/*! 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 86 The cylinder shown is supported ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 86

The cylinder shown is supported by an incompressible liquid of density \(\rho,\) and is hinged along its length. The cylinder, of mass \(M,\) length \(L,\) and radius \(R,\) is immersed in liquid to depth \(H .\) Obtain a general expression for the cylinder specific gravity versus the ratio of liquid depth to cylinder radius, \(\alpha=H / R,\) needed to hold the cylinder in equilibrium for \(0 \leq \alpha<1 .\) Plot the results.

Short Answer

Expert verified
The specific gravity \(\gamma\) of the cylinder needed to hold it in equilibrium is given by the general expression \(\gamma = \frac{M}{\rho \pi R^3 \alpha}\). This expression shows a hyperbolic relationship. The plot of \(\gamma\) versus \(H / R = \alpha\) will confirm this relationship: as the ratio of liquid depth to cylinder radius (\(\alpha\)) increases, the specific gravity needed to hold the cylinder in equilibrium increases as well.

Step by step solution

01

Understanding the problem

The specific gravity of the cylinder is defined as the ratio of the cylinder's mass to the mass of a volume of liquid equal to the volume of the cylinder. In terms of density, it's \(\gamma = \frac{M}{\rho V}\), where \(V\) represents the volume of the cylinder. The force of buoyancy pushing the cylinder upwards is equal to the weight of the liquid displaced by the cylinder, so \(\rho g V_{\text{{liquid}}} = M g\). Here \(V_{\text{{liquid}}}\) is the volume of the liquid displaced by the submerged part of the cylinder, and \(g\) is the acceleration due to gravity. The submerged volume is given by \(\pi R^2 H\). Setting the two forces equal to each other will let us solve for \(\gamma\), the specific gravity.
02

Equating forces and solving for specific gravity

Setting \(\rho g V_{\text{{liquid}}} = M g\) and inserting for \(V_{\text{{liquid}}}\) we get \(\rho g \pi R^2 H = M g\). We can cancel \(g\) from both sides, and solving for \(\gamma\), we find that \(\gamma = \frac{M}{\rho \pi R^2 H} = \frac{M}{\rho V}\), where \(V = \pi R^2 H\). This gives us an equation for the specific gravity as a function of \(H / R = \alpha\).
03

Expressing specific gravity as a function of \(\alpha\)

To express our equation in terms of \(\alpha\), we write \(H\) as \(alpha \times R\), so our equation becomes \(\gamma = \frac{M}{\rho \pi R^2 \alpha R} = \frac{M}{\rho \pi R^3 \alpha}\). This is our general expression for the specific gravity as a function of \(\alpha\).
04

Plotting the results

Plotting the results requires a suitable plotting software or tool. On your plot, the x-axis should represent the depth to radius ratio \(\alpha\) and the y-axis should represent the specific gravity \(\gamma\). You will notice that the plot of specific gravity versus \(\alpha\) shows a hyperbolic relationship, illustrating how the specific gravity required to hold the cylinder in equilibrium increases as \(\alpha\) increases.

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