To find the vertical reaction at the hinge, we analyze the forces and moments acting on the beam. The system is in static equilibrium, meaning the sum of forces and the sum of torques are zero.

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Step 1: Forces acting on the beam

1. Weight of the beam ($W_{\text{beam}}$): $W_{\text{beam}} = m_{\text{beam}} \cdot g = 15 \cdot 9.8 = 147 \, \text{N}$ The weight of the beam acts at its center, 4 m from the hinge.

2. Weight of the mass ($W_m$): $W_m = m \cdot g = 61 \cdot 9.8 = 597.8 \, \text{N}$ The weight of the mass acts at a point 3 m from the right end of the beam, or 5 m from the hinge.

3. Tension in the rope ($T$): The tension has both vertical and horizontal components:

   • Vertical: $T_y = T \sin(11^\circ)$
   • Horizontal: $T_x = T \cos(11^\circ)$

4. Reactions at the hinge:

   • Vertical reaction: $R_y$ (to be determined)
   • Horizontal reaction: $R_x$

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Step 2: Torque equilibrium

Taking torques about the hinge (counterclockwise positive), the sum of torques must be zero: $\sum \tau = 0$ $T \sin(11^\circ) \cdot 8 - W_{\text{beam}} \cdot 4 - W_m \cdot 5 = 0$

Substituting known values: $T \cdot 8 \cdot \sin(11^\circ) - 147 \cdot 4 - 597.8 \cdot 5 = 0$ $T \cdot 8 \cdot 0.1908 - 588 - 2989 = 0$ $T \cdot 1.5264 = 3577$ $T = \frac{3577}{1.5264} \approx 2343.8 \, \text{N}$

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Step 3: Vertical force equilibrium

The sum of vertical forces must be zero: $R_y + T_y - W_{\text{beam}} - W_m = 0$ Substituting known values: $R_y + T \sin(11^\circ) - 147 - 597.8 = 0$ $R_y + 2343.8 \cdot 0.1908 - 147 - 597.8 = 0$ $R_y + 447 - 147 - 597.8 = 0$ $R_y = 297.8 \, \text{N}$

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Final Answer:

The vertical reaction at the hinge is: $\boxed{R_y = 297.8 \, \text{N}}$

Would you like a detailed explanation or have any follow-up questions? Here are related questions to explore:

1. How do you calculate the horizontal reaction at the hinge?
2. How does changing the angle of the rope affect the reaction forces?
3. What would happen if the beam's mass was concentrated at one end?
4. How does torque equilibrium simplify solving static problems?
5. Why is the center of mass critical in torque calculations?

Tip: When solving for equilibrium, start with torque equations to reduce unknowns in the force equations.