Problem 325 | Equilibrium of Three-force System

$\cos \beta = \dfrac{1.5}{2}$

$\beta = 41.41^\circ$
 

$30^\circ + \beta = 71.41^\circ$
 

$\phi = 18.59^\circ + \alpha$

$\theta = 90^\circ - \alpha$
 

$\dfrac{P}{\sin 71.41^\circ} = \dfrac{2000}{\sin \phi }$

$P = \dfrac{2000 \sin 71.41^\circ}{\sin (18.59^\circ + \alpha)}$
 

$\dfrac{dP}{d\alpha} = \dfrac{-2000 \sin 71.41^\circ \cos (18.59^\circ + \alpha)}{\sin^2 (18.59^\circ + \alpha)} = 0$

$-2000 \sin 71.41^\circ \cos (18.59^\circ + \alpha) = 0$

$\cos (18.59^\circ + \alpha) = 0$

$18.59^\circ + \alpha = 90^\circ$

$\alpha = 71.41^\circ$           answer
 

$P_{min} = \dfrac{2000 \sin 71.41^\circ}{\sin (18.59^\circ + 71.41^\circ)}$

$P_{min} = 1895.65 \, \text{ lb}$           answer
 

$\phi = 18.59^\circ + 71.41^\circ = 90^\circ$

$\theta = 90^\circ - 71.41^\circ = 18.59^\circ$
 

$\dfrac{R}{\sin \theta} = \dfrac{2000}{\sin \phi}$

$\dfrac{R}{\sin 18.59^\circ} = \dfrac{2000}{\sin 90^\circ}$

$R = 637.59 \, \text{ lb}$           answer