/*! 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 52 A block that is \(a\) mm square ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 52

A block that is \(a\) mm square slides across a flat plate on a thin film of oil. The oil has viscosity \(\mu\) and the film is \(h\) mm thick. The block of mass \(M\) moves at steady speed \(U\) under the influence of constant force \(F\). Indicate the magnitude and direction of the shear stresses on the bottom of the block and the plate. If the force is removed suddenly and the block begins to slow, sketch the resulting speed versus time curve for the block. Obtain an expression for the time required for the block to lose 95 percent of its initial speed.

Short Answer

Expert verified
The magnitude of the shear stress on the block and the plate is \(渭 \times (U/h)\) and its direction is opposite to the force direction. The block decelerates at \(-渭 \times (U/h)\), and its speed decays exponentially over time. The time required for the block to lose 95 percent of its initial speed is \(h 脳 log(0.05) / 渭\).

Step by step solution

01

Determine the shear stress

The shear stress on the bottom of the block and the plate can be calculated from Newton's law of viscosity, which states that the shear stress is equal to the fluid viscosity times the velocity gradient. According to the law, the shear stress \(蟿 = \mu \times (U/h)\). Here, the direction of the shear stress will be the same as the direction of the relative velocity, which is opposite to the force direction.
02

Identify the deceleration of the block

When the force is removed, the block begins to experience deceleration due to the fluid's viscosity. The Equation of motion for the block under the conditions given would be \(F = M \times a + 渭 \times A \times (U/h)\). When the force is removed, \(F = 0\), and \(a\) becomes \(-渭 \times (U/h)\). Since the block is slowing down, the deceleration is negative by convention.
03

Sketch the speed versus time curve

The speed of the block decays exponentially with time. Initially, it is \(U\) and finally becomes zero. Therefore, the graph is a decaying exponential curve. The curve starts at the y-intercept equal to \(U\) and as time goes to infinity, the speed goes to zero, thus ending at the x-axis.
04

Calculate the time for the block to lose 95% of its speed

Since we have deceleration as a constant, we can use the kinematic equation to solve for time. Specifically, we use the equation \(U = U_o + a \times t\). We set \(U = 0.05U_0\) (because 5% of initial speed remains), replace \(a\) with \(-渭 \times (U_0/h)\), and solve for \(t\). The result is \(t = h \times log(0.05) / 渭\).

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