# Problem 004-mm | Method of Members

**Problem 004-mm**

For the structure shown in Fig. FR-004(MM), members AD, DC, and ABC are assumed to be solid rigid members; member ED is a cable. For this structure, determine the reaction at A, the tension on cable ED, and the force in member DC.

**Solution 004-mm**

$x = 5 \cos 37^\circ = 4 \, \text{ m}$

$y = 5 \sin 37^\circ = 3 \, \text{ m}$

$\Sigma M_A = 0$

$6(\frac{4}{5}T) = 2(100) + 6(800)$

$T = 1041.67 \, \text{ kN}$ *answer*

$\Sigma M_E = 0$

$6A_H = 2(100) + 6(800)$

$A_H = 833.33 \, \text{ kN}$ *answer*

$\Sigma F_V = 0$

$A_V + \frac{3}{5}T = 100 + 800$

$A_V + \frac{3}{5}(1041.67) = 900$

$A_V = 275 \, \text{ kN}$ *answer*

From FBD of member ABC

$\Sigma M_A = 0$

$6(\frac{3}{\sqrt{13}}F_{CD}) = 2(100) + 6(800)$

$F_{CD} = 1001.54$ *answer*

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