From the FBD of the whole system

$\Sigma M_F = 0$
$4A_H + 12A_V + 5(300 \times 10) = 6(1000)$

$4A_H + 12A_V = -9000$

$A_H + 3A_V = -2250$ ← Equation (1)

$\Sigma M_A = 0$
$12F_V = 4F_H + 6(1000) + 9(300 \times 10)$

$12F_V - 4F_H = 33\,000$

$3F_V - F_H = 8\,250$ ← Equation (2)

$\Sigma F_H = 0$

$A_H = F_H + 300(10)$

$A_H = F_H + 3000$ ← Equation (3)

From FBD of member BC

$\Sigma M_B = 0$
$10C_H = 5(300 \times 10)$

$C_H = 1500 ~ \text{lb}$

From the FBD of member CD

$\Sigma F_H = 0$
$D_H = 1500 ~ \text{lb}$

$From the FBD of member DF

$\Sigma M_E = 0$
$4F_H = 4(1500)$

$F_H = 1500 ~ \text{lb}$

Substitute F_{H} = 1500 lb to Equation (2)

$3F_V - 1\,500 = 8\,250$
$F_V = 3250 ~ \text{lb}$

Substitute F_{H} = 1500 lb to Equation (3)

$A_H = 1500 + 3000$
$A_H = 4500 ~ \text{lb}$

Substitute AH = 4500 lb to Equation (1)

$4500 + 3A_V = -2250$
$A_V = -2250 ~ \text{lb}$

**Answer Summary**

A_{H} = 4500 lb to the left

A_{V} = 2250 lb downward

F_{H} = 1500 lb to the right

F_{V} = 3250 lb upward
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