This problem involves calculating the weight $W$ and the tension $T$ to maintain the equilibrium of a rigid body, considering the given distances and forces.

Step 1: Equilibrium Conditions

For equilibrium:

1. Sum of vertical forces = 0: $T = W + 50 \, \text{N}$
2. Sum of moments about a point (e.g., the pivot at the rope) = 0: Taking moments about the point where the tension $T$ acts: $W \cdot 3 = 50 \cdot 6$

Step 2: Solve for $W$

From the moment equation: $W \cdot 3 = 50 \cdot 6$ $W = \frac{50 \cdot 6}{3} = 100 \, \text{N}$

Step 3: Solve for $T$

From the vertical forces equation: $T = W + 50$ $T = 100 + 50 = 150 \, \text{N}$

Final Results

• Weight $W = 100 \, \text{N}$
• Tension $T = 150 \, \text{N}$

Would you like further explanation or a diagram?

Related Questions:

1. What if the distances were reversed (3 m and 6 m for the weights)?
2. How do we consider the bar's weight if it’s not negligible?
3. What happens to the tension if an additional force acts on the center?
4. How does changing the length of the bar affect the equilibrium conditions?
5. Can you analyze a similar problem with a non-horizontal bar?

Tip:

Always verify both moment and force equilibrium equations for accuracy in such problems!