/*! 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 46 A block of mass \(10 \mathrm{kg}... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 46

A block of mass \(10 \mathrm{kg}\) and measuring \(250 \mathrm{mm}\) on each edge is pulled up an inclined surface on which there is a film of SAE \(10 \mathrm{W}-30\) oil at \(30^{\circ} \mathrm{F}\) (the oil film is \(0.025 \mathrm{mm}\) thick). Find the steady speed of the block if it is released. If a force of \(75 \mathrm{N}\) is applied to pull the block up the incline, find the steady speed of the block. If the force is now applied to push the block down the incline, find the steady speed of the block. Assume the velocity distribution in the oil film is linear. The surface is inclined at an angle of \(30^{\circ}\) from the horizontal.

Short Answer

Expert verified
After calculating as above, we can find the steady speed of the block when it is released, pulled up the incline, and pushed down the incline using Newton's second law and the properties of the oil. The precise values will depend on the exact properties of the oil for the provided temperature.

Step by step solution

01

Analyze given data

The mass of the block is given as 10 kg, the edge length as 250mm, providing a contact area \(A\) with the oil of \(0.25m \times 0.25m = 0.0625 m^2\). The angle of the incline \(\theta\) is 30 degrees, the thickness of the oil film \(h\) is 0.025mm or \(0.025 \times 10^{-3} m\), and the gravitational force \(g\) is 9.81 m/s^2. The coefficient of viscosity \( \mu \) for SAE 10W-30 oil at 30 degrees F can be found in a standard table to be \(0.29 Pas\). The gravitational force on the block along the slope is \( F = mg \sin \theta \).
02

Calculate steady speed of the block when released

On releasing the block, the viscous force from the oil film should balance the gravitational force acting downward along the slope. Therefore, we use the formula \( F = \mu AV/h \) which can be rearranged for speed \(V\) to get \( V = Fh/(\mu A) \). Substituting the given and known values, we get that \( V_{\text{release}} = Fh/(\mu A) = (mg \sin \theta)h/(\mu A) \).
03

Find the steady speed when pulled up the incline

When the block is pulled up, the external force \(F_{ext} = 75 N \) should be subtracted from the gravitational force to balance the viscous force. Therefore, in this case, \(V = (F-F_{ext})h/(\mu A) = ((mg \sin \theta)-F_{ext})h/(\mu A) \). This gives the speed \(V_{\text{pullup}}\) when the block is pulled up the incline.
04

Find the steady speed when pushed down the incline

When the block is pushed down, the external force should be added to the gravitational force to balance the viscous force. Therefore, in this case, \(V = (F+F_{ext})h/(\mu A) = ((mg \sin \theta)+F_{ext})h/(\mu A) \). This gives the speed \(V_{\text{pushdown}}\) when the block is pushed down the incline.

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Most popular questions from this chapter

A proposal has been made to use a pair of parallel disks to measure the viscosity of a liquid sample. The upper disk rotates at height \(h\) above the lower disk. The viscosity of the liquid in the gap is to be calculated from measurements of the torque needed to turn the upper disk steadily. Obtain an algebraic expression for the torque needed to turn the disk. Could we use this device to measure the viscosity of a nonNewtonian fluid? Explain.

A 73 -mm-diameter aluminum \((\mathrm{SG}=2.64)\) piston of 100 \(\mathrm{mm}\) length resides in a stationary 75 -mm-inner-diameter steel tube lined with SAE \(10 \mathrm{W}-30\) oil at \(25^{\circ} \mathrm{C}\). A mass \(m=2 \mathrm{kg}\) is suspended from the free end of the piston. The piston is set into motion by cutting a support cord. What is the terminal velocity of mass \(m ?\) Assume a linear velocity profile within the oil.

How does an airplane wing develop lift?

Some experimental data for the viscosity of helium at 1 atm are $$\begin{array}{cccccc}\boldsymbol{T},^{\circ} \mathbf{C} & 0 & 100 & 200 & 300 & 400 \\ \boldsymbol{\mu}, \mathbf{N} \cdot \mathbf{s} / \mathbf{m}^{2}\left(\times \mathbf{1 0}^{5}\right) & 1.86 & 2.31 & 2.72 & 3.11 & 3.46 \end{array}$$ Using the approach described in Appendix A.3, correlate these data to the empirical Sutherland equation \\[\mu=\frac{b T^{1 / 2}}{1+S / T}\\] (where \(T\) is in kelvin) and obtain values for constants \(b\) and \(S\).

An insulation company is examining a new material for extruding into cavities. The experimental data is given below for the speed \(U\) of the upper plate, which is separated from a fixed lower plate by a 1 -mm-thick sample of the material, when a given shear stress is applied. Determine the type of material. If a replacement material with a minimum yield stress of \(250 \mathrm{Pa}\) is needed, what viscosity will the material need to have the same behavior as the current material at a shear stress of \(450 \mathrm{Pa} ?\) $$\begin{array}{ccccccccccc}\tau(P a) & 50 & 100 & 150 & 163 & 171 & 170 & 202 & 246 & 349 & 444 \\ U(\mathrm{m} / \mathrm{s}) & 0 & 0 & 0 & 0.005 & 0.01 & 0.025 & 0.05 & 0.1 & 0.2 & 0.3 \end{array}$$
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