Problem 001-mm | Method of Members

$b = 16 \sin 30^\circ = 8 \text{ m}$

$c = a \tan 37^\circ = 13.86 \tan 37^\circ = 10.44 \text{ m}$
 

Tension on cable CD
$\Sigma M_A = 0$

$(T \sin 53^\circ)(8 + 10.44) = 8000(13.86) + 12\,000(4)$

$T = 10\,788.47 \, \text{ lb}$           answer
 

Reaction at A
$\Sigma M_D = 0$

$A_x(8 + 10.44) = 8000(13.86) + 12\,000(4)$

$A_x = 8616.05 \, \text{ lb}$
 

$\Sigma F_V = 0$

$A_y + T \cos 53^\circ = 8000 + 12\,000$

$A_y + 10\,788.47 \cos 53^\circ = 8000 + 12\,000$

$A_y = 13\,507.34 \, \text{ lb}$
 

$R_A = \sqrt{{A_y}^2 + {A_x}^2}$

$R_A = \sqrt{8616.05^2 + 13\,507.34^2}$

$R_A = 16\,021.38 \, \text{ lb}$
 

$\tan \theta_{Ax} = \dfrac{A_y}{A_x} = \dfrac{13\,507.34}{8616.05}$

$\theta_{Ax} = 57.47^\circ$
 

Thus, R_A = 16 021.38 lb at θ_Ax = 57.47° with the horizontal.           answer
 

Force on member CD
$\tan \beta = \dfrac{8}{13.86 - 8}$

$\beta = 53.78^\circ$
 

$\Sigma M_A = 0$

$(F_{BC} \sin \beta)(8) = 12\,000(4)$

$(F_{BC} \sin 53.78^\circ)(8) = 12\,000(4)$

$F_{BC} = 7437.21 \, \text{ lb tension}$           answer