Here is the Free Body Diagram (FBD) for the cantilever beam with the applied forces and reactions as described.

2. Summing the Forces:

• The sum of the vertical forces must be zero: $\Sigma F_y = R_y - 50 \, \text{lb} - 50 \, \text{lb} = 0$ $R_y = 100 \, \text{lb}$ Thus, the vertical reaction at the fixed end is $R_y = 100 \, \text{lb}$.

3. Summing the Moments about the Fixed End:

• Take moments about the fixed end (assuming counterclockwise moments are positive): $\Sigma M = 0$ Moment due to the force at 5 feet: $M_1 = 50 \, \text{lb} \times 5 \, \text{ft} = 250 \, \text{lb-ft} \, \text{(clockwise)}$ Moment due to the force at 8 feet: $M_2 = 50 \, \text{lb} \times 8 \, \text{ft} = 400 \, \text{lb-ft} \, \text{(clockwise)}$ The total moment at the fixed end must balance these: $M = M_1 + M_2 = 250 + 400 = 650 \, \text{lb-ft}$ Thus, the reaction moment at the fixed end is $M = 650 \, \text{lb-ft}$.

Final Reactions:

• Vertical Reaction: $R_y = 100 \, \text{lb}$
• Moment Reaction: $M = 650 \, \text{lb-ft}$

Would you like any further clarifications on this?

Here are some related questions for better understanding:

1. How does the location of the applied forces affect the moment at the fixed end?
2. What happens to the reactions if the forces are moved closer to the fixed support?
3. How would the reactions change if one of the forces were removed?
4. What are the differences in reactions between a simply supported beam and a cantilever beam under the same loading conditions?
5. How do distributed loads affect the calculation of moments and reactions in cantilever beams?

Tip: Always double-check the units when calculating moments to ensure they are consistent across all terms.