/*! 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 47 A door \(1 \mathrm{m}\) wide and... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 47

A door \(1 \mathrm{m}\) wide and \(1.5 \mathrm{m}\) high is located in a plane vertical wall of a water tank. The door is hinged along its upper edge, which is \(1 \mathrm{m}\) below the water surface. Atmospheric pressure acts on the outer surface of the door. (a) If the pressure at the water surface is atmospheric, what force must be applied at the lower edge of the door in order to keep the door from opening? (b) If the water surface gage pressure is raised to 0.5 atm, what force must be applied at the lower edge of the door to keep the door from opening? (c) Find the ratio \(F / F_{0}\) as a function of the surface pressure ratio $$p_{s} / p_{\text {atm }^{*}}\left(F_{0}\right.$$ is the force required when \(\left.p_{s}=p_{\text {atm }} .\right)\)

Short Answer

Expert verified
The forces required to keep the door from opening when the pressure at the water surface is atmospheric and 1.5 times atmospheric can be calculated using the principles of fluid mechanics. The ratio \(F/F_0\) as a function of the surface pressure ratio \(P_s/P_{atm}\) can be expressed in a symbolic form.

Step by step solution

01

Calculate the pressure difference across the door for part (a)

Use the formula for hydrostatic pressure \(P = P_0 + \rho gh\) to calculate the pressure at the top and the bottom of the door. The top of the door is at a depth of 1m and the bottom at a depth of 2.5m. Use the values \(\rho = 1000 \, kg/m^3, g = 9.8 \, m/s^2\) for water and \(P_0 = 101325 \, Pa\) for atmospheric pressure. Then find the difference to get the varying pressure across the door.
02

Calculate the net force acting on the door due to water pressure for part (a)

The pressure force on an area is given by \(F = P \times A\), where \(A\) is the area. In this case, the pressure is not the same across the area, so we'll integrate the pressure from the top to the bottom of the door. Remember the area \(A = width \times height = 1 \, m \times 1.5 \, m\).
03

Repeat steps 1 and 2 for part (b)

Here, the pressure at the surface is 1.5 times atmospheric pressure, so \(P_0 = 1.5 \times 101325 \, Pa\). Repeat the previous steps to get the force required to keep the door from opening when the pressure at the water surface is raised.
04

Set up a ratio of the forces for part (c)

We define \(F_0\) as the force required to keep the door from opening when the pressure at the water surface is atmospheric, and \(F\) as the force required to keep the door from opening for any given surface pressure \(P_s\). Set up the ratio \(F/F_0\) as a function of the surface pressure ratio \(P_s/P_{atm}\), where \(P_{atm}\) is the atmospheric pressure. We don't need the actual values of \(F\) and \(F_0\); this is an symbolic expression based on the formulas from steps 1-3.

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Most popular questions from this chapter

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