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From the figure,in \triangle A B C,\begin{aligned}& A B=500 \mathrm{~mm}, \quad B C=30 \mathrm{~mm}, \\\therefore & \tan \beta=\frac{B C}{A B}=\frac{30 \mathrm{~mm}}{500 \mathrm{~mm}}, \\& \beta=\tan ^{-1}\left(\frac{30}{500}\right) \Rightarrow \beta=3.434^{\circ} \text { or } 0.0599 \mathrm{rad}\end{aligned}Let the eccentricily of load (500kN/m) be ee-m,\therefore The value of Moment due to eccentricity of 500 \mathrm{uN} / \mathrm{m} should be equell to 50 \mathrm{uN} \cdot \mathrm{m} / \mathrm{m},ie', 500 \mathrm{uN} / \mathrm{m} \times e-m=50 \mathrm{kN} \cdot \mathrm{m} / \mathrm{m}.: e=\frac{1}{10} \mathrm{~m} \quad \Rightarrow e=10 \mathrm{~cm} 3(i) Indination ( \beta )Let the resultant force due to 50 \mathrm{uN} / \mathrm{m} and 600 \mathrm{kN} / \mathrm{m} lee R,The value of R is as, R=\sqrt{600^{2}+50^{2}}=602.0797 \mathrm{ln} / \mathrm{m}The angle of un clination (\beta) of R is as,\begin{array}{l}\therefore \cos \beta=\frac{600 \mathrm{uN} / \mathrm{m}}{R}=\frac{600}{602.0797} \\\because \beta=4.764^{\circ} \text { or } 0.0831 \text { rad }\end{array}(ii) Eccentricily (e): The eccentricity (e):The moment due is eccentricitg (e) of lond 600 \mathrm{kN} / \mathrm{m} is equal to 804 \mathrm{~N} \mathrm{~m} / \mathrm{m},\begin{array}{l}: 600 \mathrm{wN} / \mathrm{m} \times e-\mathrm{m}=80 \mathrm{uN} \mathrm{m} / \mathrm{m} . \\: e=0.133 \mathrm{~m} \text { or } 13.3 \mathrm{~cm}\end{array} ...