A mast weighing 500 lb is hinged at its bottom to a fixed support at point O (see Figure 3.5.2). The mast is 70 ft long and has a uniform mass distribution, so its center of mass is 35 ft from O. The winch applies a force f = 380 lb to the cable. The mast is supported initially at the 30° angle, and the cable at A is initially horizontal. Derive the equation of motion of the mast. You may assume that the pulley inertias are negligible and that the pulley diameter d is very small compared to the other dimensions.
Part (b) of the figure shows the geometry of the mast at some angle θ > 30°, with the diameterd neglected. From the law of sines,
\sin\phi=P{\frac{\sin(180^{\circ}-\mu-\theta)}{Q}}=P{\frac{\sin(\mu+\theta)}{Q}}From the law of cosines,
Q={\sqrt{P^{2}+L^{2}-2P L\cos(180^{\circ}-\mu-\theta)}}={\sqrt{P^{2}+L^{2}+2P L\cos(\mu+\theta)}} (1)
where H = 20 ft, W = 5 ft, and
\mu=\tan^{-1}\left({\frac{H}{W}}\right)=\tan^{-1}\left({\frac{20}{5}}\right)=76^{\circ}=1.33\,\mathrm{rad}P = {\sqrt{H^{2}+W^{2}}}=20.6~{\mathfrak{m}}
The moment equation about the fixed point O is
I_{O}\ddot{\theta}=-m g R\cos\theta+\frac{f L P}{Q}\sin(\mu+\theta) (2)
The moment of inertia is
I_{O}={\frac{1}{3}}m(70)^{2}={\frac{1}{3}}{\frac{500}{32.2}}70^{2}=25,400\;\mathrm{slug-ft}^{2}The force f at point A is twice the applied force of 380 lb, because of the action of the small pulley. Thus f = 760 lb. With the given values, the equation of motion becomes
25,400\ \ddot{\theta}=-17,500\cos\theta+\frac{626,000}{Q}\sin(1.33+\theta) (3)
where
Q={\sqrt{2020+1650\cos(1.33+\theta)}} (4)
Equation (3) cannot be solved in closed-form to find θ(t), so it must be solved numerically using the methods to be introduced in Chapter 5. Consider how much more complex the model would have to be if we could not neglect the pulley inertia. We could no longer take the cable tension to be the same on each side of the pulley, and we would need to write two additional equations of motion, one for the pulley rotation and one for its translation.