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Chapter 15: Problem 2

Three Liquids Three liquids that will not mix are poured into a cylindrical container. The volumes and densitics of the liquids are \(0.50 \mathrm{~L}, 2.6 \mathrm{~g} / \mathrm{cm}^{3} ; 0.25 \mathrm{~L}, 1.0 \mathrm{~g} / \mathrm{cm}^{3} ;\) and \(0.40 \mathrm{~L}, 0.80 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the force on the bottom of the container due to these liquids? One liter \(=1 \mathrm{~L}=1000 \mathrm{~cm}^{3}\). (Ignore the contribution due to the atmosphere.)

Short Answer

Expert verified
The total force on the bottom of the container is 18.33 N.

Step by step solution

01

- Convert Volumes to Cubic Centimeters

Convert the volumes of the liquids from liters to cubic centimeters using the conversion factor: 1 liter = 1000 cm鲁.For liquid 1: Volume = 0.50 L = 0.50 脳 1000 cm鲁 = 500 cm鲁For liquid 2: Volume = 0.25 L = 0.25 脳 1000 cm鲁 = 250 cm鲁For liquid 3: Volume = 0.40 L = 0.40 脳 1000 cm鲁 = 400 cm鲁
02

- Calculate Mass of Each Liquid

Use the formula for mass, which is mass = density 脳 volume, to calculate the mass of each liquid.For liquid 1: Mass = 2.6 g/cm鲁 脳 500 cm鲁 = 1300 gFor liquid 2: Mass = 1.0 g/cm鲁 脳 250 cm鲁 = 250 gFor liquid 3: Mass = 0.80 g/cm鲁 脳 400 cm鲁 = 320 g
03

- Calculate the Weight of Each Liquid

Convert the mass of each liquid to weight using the relation weight = mass 脳 gravitational acceleration. Assume gravitational acceleration g = 9.8 m/s虏.For liquid 1: Weight = 1300 g 脳 9.8 m/s虏 = 1300 脳 0.0098 kg鈰卪/s虏 = 12.74 NFor liquid 2: Weight = 250 g 脳 9.8 m/s虏 = 250 脳 0.0098 kg鈰卪/s虏 = 2.45 NFor liquid 3: Weight = 320 g 脳 9.8 m/s虏 = 320 脳 0.0098 kg鈰卪/s虏 = 3.14 N
04

- Calculate Total Force on the Bottom of the Container

Add up the weights of all three liquids to get the total force exerted on the bottom of the container.Total Force = Weight of liquid 1 + Weight of liquid 2 + Weight of liquid 3Total Force = 12.74 N + 2.45 N + 3.14 N = 18.33 N

