/*! 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 51 A block \(0.1 \mathrm{m}\) squar... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 51

A block \(0.1 \mathrm{m}\) square, with \(5 \mathrm{kg}\) mass, slides down a smooth incline, \(30^{\circ}\) below the horizontal, on a film of SAE 30 oil at \(20^{\circ} \mathrm{C}\) that is \(0.20 \mathrm{mm}\) thick. If the block is released from rest at \(t=0,\) what is its initial acceleration? Derive an expression for the speed of the block as a function of time. Plot the curve for \(V(t)\). Find the speed after 0.1 s. If we want the mass to instead reach a speed of \(0.3 \mathrm{m} / \mathrm{s}\) at this time, find the viscosity \(\mu\) of the oil we would have to use.

Short Answer

Expert verified
From the given problem and the five steps described above, we can calculate the initial acceleration of the block, derive an expression for its speed as a function of time, find its speed after 0.1s and determine the viscosity of the oil for the block to reach a certain speed. For mathematical solutions, substituting the provided or known values into the relevant formula in each step is necessary.

Step by step solution

01

Calculation of the Forces

First we identify the forces acting on the block. The two main forces we are interested in here are the gravitational force acting downwards and the drag force due to the oil. The gravitational force, \(F_{gravity}\), can be calculated using the component of gravitational acceleration, \(g\), acting along the incline. So, \(F_{gravity} = mg \sin (\theta)\) where \(m = 5 kg\) is the mass of the block, \(g= 9.8 m/s^2\) is the gravitational acceleration, and \(\theta = 30^{\circ}\) is the angle of inclination.
02

Initial Acceleration

For the initial acceleration, as the block is initially at rest, the velocity \(v = 0\). Therefore, the drag force \(F_{drag}\) = 0 (as it is proportional to velocity). So, the net force on the block is the gravitational force along the incline, leading to the initial acceleration \(a_0 = \frac{F_{gravity}}{m} = g \sin (\theta)\).
03

Expression for Speed

For the drag force, we use Stokes' Law which states that it is proportional to the velocity and can be given as \(F_{drag} = 6 \pi \mu r v\), where \(\mu\) is the dynamic viscosity, \(r\) is radius of the block and \(v\) is the velocity. But for a square block, as given in the problem, we use \(F_{drag} = \mu Av \) where A = \(0.1 m^2\) is area of the block. Equate this drag force to the force due to gravity along the incline leads us to the equation of motion, which can be solved analytically using standard methods to get velocity as a function of time, \(v(t)\).
04

Speed after 0.1s

Substitute \(t = 0.1s\) in the equation \(v(t)\) obtained above to find the speed.
05

Calculation of Viscosity

With the desire to make the block reach a speed of \(0.3 m/s\) after the same time interval, \(t = 0.1s\), we have to calculate a new viscosity. Plugging \(v = 0.3 m/s\) and \(t = 0.1s\) into the equation for the velocity, we can solve for the new viscosity \(\mu\).

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Most popular questions from this chapter

A concentric-cylinder viscometer is shown. Viscous torque is produced by the annular gap around the inner cylinder. Additional viscous torque is produced by the flat bottom of the inner cylinder as it rotates above the flat bottom of the stationary outer cylinder. Obtain an algebraic expression for the viscous torque due to flow in the annular gap of width \(a\). Obtain an algebraic expression for the viscous torque due to flow in the bottom clearance gap of height \(b\). Prepare a plot showing the ratio, \(b / a,\) required to hold the bottom torque to 1 percent or less of the annulus torque, versus the other geometric variables. What are the design implications? What modifications to the design can you recommend?

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.

Fluids of viscosities \(\mu_{1}=0.1 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) and \(\mu_{2}=0.15 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) are contained between two plates (each plate is \(1 \mathrm{m}^{2}\) in area). The thicknesses are \(h_{1}=0.5 \mathrm{mm}\) and \(h_{2}=0.3 \mathrm{mm},\) respectively. Find the force \(F\) to make the upper plate move at a speed of \(1 \mathrm{m} / \mathrm{s}\). What is the fluid velocity at the interface between the two fluids?

Tape is to be coated on both sides with glue by drawing it through a narrow gap. The tape is 0.015 in. thick and 1.00 in. wide. It is centered in the gap with a clearance of 0.012 in. on each side. The glue, of viscosity \(\mu=0.02\) slug \(/(\mathrm{ft} \cdot \mathrm{s})\) completely fills the space between the tape and gap. If the tape can withstand a maximum tensile force of 25 lbf, determine the maximum gap region through which it can be pulled at a speed of 3 ft/s.

The cone and plate viscometer shown is an instrument used frequently to characterize non-Newtonian fluids. It consists of a flat plate and a rotating cone with a very obtuse angle (typically \(\theta\) is less than 0.5 degrees). The apex of the cone just touches the plate surface and the liquid to be tested fills the narrow gap formed by the cone and plate. Derive an expression for the shear rate in the liquid that fills the gap in terms of the geometry of the system. Evaluate the torque on the driven cone in terms of the shear stress and geometry of the system.
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