/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A large aquarium of height \(5.0... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A large aquarium of height \(5.00 \mathrm{~m}\) is filled with fresh water to a depth of \(2.00 \mathrm{~m}\). One wall of the aquarium consists of thick plastic \(8.00 \mathrm{~m}\) wide. By how much does the total force on that wall increase if the aquarium is next filled to a depth of \(4.00 \mathrm{~m} ?\)

Short Answer

Expert verified
The total force increases by 470,880 N when the water level rises to 4.00 m.

Step by step solution

01

Understand the Problem

We need to calculate the difference in force exerted on the wall of the aquarium when the water level increases from a depth of \(2.00 \, \text{m}\) to \(4.00 \, \text{m}\). We know that the pressure at a depth in a fluid is given by \(P = \rho gh\), where \(\rho\) is the density of the fluid, \(g\) is the acceleration due to gravity, and \(h\) is the depth.
02

Calculate Initial Force at Depth 2m

The initial pressure at the bottom of the 2.00 m depth is \(P_1 = \rho g h_1\) where \(h_1 = 2.00 \, \text{m}\). The force on the wall is the pressure times the area over which it acts. Since pressure increases linearly with depth, we use average depth for calculation: \(h_{avg1} = \frac{h_1}{2} = 1.00 \, \text{m}\). Hence, \(F_1 = \rho g h_{avg1} \times \text{width} \times h_1\).
03

Calculate Final Force at Depth 4m

Once the depth is increased to 4.00 m, the pressure at the bottom becomes \(P_2 = \rho g h_2\), where \(h_2 = 4.00 \, \text{m}\). Using the average depth \(h_{avg2} = \frac{h_2}{2} = 2.00 \, \text{m}\), the force is \(F_2 = \rho g h_{avg2} \times \text{width} \times h_2\).
04

Derive Force Increase

The increase in force is the difference between the final and initial forces, \(\Delta F = F_2 - F_1\). Substituting values, use \(g = 9.81 \, \text{m/s}^2\), and \(\rho = 1000 \, \text{kg/m}^3\) (density of fresh water). Calculate \(F_1\) and \(F_2\) using the expressions from Steps 2 and 3, and then find \(\Delta F\).
05

Calculate Values

Plugging the given values into the expressions, calculate \(F_1 = 1000 \cdot 9.81 \cdot 1.00 \cdot 8.00 \cdot 2.00 = 156,960 \, \text{N}\) and \(F_2 = 1000 \cdot 9.81 \cdot 2.00 \cdot 8.00 \cdot 4.00 = 627,840 \, \text{N}\). Thus, \(\Delta F = 627,840 - 156,960 = 470,880 \, \text{N}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Fluid Mechanics
Fluid mechanics is the branch of physics that deals with the study of fluids (liquids and gases) and the forces acting on them. It involves understanding how fluids behave when they are at rest and in motion. In the context of the given problem, we are particularly concerned with hydrostatics, which is the study of fluids at rest.

In hydrostatics, the pressure exerted by a fluid at rest is due to the weight of the fluid acting on a surface. This pressure is isotropic, meaning it is exerted equally in all directions. The key formula used in problems involving hydrostatics is the pressure-depth relationship:
  • Pressure (\(P\)) within a fluid column of height (\(h\)) is calculated as \(P = \rho gh\).
  • Here, \(\rho\) represents fluid density, \(g\) is the gravitational acceleration, and \(h\) is the depth of fluid.
Understanding these concepts helps in analyzing situations like the one in the exercise, where fluid pressure changes as the depth of water in an aquarium increases.
Force Calculation
Force due to fluid pressure on a surface can be calculated using the relationship between pressure and area. In the aquarium problem, the force on the wall is determined by the area on which the fluid's pressure acts.