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Key Concepts

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

density and volume relationship
Understanding the relationship between density and volume is crucial in physics. Density is a measure of how much mass is packed into a given volume. It is usually expressed in units like grams per cubic centimeter \text{\(g/cm^3\)}.
In our exercise, the density of each liquid is multiplied by its volume to find the mass. To illustrate:
  • For liquid 1, a density of 2.6 g/cm鲁 and a volume of 500 cm鲁 results in a mass of 1300 g.
  • For liquid 2, a density of 1.0 g/cm鲁 and a volume of 250 cm鲁 results in a mass of 250 g.
  • For liquid 3, a density of 0.80 g/cm鲁 and a volume of 400 cm鲁 results in a mass of 320 g.
Always remember, the formula connecting these quantities is: \text{\( \text{mass} = \text{density} \times \text{volume} \)}. This relationship helps us convert density and volume into the more tangible concept of mass, which is essential in calculating further properties like weight and force.
mass and weight conversion
It's important to distinguish between mass and weight. Mass is the amount of matter in an object and is measured in grams or kilograms.
Weight, on the other hand, is the force exerted by gravity on that mass, measured in Newtons (N). The conversion from mass to weight involves multiplying the mass by the gravitational acceleration, which is \text{\( 9.8 \text{ m/s}^2 \)} on Earth.
In our exercise, we converted the mass of each liquid to weight:
  • Liquid 1 has a mass of 1300 g. By multiplying with 9.8 m/s虏, we get a weight of 12.74 N.
  • Liquid 2 has a mass of 250 g. By multiplying with 9.8 m/s虏, we get a weight of 2.45 N.
  • Liquid 3 has a mass of 320 g. By multiplying with 9.8 m/s虏, we get a weight of 3.14 N.
The general conversion formula used is: \text{\( \text{weight} = \text{mass} \times \text{gravitational acceleration} \)}. This conversion is necessary because forces are described by weight rather than mass in many physical scenarios.
gravitational force
Gravitational force is a fundamental concept in physics. It is the attractive force that pulls two masses towards each other.
On Earth, this force is what gives us weight and it is calculated by multiplying an object's mass by the acceleration due to gravity.
In the context of our exercise, each liquid exerts a force on the bottom of the container. This force is due to gravity pulling down on the mass of the liquids.
Thus, combining the weights gives us the total force exerted by all three liquids:
  • Weight of liquid 1: 12.74 N
  • Weight of liquid 2: 2.45 N
  • Weight of liquid 3: 3.14 N
  • Total Force: 12.74 N + 2.45 N + 3.14 N = 18.33 N
This total weight (or gravitational force) is what the container's bottom experiences. Understanding gravitational force involves recognizing its impact on everyday objects, which helps us comprehend how forces operate and interact.

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Most popular questions from this chapter

Venturi Meter A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (Fig. \(15-48\) ); the cross-sectional area \(A\) of the entrance and exit of the meter matches the pipe's cross-sectional arca At the entrance and cxit, the fluid flows through the pipe with speed \(v_{A}=\left|\vec{v}_{A}\right| .\) But it flows through a narrow "throat" of cross-sectional area \(B\) with speed \(v_{B}=\left|\vec{v}_{B}\right| .\) A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's speed is accompanied by a change \(\Delta P\) in the fluid's pressure, which causes a height difference \(h\) of the liquid in the two arms of the manometer. (Here \(\Delta P\) means pressure in the throat minus pressure in the pipe.) (a) By applying Bernoulli's equation and the equation of continuity to points 1 and 2 in Fig. \(15-48\), show that $$ \vec{v}_{A}=\sqrt{\frac{2 B^{2} \Delta P}{\rho\left(B^{2}-A^{2}\right)}} $$ where \(\rho\) is the density of the fluid. (b) Suppose that the fluid is fresh water, that the cross-sectional areas are \(64 \mathrm{~cm}^{2}\) in the pipe and \(32 \mathrm{~cm}^{2}\) in the throat, and that the pressure is \(55 \mathrm{kPa}\) in the pipe and \(41 \mathrm{kPa}\) in the throat. What is the rate of water flow in cubic meters per second?

Garden Hose A garden hose with an internal diameter of \(1.9 \mathrm{~cm}\) is connected to a (stationary) lawn sprinkler that consists merely of an enclosure with 24 holes, each \(0.13 \mathrm{~cm}\) in diameter. If the water in the hose has a speed of \(0.91 \mathrm{~m} / \mathrm{s}\), at what speed does it leave the sprinkler holes?

Syringe Find the pressure increase in the fluid in a syringe when a nurse applies a force of \(42 \mathrm{~N}\) to the syringe's circular piston, which has a radius of \(1.1 \mathrm{~cm}\).

Swimming Pool A swimming Problem 10. pool has the dimensions \(24 \mathrm{~m} \times 9.0\) \(\mathrm{m} \times 2.5 \mathrm{~m}\). When it is filled with water, what is the force (resulting from the water alone) on (a) the bottom, (b) each short side, and (c) each long side? (d) If you are concerned with the possibility that the concrete walls and floor will collapse, is it appropriate to take the atmospheric pressure into account? Why?

Block of Wood A block of wood floats in fresh water with twothirds of its volume submerged. In oil the block floats with \(0.90\) of its volume submerged. Find the density of (a) the wood and (b) the oil.
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