The moment at C due to reaction R

_{A} is R

_{A}L and the moment at C due to uniform load w

_{o} is –w

_{o}b(0.5b) = -½w

_{o}b

^{2}.

The deviation at A from tangent line through C is zero. Thus,

$EI \, t_{A/C} = (Area_{AC}) \bar{X}_A = 0$

$\frac{1}{2}L(R_AL)(\frac{2}{3}L) - \frac{1}{3}b(\frac{1}{2}w_ob^2)(L - \frac{1}{4}b) = 0$

$\frac{1}{3}R_AL^3 - \frac{1}{6}w_ob^3(L - \frac{1}{4}b) = 0$

$\frac{1}{3}R_AL^3 - \frac{1}{24}w_ob^3(4L - b) = 0$

$\frac{1}{3}R_AL^3 = \frac{1}{24}w_ob^3(4L - b)$

$R_A = \dfrac{w_ob^3}{8L^3}(4L - b)$ *answer*