1. **State the problem:** Calculate the sum of the moments about the fixed end of the beam. 2. **Identify forces and distances:** - Force $F_1 = 15$ N upward at distance $d_1 = 1.4$ m - Force $F_2 = 10$ N downward at distance $d_2 = 2.6$ m (1.4 + 1.2) - Force $F_3 = 10$ N at $80^\circ$ downward to the right at distance $d_3 = 4.2$ m (1.4 + 1.2 + 1.6) 3. **Moment formula:** $$ M = F \times d \times \sin(\theta) $$ where $\theta$ is the angle between the force direction and the lever arm (beam). 4. **Calculate moments:** - $M_1 = 15 \times 1.4 = 21$ Nm (counterclockwise, positive) - $M_2 = 10 \times 2.6 = 26$ Nm (clockwise, negative) - For $F_3$, vertical component is $F_3 \sin 80^\circ$ downward, so moment: $$ M_3 = 10 \times 4.2 \times \sin 80^\circ $$ Calculate $\sin 80^\circ \approx 0.9848$: $$ M_3 = 10 \times 4.2 \times 0.9848 = 41.3 \text{ Nm (clockwise, negative)} $$ 5. **Sum moments:** $$ \sum M = M_1 - M_2 - M_3 = 21 - 26 - 41.3 = -46.3 \text{ Nm} $$ 6. **Interpretation:** Negative sign means net moment is clockwise. **Final answer:** $$ \boxed{-46.3 \text{ Nm (clockwise)}} $$