$\Sigma M_{R_2} = 0$

$LR_1 = \frac{1}{3} LF$

$R_1 = \frac{1}{3}\,( \, \frac{1}{2}Lw_o \, )$

$R_1 = \frac{1}{6} Lw_o$

$\Sigma M_{R_1} = 0$

$LR_2 = \frac{2}{3} LF$

$R_2 = \frac{2}{3} \, ( \, \frac{1}{2} Lw_o \, )$

$R_2 = \frac{1}{3} Lw_o$

$\dfrac{y}{x} = \dfrac{w_o}{L}$

$y = \dfrac{w_o}{L}x$

$F_x = \frac{1}{2} xy = \frac{1}{2}x \left( \dfrac{w_o}{L}x \right)$

$F_x = \dfrac{w_o}{2L}x^2$

$V = R_1 - F_x$

$V = \frac{1}{6} Lw_o - \dfrac{w_o}{2L}x^2$

$M = R_1x - F_x\,(\frac{1}{3}x)$

$M = \frac{1}{6}Lw_ox - \dfrac{w_o}{2L}x^2\,(\frac{1}{3}x)$

$M = \frac{1}{6}Lw_ox - \dfrac{w_o}{6L}x^3$

**To draw the Shear Diagram:**

V = 1/6 Lw_{o} - w_{o}x^{2}/2L is a second degree curve; at x = 0, V = 1/6 Lw_{o} = R_{1}; at x = L, V = -1/3 Lw^{o} = -R_{2}; If a is the location of zero shear from left end, 0 = 1/6 Lw_{o} - w_{o}x^{2}/2L, x = 0.5774L = a; to check, use the squared property of parabola:

a^{2}/R_{1} = L^{2}/(R_{1} + R_{2})

a^{2}/(1/6 Lw_{o}) = L^{2}/(1/6 Lw_{o} + 1/3 Lw_{o})

a^{2} = (1/6 L^{3}w_{o})/(1/2 Lw_{o}) = 1/3 L_{2}

a = 0.5774L

**To draw the Moment Diagram:**

M = 1/6 Lw_{o}x - w_{o}x^{3}/6L is a third degree curve; at x = 0, M = 0; at x = L, M = 0; at x = a = 0.5774L, M = M_{max}.

M_{max} = 1/6 Lw_{o}(0.5774L) - w_{o}(0.5774L)^{3}/6L

M_{max} = 0.0962L^{2}w_{o} - 0.0321L^{2}w_{o}

M_{max} = 0.0641L^{2}w_{o}