Question
What force is needed to pull one microscope slide over another at a speed of 1.00 cm/s, if there is a 0.500-mm-thick layer of 20C20^\circ\textrm{C} water between them and the contact area is 8.00 cm28.00\textrm{ cm}^2?
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Final Answer

1.60×105 N1.60\times 10^{-5}\textrm{ N}

Solution video

OpenStax College Physics for AP® Courses, Chapter 12, Problem 30 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We're told that a microscope slide of a certain area of eight centimeters squared is pulled at a speed of one centimeter per second. And it's separated by a distance of 0.5 millimeters from another microscope slide with water in between them and we'll need to know what the viscosity of the water is in order to find the force needed to pull this top slide at that speed. And we look up in our handy table 12.1, for water and at 20 degrees Celsius has the viscosity of 1.002 Milla pascals seconds and the Milla means times ten to the minus three. So I'm gonna write this down as 1.002 times ten to the minus three pascals seconds. Doing unit conversion at this stage of writing down the things we know. So this eight square centimeter area of the microscope slide, we convert it into square meters by multiplying by one meter for every one hundred centimeters. And we do that twice. And that's 8.00 times ten to the minus four square meters. The distance separating the two plates, which we call capital L is 0.5 times ten to the minus three meters. Because that's what the prefix milla means times ten to the minus three. And the prefix centa means times ten to the minus two. We have this one times ten to the minus two meters per second. So viscosity of a fluid is defined in terms of this force and the distance between the plates and the speed that is being pulled in in the area of the plate. And we can rearrange it to solve for F by multiplying both sides by V A over L. And that'll isolate F and then we have F then is the speed times the area, times the viscosity over L. So that's 1.00 times ten to the minus two meters per second times eight times ten to the minus four square metres times 1.002 times ten to the minus three pascal seconds divided by 0.500 times ten to the minus three metres. And we get 1.60 times ten to the minus five newtons of force required.