**Problem 001-mm**

The structure shown in Fig F-001(MM) is pinned together at points A, B, and C and held in equilibrium by the cable CD. A load of 12,000 lb is acting at the midpoint of member AB, and a load of 8000 lb is applied at point C. Determine the reaction at A, the internal force in member BC, and the tension on cable CD.

**Solution 001-mm**

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

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

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*