Answer to Question #249918 in Calculus for Phenominal1

Question

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

Answer 1

The force F (in pounds) acting at an angle $\theta$ with the horizontal that is needed to drag a crate weighing W pounds along a horizontal surface at a constant velocity is given by $F=\dfrac{\mu W}{\cos \theta+\mu \sin \theta}$ where $\mu$ is a constant called the coefficient of sliding friction between the crate and the surface. Suppose that the crate weighs $150 \mathrm{lb}$ and that $\mu=0.3$

- Find $d F / d \theta$ when $\theta=30^{\circ}$ . Express the answer in units of pounds/degree.

$\theta = 30° × \dfracπ{180°}\ rad= \dfracπ{6}\text{ rad}$

$F= \dfrac{0.3× 150}{(\cosθ+0.3\sinθ)}$

$F= \dfrac{45}{(\cosθ+0.3\sinθ)}$

$\dfrac{dF}{dθ}=\dfrac{[(\cosθ+0.3\sinθ)(\dfrac{d}{dθ}45)]-[45(\dfrac d{dθ}\cosθ + 0.3\dfrac d{dθ }\sinθ)] }{(\cosθ+0.3\sinθ)^2 }$

$\dfrac{dF}{dθ}=\dfrac{[-45(-\sinθ+0.3\cosθ)]}{(\cosθ+0.3\sinθ)^2}$

$\dfrac{dF}{dθ} =\dfrac{(45\sinθ-13.5\cosθ)}{(\cosθ+0.3\sinθ)^2}$