Answer to Question #252401 in Calculus for Steve

Question #252401

The force F (in pounds) acting at an angle θ with the horizontal that is needed to drag a crate weighing W pounds along a horizontal surface at a constant velocity is given by

F = μW/(cosθ +μsinθ)


where μ is a constant called the coefficient of sliding friction between the crate and the surface (see the accompanying figure). Suppose that the crate weighs 150 lb and that μ = 0.3.

(a) Find dF /dθ when θ =30°. Express the answer in units of pounds/degree.


(b) Find dF /dt when θ =30° if θ is decreasing at the rate of 0.5°/s at this instant.


1
Expert's answer
2021-10-18T13:51:04-0400

"F = \\frac{\u03bcW}{(cos\u03b8 +\u03bcsin\u03b8)}\\\\\n\\therefore \\frac{dF}{d\\theta}=\\mu W\\times(\\frac{-1}{(cos\u03b8 +\u03bcsin\u03b8)^2})\\times(-sin\\theta +\\mu cos \\theta)\\\\\n(a) \\ (\\frac{dF}{d\\theta})_{\\theta=30\\degree}=(0.3\\times 150)\\times (\\frac{-1}{(cos 30\\degree+(0.3)sin 30\\degree)^2})\\times (-sin30\\degree+(0.3)cos30\\degree)\\\\\n=45\\times (\\frac{-1}{1.03})\\times (-0.24)\\\\\n=10.49 \\ pounds\/degree"


"(b) \\ (\\frac{dF}{dt})=(0.3\\times 150)\\times (\\frac{-1}{(cos \\theta+(0.3)sin \\theta)^2})\\times (-sin\\theta+(0.3)cos\\theta)\\times (\\frac{d\\theta}{dt})\\\\\n=45\\times \\frac{sin\\theta-(0.3)cos\\theta}{(cos \\theta+(0.3)sin \\theta)^2})\\times (-0.5)\\\\\n=-22.5(\\frac{sin\\theta-(0.3)cos\\theta}{(cos \\theta+(0.3)sin \\theta)^2})"


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