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(a)Resultant of load \left(1800 \frac{\mathrm{N}}{\mathrm{m}}\right. acts at \frac{1}{3}(1.5) fiom A=\frac{1.5}{3}=0.5 \mathrm{~m}A_{x}Resultant of load \left(600 \frac{\mathrm{N}}{\mathrm{m}}\right) acts \frac{1}{3}(2.5) from B=4-\frac{2.5}{3}=3.17 \mathrm{~m}moment about A \quad \Sigma M_{A}=0\begin{array}{l}B_{y}(4)-\frac{1}{2}(600)(2.5)(3.17)-\frac{1}{2}(1800)(1.5)(0.5) \\=0 \\B y=763.125 \mathrm{~N} \\\sum F_{y}=0 \Rightarrow A y+B y-\frac{1}{2}(1800)(1.5) \\-\frac{1}{2}(600)(2.5)=0 \\A_{y}+763-125-2100=0 \\A y=1336.875 \mathrm{~N} \\\sum f_{x}=0 \Rightarrow A_{x}=0 \\\end{array}(b) Resultant of load ( \left.40 \frac{\mathrm{lb}}{\mathrm{in}}\right) acts at \frac{10}{2} fiom A=5 \mathrm{~m} \mathrm{fiom} AReswlant of load \left(90-40=50 \frac{(b}{\text { in }}\right) acts at \left(\frac{6}{3}\right) from B=2 \mathrm{~m} fiom B=8 \mathrm{~m} fiom A\begin{array}{l} \Sigma F_{y}=0 \Rightarrow 40(10)+\frac{1}{2}(50)(6)-B y=0 \\\Rightarrow B_{y}=550(b) \\\Sigma F_{x}=0 \Rightarrow B x=0 \\\Sigma M=0 \Rightarrow 40(10)(5)+\frac{1}{2}(50)(6)(2)-M_{B}=0 \\\left.M_{B}=0300 \mathrm{~b} . \mathrm{in}\right)\end{array} ...