To solve this problem, we need to determine the force in rod A when a load of $P = 35 \, \text{kN}$ is applied.

Step-by-Step Solution

1. Understanding the Setup

• A rigid beam is pinned at point $O$.
• The beam is supported by two rods:
  • Rod $A$ is located at a distance of $L_1 = 2.5 \, \text{m}$ from point $O$.
  • Rod $B$ is located at a distance of $L_2 = 3.0 \, \text{m}$ from point $O$.
• A downward force $P = 35 \, \text{kN}$ is applied at a point 3 meters to the left of $O$.

2. Identify Reaction Forces

The beam is in static equilibrium, meaning the sum of all forces and moments around point $O$ must be zero.

Let:

• $F_A$ be the force in rod $A$.
• $F_B$ be the force in rod $B$.

3. Sum of Vertical Forces

$F_A + F_B = P = 35 \, \text{kN}$

4. Sum of Moments about Point $O$

Taking moments around point $O$ (counterclockwise moments are positive):

$- F_A \cdot 2.5 + F_B \cdot 3 = P \cdot 3$

Substitute $P = 35 \, \text{kN}$:

$- F_A \cdot 2.5 + F_B \cdot 3 = 35 \cdot 3$

$- F_A \cdot 2.5 + F_B \cdot 3 = 105$

5. Solve the System of Equations

We have two equations:

1. $F_A + F_B = 35$
2. $-2.5F_A + 3F_B = 105$

Substitute $F_B = 35 - F_A$ into the second equation:

$-2.5F_A + 3(35 - F_A) = 105$

Expand and simplify:

$-2.5F_A + 105 - 3F_A = 105$

$-5.5F_A = 0$

$F_A = 0$

6. Substitute Back to Find $F_B$

$F_B = 35 - F_A = 35 - 0 = 35 \, \text{kN}$

Thus, the force in rod A is $0 \, \text{kN}$, and the force in rod B is $35 \, \text{kN}$.

Conclusion

The correct answer appears not to be listed among the options provided. Please confirm if the problem has any additional constraints or different interpretations.