Chapter 4, Problem 6P

A 300-lb garage door is being opened by pulling on the cable as shown. As the door is lifted the force, F, in the cable, as a function of the angle ?, is given by:

$F=\frac{300\cdot 4.5\sin \theta }{3\cos (\alpha -\theta )}$

where

$\sin \alpha =\frac{1+3\cos \theta }{\sqrt{{(1+3\cos \theta )}^2+{(3-3\sin \theta )}^2}}$

Calculate F for ?=0° through 90° with increments of 10°. Display the results in a two-column table.