When calculating force exerted by a fluid, remember the following:
  • Force (\(F\)) is the product of pressure (\(P\)) and area (\(A\)). In formula terms, \(F = P \times A\).
  • The pressure is calculated using the depth of the fluid column, as discussed in the Fluid Mechanics section: \(P = \rho gh\).
Since the pressure varies with depth, the average pressure calculation is used to find the force exerted on surfaces submerged in the fluid. This involves considering the mean depth of the fluid column to simplify the calculations for uniform surfaces such as the walls of an aquarium.
Pressure Difference
Understanding pressure difference is crucial when calculating how forces change in response to variations in water depth, like in our aquarium scenario. Hydrostatic pressure differences arise because pressure in a liquid depends on depth. A deeper fluid means higher pressure at the bottom section associated with increased force exerted by the water.

The pressure difference in a fluid column is directly proportional to its height, which translates into calculated force difference like this:
  • For initial and final states, compute average depths: \(h_1 = 2.00 \text{ m}\) and \(h_2 = 4.00 \text{ m}\), leading to mean depths of \(1.00 \text{ m}\) and \(2.00 \text{ m}\), respectively.
  • The total change in force can be computed by calculating the forces for each depth and noting the difference: \(\Delta F = F_2 - F_1\).
This calculation demonstrates the impact of increasing fluid depth on the resultant forces, further exemplifying the core principle that deeper fluids exert more pressure and consequently more force.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

About one-third of the body of a person floating in the Dead Sca will be above the waterline. Assuming that the human body density is \(0.98 \mathrm{~g} / \mathrm{cm}^{3}\), find the density of the water in the Dead Sea. (Why is it so much greater than \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\) ?)

An astronaut exercising on a treadmill maintains a pulse rate of 150 per minute. If he exercises for \(1.00 \mathrm{~h}\) as measured by a clock on his spaceship, with a stride length of \(1.00 \mathrm{~m} / \mathrm{s},\) while the ship travels with a speed of \(0.900 \mathrm{c}\) relative to a ground station, what are (a) the pulse rate and (b) the distance walked as measured by someone at the ground station?

A spaceship approaches Earth at a speed of \(0.42 c .\) A light on the front of the ship appears red (wavelength \(650 \mathrm{nm}\) ) to passengers on the ship. What (a) wavelength and (b) color (blue, green, or yellow) would it appear to an observer on Earth?

Temporal separation between two events. Events \(A\) and \(B\) occur with the following spacctime coordinates in the reference frames of Fig. 37 - 25 : according to the unprimed frame, \(\left(x_{A}, t_{A}\right)\) and \(\left(x_{B}, t_{B}\right) ;\) according to the primed frame, \(\left(x_{A}^{\prime}, t_{A} ^{\prime}\right)\) and \(\left(x_{B}^{\prime}, r_ {B}^{\prime}\right) .\) In the unprimed frame, \(\Delta t=t_{n}-t_{A}=1.00 \mu \mathrm{s}\) and \(\Delta x=x_{B}-x_{A}=240 \mathrm {~m}\) (a) Find an expression for \(\Delta t^{\prime}\) in terms of the speed parameter \(\beta\) and the given data. Graph \(\Delta r^{\prime}\) versus \(\beta\) for the following two ranges of \(\beta\) (b) 0 to 0.01 and (c) 0.1 to 1 . (d) At what value of \(\beta\) is \(\Delta t^{\prime}\) minimum and (c) what is that minimum? (f) Can one of these events cause the other? Explain.

A liquid of density \(900 \mathrm{~kg} / \mathrm{m}^{3}\) flows through a horizontal pipe that has a cross-sectional area of \(1.90 \times 10^{-2} \mathrm{~m}^{2}\) in region \(A\) and a cross-sectional area of \(9.50 \times 10^{-2} \mathrm{~m}^{2}\) in region \(B\). The pressure difference between the two regions is \(7.20 \times 10^{3} \mathrm{~Pa}\). What are (a) the volume flow rate and (b) the mass flow rate?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.