Beam Forces 56D3E9

1. **State the problem:** We have a horizontal beam from $0$ cm to $40$ cm with five forces applied at different points and directions. We want to analyze the forces step by step. 2. **List the forces and their components:** - At $0$ cm: $10$ N downward (red). - At $10$ cm: $20$ N at $30^\circ$ below horizontal, down-right (blue). - At $20$ cm: $50$ N upward (green). - At $30$ cm: $20$ N at $65^\circ$ above horizontal, up-right (purple). - At $40$ cm: $10$ N at $45^\circ$ below horizontal, down-right (orange). 3. **Resolve angled forces into horizontal and vertical components:** For force at $10$ cm: $$F_{10x} = 20 \cos 30^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32\,N$$ $$F_{10y} = -20 \sin 30^\circ = -20 \times \frac{1}{2} = -10\,N$$ (negative because downward) For force at $30$ cm: $$F_{30x} = 20 \cos 65^\circ = 20 \times 0.4226 = 8.45\,N$$ $$F_{30y} = 20 \sin 65^\circ = 20 \times 0.9063 = 18.13\,N$$ (positive because upward) For force at $40$ cm: $$F_{40x} = 10 \cos 45^\circ = 10 \times \frac{\sqrt{2}}{2} = 7.07\,N$$ $$F_{40y} = -10 \sin 45^\circ = -10 \times \frac{\sqrt{2}}{2} = -7.07\,N$$ (negative because downward) 4. **Sum horizontal forces:** Only forces at $10$ cm, $30$ cm, and $40$ cm have horizontal components: $$\sum F_x = F_{10x} + F_{30x} + F_{40x} = 17.32 + 8.45 + 7.07 = 32.84\,N$$ 5. **Sum vertical forces:** Forces at $0$ cm, $10$ cm, $20$ cm, $30$ cm, and $40$ cm vertical components: $$\sum F_y = (-10) + (-10) + 50 + 18.13 + (-7.07) = 41.06\,N$$ 6. **Calculate net force magnitude and direction:** $$F_{net} = \sqrt{(32.84)^2 + (41.06)^2} = \sqrt{1078.6 + 1685} = \sqrt{2763.6} \approx 52.56\,N$$ $$\theta = \tan^{-1} \left( \frac{41.06}{32.84} \right) = \tan^{-1}(1.25) \approx 51.34^\circ$$ above horizontal. 7. **Summary:** The net force on the beam is approximately $52.56$ N directed $51.34^\circ$ above the horizontal to the right.