/*! 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 An electrical short cuts off all... [FREE SOLUTION] | 91Ó°ÊÓ

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An electrical short cuts off all power to a submersible diving vehicle when it is 30 m below the surface of the ocean. The crew must push out a hatch of area 0.75 \(\mathrm{m}^{2}\) and weight 300 \(\mathrm{N}\) on the bottom to escape. If the pressure inside is 1.0 atm, what downward force must the crew exert on the hatch to open it?

Short Answer

Expert verified
The crew must exert a force of approximately 226,498.125 N.

Step by step solution

01

Understand the Problem

We need to calculate the downward force required to open a hatch on a submersible diving vehicle under certain conditions. We have the depth at 30 m, hatch area at 0.75 m², hatch weight at 300 N, and inside pressure at 1.0 atm.
02

Convert Atmospheric Pressure to Pascals

Convert the inside pressure of 1.0 atm to pascals. Recall that 1 atm is equivalent to approximately 101,325 Pa. So, the inside pressure is 101,325 Pa.
03

Calculate External Pressure

The external pressure at 30 meters depth equals the sum of atmospheric pressure at the surface and the hydrostatic pressure due to the water above. Hydrostatic pressure is calculated with: \[ P_{ ext{hydrostatic}} = \rho \cdot g \cdot h \]where \(\rho\) is the density of seawater ( \(\approx 1025 \; \text{kg/m}^3\)), \(g\) is the acceleration due to gravity (9.81 m/s²) and \(h\) is the depth (30 m).
04

Calculate Hydrostatic Pressure

Using the equation for hydrostatic pressure:\[ P_{ ext{hydrostatic}} = 1025 \cdot 9.81 \cdot 30 \approx 301,597.5 \; \text{Pa} \]
05

Calculate Total External Pressure

The total external pressure acting on the hatch is the sum of atmospheric pressure and hydrostatic pressure:\[ P_{ ext{external}} = 101,325 \; \text{Pa} + 301,597.5 \; \text{Pa} = 402,922.5 \; \text{Pa} \]
06

Calculate Net Pressure on the Hatch

The net pressure (\(\Delta P\)) is the difference between external and internal pressure:\[ \Delta P = 402,922.5 \; \text{Pa} - 101,325 \; \text{Pa} = 301,597.5 \; \text{Pa} \]
07

Calculate Net Force Required

The net force exerted by this pressure difference is given by:\[ F_{\text{net}} = \Delta P \cdot A \]where \(A\) is the area of the hatch. Substituting the given values:\[ F_{\text{net}} = 301,597.5 \cdot 0.75 = 226,198.125 \; \text{N} \]
08

Account for the Weight of the Hatch

The crew also needs to overcome the weight of the hatch. The total force that the crew must exert is the sum of the net force due to the pressure difference and the weight of the hatch:\[ F_{\text{crew}} = F_{\text{net}} + \text{Weight of the hatch} = 226,198.125 + 300 = 226,498.125 \; \text{N} \]
09

Final Result

The downward force the crew must exert on the hatch to open it is approximately 226,498.125 N.

