Answer to Question #252401 in Calculus for Steve

Question

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.

Answer 1

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

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