# Problem 005-mm | Method of Members

**Problem 005-cb**

For the cabled structure in Fig. 005(FR-CB), member ABC which is assumed to be rigid is pinned at A and held in equilibrium by cable CD. For this structure, determine the reaction at A and the tension in the cable.

**Solution 005-cb**

$b = 12 \sin 60^\circ = 6\sqrt{3} \, \text{ ft}$

$c = 8 \sin 60^\circ = 4\sqrt{3} \, \text{ ft}$

$d = b \tan 37^\circ = 6\sqrt{3} \tan 37^\circ \, \text{ ft}$

$\Sigma M_A = 0$

$12T_n = 10\,000c + 8000b$

$12(T \sin 67^\circ) = 10\,000(4\sqrt{3}) + 8000(6\sqrt{3})$

$T = 13\,798.63 \, \text{ lb}$ *answer*

$\Sigma M_D = 0$

$A_H(a + d) = 10\,000c + 8000b$

$A_H(6 + 6\sqrt{3} \tan 37^\circ) = 10\,000(4\sqrt{3}) + 8000(6\sqrt{3})$

$A_H = 11\,020.08 \, \text{ lb}$ *answer*

$\Sigma F_V = 0$

$A_V + T \sin 37^\circ = 10\,000 + 8000$

$A_V + 13\,798.63 \sin 37^\circ = 10\,000 + 8000$

$A_V = 9695.78 \, \text{ lb}$ *answer*