume by 2 to find the volume of each leg:
  • \( V_{each leg} = \frac{0.01916 \, \text{m}^3}{2} = 0.00958 \, \text{m}^3 \)
Mass Calculation Process
Mass is the measure of the amount of matter in an object. Once we have the volume of the legs, mass calculation becomes straightforward with density. Using the formula for mass which is the product of density and volume:
  • \( m_{leg} = \rho_{legs} \cdot V_{each leg} \)
  • \( m_{leg} = 1050 \, \text{kg/m}^3 \times 0.00958 \, \text{m}^3 \)
  • \( m_{leg} = 10.059 \, \text{kg} \)
Thus, each leg has an approximate mass of 10.059 kg. Be sure to apply the correct density and volume values to achieve accurate mass calculations. This clear understanding of mass correlates the weight experienced in air and water, showing the relationship between buoyancy, volume, and mass.

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Most popular questions from this chapter

Two drinking glasses, 1 and 2 , are filled with water to the same depth. Glass 1 has twice the diameter of glass \(2 .\) (a) Is the weight of the water in glass 1 greater than, less than, or equal to the weight of the water in glass \(2 ?\) (b) Is the pressure at the bottom of glass 1 greater than, less than, or equal to the pressure at the bottom of glass \(2 ?\)

BIO Vasodilation When the body requires an increased blood flow rate in a particular organ or muscle, it can accomplish this by increasing the diameter of arterioles in that area. This is referred to as vasodilation. What percentage increase in the diameter of an arteriole is required to double the volume flow rate of blood, all other factors remaining the same?

On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outside the window moves with a speed of approximately \(170 \mathrm{m} / \mathrm{s}\) shortly after takeoff, and that the air inside the plane is at atmospheric pressure. (a) Find the pressure difference between the inside and outside of the window. (b) If the window is \(25 \mathrm{cm}\) by \(42 \mathrm{cm},\) find the force exerted on the window by air pressure.

A horizontal pipe carries oil whose coefficient of viscosity is \(0.00012 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\). The diameter of the pipe is \(5.2 \mathrm{cm},\) and its length is \(55 \mathrm{m}\) (a) What pressure difference is required between the ends of this pipe if the oil is to flow with an average speed of \(1.2 \mathrm{m} / \mathrm{s}\) ? (b) What is the volume flow rate in this case?

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