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

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

$F_x = \frac{1}{2}xy$

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

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

**Shear equation:**

$V = -\dfrac{w_o}{2L}x^2$

**Moment equation:**

$M = -\frac{1}{3}x F_x = -\dfrac{1}{3}x \left( \dfrac{w_o}{2L}x^2 \right)$

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

**To draw the Shear Diagram:**

- V = - w
_{o} x^{2} / 2L is a second degree curve; at x = 0, V = 0; at x = L, V = -½ w_{o}L.

**To draw the Moment Diagram:**

- M = - w
_{o} x^{3} / 6L is a third degree curve; at x = 0, M = 0; at x = L, M = - 1/6 w_{o}L^{2}.