Question

replace the loading shown in (figure 1) by an equivalent resultant force and couple moment acting at point a. suppose that w = 350 n/m.
part b
determine the couple moment. express your answer to three significant figures and include the appropriate units. assume the positive direction is counterclockwise.

Explanation:

Step1: Calculate resultant force of the distributed - load

The distributed - load can be divided into a rectangular part and a triangular part. The rectangular part has a force $F_1 = w\times3$ and the triangular part has a force $F_2=\frac{1}{2}\times w\times3$. Given $w = 350$ N/m.
$F_1=350\times3 = 1050$ N, $F_2=\frac{1}{2}\times350\times3=525$ N. The resultant force $F = F_1 + F_2=1050 + 525=1575$ N.

Step2: Calculate the moment of each part about point A

The moment of the rectangular part about point A, $M_1=F_1\times\frac{3}{2}=1050\times1.5 = 1575$ N·m. The moment of the triangular part about point A, $M_2=F_2\times\frac{3\times2}{3}=525\times2 = 1050$ N·m.

Step3: Calculate the total couple moment about point A

The couple moment $(M_R)_A = M_1+M_2$.
$(M_R)_A=1575 + 1050=2625$ N·m.

Answer:

Value: $2630$ (rounded to three significant figures)
Units: N·m