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Key Concepts

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

Pressure Calculation
Pressure is a fundamental concept in fluid mechanics that helps us understand how forces work in fluids. In simple terms, pressure is defined as the force exerted over an area. It is expressed mathematically as:
  • \[ P = \frac{F}{A} \]
where \( P \) is the pressure, \( F \) is the force, and \( A \) is the area over which the force is distributed.
In this exercise, we are dealing with two primary pressures: the atmospheric pressure and the hydrostatic pressure due to water at a certain depth.
Atmospheric pressure at sea level is approximately 101,325 pascals (Pa), which is equivalent to 1 atmosphere (atm). When calculating pressures underwater, we must consider both this atmospheric pressure and additional pressures due to the depth of water, which brings us to the concept of hydrostatic pressure.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. In simpler words, it refers to the pressure felt at a specific depth in a fluid; this pressure increases with depth.The fundamental formula for calculating hydrostatic pressure is:
  • \[ P_{\text{hydrostatic}} = \rho \cdot g \cdot h \]
where \( \rho \) (rho) represents the fluid's density, \( g \) is the acceleration due to gravity, and \( h \) is the depth below the surface.
In the given problem, we use the density of seawater, which is approximately 1025 kg/m³, the standard gravity rate of 9.81 m/s², and the depth of 30 meters to find the hydrostatic pressure.
This calculated pressure adds to the atmospheric pressure to give a total external pressure, exerted on the submersible's hatch. Clearly understanding these pressures allows us to move forward to see how they influence the force calculations needed to open the hatch.
Force Exertion Calculation
Once we have understood the pressures involved, calculating the force exertion becomes straightforward. The goal is to find the amount of force needed to overcome both the pressure exerted on the hatch by the water above and the weight of the hatch itself.First, we calculate the net pressure on the hatch, which is the difference between the external pressure (sum of atmospheric and hydrostatic pressures) and the internal pressure inside the vehicle. This is expressed as:
  • \[ \Delta P = P_{\text{external}} - P_{\text{internal}} \]
To find the force resulting from this net pressure, we multiply it by the area of the hatch (0.75 m²):
  • \[ F_{\text{net}} = \Delta P \cdot A \]
Finally, since the crew also has to contend with the hatch's weight (300 N), the total downward force needed is the sum of this weight and the net force due to pressure difference:
  • \[ F_{\text{crew}} = F_{\text{net}} + \text{Weight of the hatch} \]
This calculation shows that the total force required by the crew to open the hatch ensures safety during emergencies while dealing with complex underwater scenarios.

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Most popular questions from this chapter

Standing on your head. (a) When you stand on your head, what is the difference in pressure of the blood in your brain compared with the pressure when you stand on your feet if you are 1.85 m tall? The density of blood is 1060 \(\mathrm{kg} / \mathrm{m}^{3}\) . (b) What effect does the increased pressure have on the blood vessels in your brain?

Blood pressure. Systemic blood pressure is defined as the ratio of two pressures, both expressed in millimeters of mercury. Normal blood pressure is about \(\frac{120 \mathrm{mm}}{80 \mathrm{mm}},\) which is usually just stated as \(\frac{120}{80}\) . (See also Problem \(24 . )\) What would normal systemic blood pressure be if, instead of millimeters of mercury, we expressed pressure in each of the following units, but continued to use the same ratio format? (a) atmospheres, (b) torr, (c) Pa, (d) \(\mathrm{N} / \mathrm{m}^{2},\) (e) psi.

A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7200 \(\mathrm{cm}^{3} / \mathrm{s}\) . At one point in the pipe, where the radius is \(4.00 \mathrm{cm},\) the water's absolute pressure is \(2.40 \times 10^{5}\) Pa. At a second point in the pipe, the water passes through a constriction where the radius is 2.00 \(\mathrm{cm} .\) What is the water's absolute pressure as it flows through this constriction?

Landing on Venus. One of the great difficulties in landing on Venus is dealing with the crushing pressure of the atmosphere, which is 92 times the earth's atmospheric pressure. (a) If you are designing a lander for Venus in the shape of a hemisphere 2.5 \(\mathrm{m}\) in diameter, how many newtons of inward force must it be prepared to withstand due to the Venusian atmosphere? (Don't forget about the bottom!) (b) How much force would the lander have to withstand on the earth?

You are designing a machine for a space exploration vehicle. It contains an enclosed column of oil that is 1.50 m tall, and you need the pressure difference between the top and the bottom of this column to be 0.125 atm. (a) What must be the density of the oil? (b) If the vehicle is taken to Mars, where the acceleration due to gravity is \(0.379 g,\) what will be the pressure difference (in earth atmospheres) between the top and bottom of the oil column?

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