/*! 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 5 If the container filled with liq... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(g / 2\). What is the value of vertical acceleration of box for given horizontal acceleration \((g / 2)\), so that no part of the bottom of the box is exposed? (1) \(g / 2\) upwards (2) \(\mathrm{g} / 4\) downwards (3) \(g / 4\) upwards (4) not possible

Short Answer

Expert verified
The vertical acceleration required is (1) \(g / 2\) upwards.

Step by step solution

01

Understand the Problem

We need to identify the vertical acceleration of the box such that no part of the bottom surface becomes uncovered when the box experiences horizontal and vertical acceleration. The conditions given are a horizontal acceleration of \( \frac{g}{2} \) and a vertical downward acceleration of \( \frac{g}{2} \).
02

Consider the Forces Acting on the Liquid

When the box is accelerated horizontally by \( \frac{g}{2} \), the surface of the liquid tilts forming an angle with the horizontal. The acceleration produces effective gravity \( g_{\text{eff}} \) combining the horizontal and vertical accelerations. This changes the hydrostatic pressure distribution.
03

Calculate Effective Gravity

The effective gravity, \( g_{\text{eff}} \), is calculated as the resultant of the horizontal and vertical accelerations using:\[g_{\text{eff}} = \sqrt{(g_{\text{vertical}})^2 + (g_{\text{horizontal}})^2}\]For our problem, we have vertical acceleration \( g_{\text{vertical}} = \frac{g}{2} \) and horizontal acceleration \( g_{\text{horizontal}} = \frac{g}{2} \).
04

Determine the Effect on Liquid Surface

With equal horizontal and vertical accelerations of \( \frac{g}{2} \), the effective gravity is:\[g_{\text{eff}} = \sqrt{\left(\frac{g}{2}\right)^2 + \left(\frac{g}{2}\right)^2} = \sqrt{\frac{g^2}{4} + \frac{g^2}{4}} = \sqrt{\frac{g^2}{2}} = \frac{g}{\sqrt{2}}\]This means the liquid tilts at an angle of 45 degrees relative to the bottom, but since the question requires vertical acceleration adjustment, consider changes to keep the liquid covering the entire base.
05

Adjustment for Full Base Coverage

To ensure the base remains covered, adjust the vertical acceleration to counteract the horizontal tilt entirely. Maintaining equal upward and downward adjustments means the vertical component must balance out the horizontal tendency, suggesting equal but opposite effective results. Option 1, \( \frac{g}{2} \) upward, matches these conditions.

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Key Concepts

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

Hydrostatic Pressure
When dealing with fluids in a container, hydrostatic pressure is a fundamental concept to understand. This pressure is the force exerted by a fluid at rest, due to gravity, on the walls of its container. It depends on the fluid's density, the area over which it is acting, and the fluid's depth relative to a point of reference.
For a fluid with density \( \rho \), at a depth \( h \), the hydrostatic pressure \( P \) is calculated as:\[ P = \rho gh \]Where \( g \) is the acceleration due to gravity.
In our scenario with a horizontally and vertically accelerating box, the way pressure is distributed changes due to these motions. The hydrostatic pressure experienced at any point in the liquid depends not only on vertical forces but also on the new "effective" gravity experienced by the liquid.
Effective Gravity
Effective gravity is a concept used to describe how both horizontal and vertical accelerations combine to affect objects, particularly fluids. When a container accelerates, effective gravity changes the perceived direction and intensity of gravitational force relative to the fluid.
This resultant force is not just the typical gravitational pull one experiences. Instead, it combines the normal gravitational pull with the acceleration experienced.
Mathematically, if a container accelerates horizontally and vertically, the effective gravity \( g_{\text{eff}} \) can be calculated as:\[g_{\text{eff}} = \sqrt{(g_{\text{vertical}})^2 + (g_{\text{horizontal}})^2}\]In our example, the box is subject to both \( \frac{g}{2} \) horizontal and vertical accelerations. This affects the direction where the downward force seems to come from and alters how the fluid inside behaves.
Liquid Surface Tilt
When a container with fluid inside accelerates, the liquid's surface tilts, rather than maintaining a flat, horizontal line as when at rest. This inclination is a direct consequence of effective gravity, which redefines the forces acting on every point along the liquid's surface.
The tilt of the liquid depends on the relative magnitude of horizontal and vertical accelerations. For equal horizontal and vertical accelerations (in this case, both \( \frac{g}{2} \)), the surface experiences an equivalent "angle of tilt," causing the container to appear as though it leans inward both from the top and bottom.
To keep the liquid covering the full base, any tilt must be countered. This can be achieved by adjusting other forces, such as vertical acceleration to maintain equilibrium, ensuring each part of the base remains submerged under the liquid.

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Most popular questions from this chapter

A block is partially immersed in a liquid and the vessel is accelerating upwards with an acceleration " \(a "\). The block is observed by two observers \(O_{1}\) and \(O_{2}\) one at rest and the other accelerating with an acceleration " \(a\) " upward. The total buoyant force on the block is (1) same for \(O_{1}\) and \(O_{2} \) (2) greater for \(O_{1}\) than \(O\). (3) greater for \(O_{2}\) than \(O_{1}\) (4) data is not sufficient

In a horizontal pipeline of uniform cross section the pressure fall by \(8 \mathrm{~N} / \mathrm{m}^{2}\) between two points separated by \(1 \mathrm{~km}\). If oil of density \(800 \mathrm{~kg} / \mathrm{m}^{3}\) flows through the pipe, find the change in KE per kg of oil at these points. (1) \(10^{-2} \mathrm{~J} / \mathrm{kg}\) (2) \(10^{-3} \mathrm{~J} / \mathrm{kg}\) (3) \(10^{-1} \mathrm{~J} / \mathrm{kg}\) (4) \(10^{-1} \mathrm{~J} / \mathrm{kg}\)

Two unequal blocks placed over each other of different densities \(\sigma_{1}\) and \(\sigma_{2}\) are immersed in a fluid of density of \(\sigma\). The block of density \(\sigma_{1}\) is fully submerged and the block of density \(\sigma_{2}\) is partly submerged so that ratio of their masses is \(1 / 2\) and \(\sigma / \sigma_{1}=2\) and \(\sigma / \sigma_{2}=0.5\). Find the degree of submergence of the upper block of density \(\sigma_{2}\). (1) \(50 \%\) submerged (2) \(25 \%\) submerged (3) \(75 \%\) submerged (4) Fully submerged

A horizontal oriented tube \(A B\) of length \(l=2.5 \mathrm{~m}\) rotates with a constant angular velocity \(\omega=5 \mathrm{rad} / \mathrm{s}\) about a stationary vertical axis \(O O\) ' passing through the end \(A\). Initially the tube is filled with an ideal fluid. The end \(A\) of the tube is open, the closed end \(B\) has a very small orifice. Find the velocity of the fluid (in \(\mathrm{m} / \mathrm{s}\) ) relative to the tube when the liquid column length in tube reduce to \(h=1 \mathrm{~m}\).

For Problems \(16-17\) A rod of length \(6 \mathrm{~m}\) has a mass \(12 \mathrm{~kg}\). It is hinged at one end \(A\) at a distance of \(3 \mathrm{~m}\) below the water surface. The specific gravity of the material of rod is \(0.5\). What weight must be attached to the other end \(B\) so that \(5 \mathrm{~m}\) of the rod is immersed in water? (1) \(7 \mathrm{kgf}\) (2) \(\frac{17}{3} \mathrm{kgf}\) (3) \(\frac{7}{5} \mathrm{kgf}\) (4) \(\frac{7}{2} \mathrm{kgf}\)

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