Solution to Problem 628 | Moment Diagrams by Parts

$10R_1 + 400 = \frac{1}{2}(6)(200)(2)$

$R_1 = 80 \, \text{lb}$
 

$\Sigma M_{R1} = 0$

$10R_2 = 400 + \frac{1}{2}(6)(200)(8)$

$R_2 = 520 \, \text{lb}$
 

$(Area_{AB})\,\bar{X}_A = 400(8)(6) + \frac{1}{2}(10)(800)(\frac{20}{3}) - \frac{1}{4}(6)(1200)(\frac{44}{5})$

$(Area_{AB})\,\bar{X}_A = 30\,026.67 \, \text{lb}\cdot\text{ft}^3$           answer
 

$(Area_{AB})\,\bar{X}_B = 400(8)(4) + \frac{1}{2}(10)(800)(\frac{10}{3}) - \frac{1}{4}(6)(1200)(\frac{6}{5})$

$(Area_{AB})\,\bar{X}_B = 23\,973.33 \, \text{lb}\cdot\text{ft}^3$           answer