/*! 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 1 For Problems 1-5 If the contai... [FREE SOLUTION] | 91Ó°ÊÓ

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For Problems 1-5 If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid for calculation of pressure, effective \(g\) is used. A closed box with horizontal base \(6 \mathrm{~m}\) by \(6 \mathrm{~m}\) and a height \(2 \mathrm{~m}\) is half filled with liquid. It is given a constant horizontal acceleration \(g / 2\) and vertical downward acceleration \(g / 2\). The angle of the free surface with the horizontal is equal to (1) \(30^{\circ}\) (2) \(\tan ^{-1} \frac{2}{3}\) (3) \(\tan ^{-1} \frac{1}{3}\) (4) \(45^{\circ}\)

Short Answer

Expert verified
The angle is \( \tan^{-1}\frac{1}{3} \), option (3).

Step by step solution

01

Understanding the problem

We have a closed box containing liquid, which is half-filled. The box is subjected to both horizontal and vertical accelerations. The problem is asking for the angle between the free surface of the liquid and the horizontal plane due to these accelerations.
02

Analyzing the accelerations

The box is subjected to a horizontal acceleration of \( a_x = \frac{g}{2} \) and a vertical (downward) acceleration of \( a_y = \frac{g}{2} \). These accelerations will affect the inclination of the liquid's free surface.
03

Calculating the effective gravity

The effective gravity acting on the liquid is the vector sum of the standard gravity \( g \) and any additional vertical acceleration. Since we are given a downward acceleration equal to \( \frac{g}{2} \), the effective gravity in the vertical direction is \( g_{eff} = g + \frac{g}{2} = \frac{3g}{2} \).
04

Calculating the angle of tilt

The free surface of the liquid will tilt at an angle \( \theta \) such that \( \tan(\theta) = \frac{a_x}{g_{eff}} \). Plugging in the values, we get \( \tan(\theta) = \frac{\frac{g}{2}}{\frac{3g}{2}} = \frac{1}{3} \).
05

Choosing the correct answer

The angle \( \theta \) is given by \( \theta = \tan^{-1}\left(\frac{1}{3}\right) \). Comparing this to the options, the correct answer is (3) \( \tan^{-1}\frac{1}{3} \).

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Key Concepts

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

Effective Gravity
In fluid mechanics, the concept of effective gravity is crucial when analyzing fluids under various acceleration conditions. In simple terms, effective gravity is the result of the natural gravitational force acting on an object combined with any other acceleration present. Imagine you are in a car that accelerates suddenly forward; you feel pushed back into your seat. Similarly, when a container filled with liquid is both horizontally and vertically accelerated, these accelerations combine with the standard gravitational force to create what's known as effective gravity. It is essentially the net acceleration the fluid "feels" in any scenario.
To calculate effective gravity, you combine the gravitational acceleration vector, typically directed downwards, with any additional acceleration vectors. In the given exercise, we have standard gravity acting downward and an extra downward acceleration of \( \frac{g}{2} \). Thus, effective gravity is \( \frac{3g}{2} \) in the vertical direction. This effective value is essential for determining how fluids behave in these conditions.
Pressure in Fluids
Pressure in fluids is a fundamental concept describing how force is distributed in a liquid. Pressure is defined as force exerted per unit area and it plays a pivotal role in determining fluid behavior, especially when containers are subjected to external forces. For liquids in an accelerated container, effective gravity influences the pressure distributions.
  • Standard gravity creates hydrostatic pressure, which increases with depth due to the weight of the fluid above. This is described by the formula \( P = \rho g h \), where \( P \) is pressure, \( \rho \) is fluid density, \( g \) is gravitational acceleration, and \( h \) is depth.
  • When external accelerations, such as in our exercise, are introduced, the pressure distribution changes. Both horizontal and vertical accelerations affect the distribution of pressure throughout the fluid.
Calculating pressures under varying accelerations requires accounting for the effective gravity in the system. This involves understanding how these additional accelerations tilt the liquid and change its depth, thereby affecting fluid pressure at different points.
Acceleration Effects
Acceleration significantly impacts fluid behavior. In a stable scenario without acceleration, water in a cup, for example, stays level due to gravity. But, if the cup starts accelerating in any direction, the water surface tilts.
  • Horizontal acceleration, as seen in the exercise, causes the liquid surface to slant. Imagine quickly moving forward; the water shifts back.
  • Vertical acceleration changes effective gravity magnitude. In the exercise, a downward acceleration is added to gravity, enhancing its effect.
Tracking how fluid surfaces change under acceleration involves vector addition of gravitational and acceleration forces. The resultant effect determines fluid equilibrium and surface orientation. Specifically, the tilt angle in fluid is found via the tangent function of horizontal to effective vertical acceleration, demonstrating these combined acceleration effects.
Inclined Liquid Surface
An inclined liquid surface is what you might observe when a fluid in a container is subjected to simultaneous horizontal and vertical forces, like in our exercise problem. The surface of the liquid aligns itself perpendicularly to the resultant force or effective gravity. How the surface inclines tells us a lot about the accelerations acting on it.
The task at hand is to determine the angle that the inclined surface makes with the horizontal. This can be computed using trigonometry. As described in the exercise, the tangent of the inclination angle \( \theta \) is found by dividing the horizontal acceleration by the effective gravity. Therefore, we have \( \tan(\theta) = \frac{\frac{g}{2}}{\frac{3g}{2}} = \frac{1}{3} \).
  • This calculation shows that the more prominent the horizontal acceleration, the steeper the tilt.
  • An angle \( \theta = \tan^{-1}(\frac{1}{3}) \) implies a moderate slope, coinciding with our exercise answer.
Inclined liquid surfaces ultimately adjust to restore equilibrium in the most stable position possible, taking into account all acting forces.

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Most popular questions from this chapter

If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid for calculation of pressure, effective \(g\) is used. A closed box with horizontal base \(6 \mathrm{~m}\) by \(6 \mathrm{~m}\) and a height \(2 \mathrm{~m}\) is half filled with liquid. It is given a constant horizontal acceleration \(g / 2\) and vertical downward acceleration \(g / 2\). Length of exposed portion of top of box is equal to (1) \(2 \mathrm{~m}\) (2) \(3 \mathrm{~m}\) (3) \(4 \mathrm{~m}\) (4) \(2.5 \mathrm{~m}\)

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