238 Finding the resultant of trapezoidal loading

$F_1 = 6(220) = 1320 \, \text{ N}$

$F_2 = \frac{1}{2}(6)(670) = 2010 \, \text{ N}$
 

 

$R = F_1 + F_2 = 1320 + 2010$

$R = 3330 \, \text{ N}$
 

$Rd = 3F_1 + 4F_2$

$3330d = 3(1320) + 4(2010)$

$3330d = 12000$

$d = 3.6 \, \text{ m}$
 

Thus, R = 3330 N downward at 3.6 m to the right of A.           answer