A uniform plank of length 3.00 m and weight 44.0 N rests on two pivots, Y and X. Pivot Y is 0.30 m from the left end and pivot X is 0.50 m from the right end. Calculate the reaction force at pivot X.

The length of the plank $L = 3.00 \text{ m}$.
The weight of the plank $W = 44.0 \text{ N}$. Since the plank is uniform, its weight acts at its center, which is at $\frac{L}{2} = \frac{3.00 \text{ m}}{2} = 1.50 \text{ m}$ from either end.
Pivot Y is $30.0 \text{ cm} = 0.30 \text{ m}$ from the left end.
Pivot X is $50.0 \text{ cm} = 0.50 \text{ m}$ from the right end.

To find the reaction force at X ($R_X$), we take moments about pivot Y. This eliminates the reaction force at Y ($R_Y$) from the equation.

• Distance of the plank's weight ($W$) from Y: The center of the plank is at $1.50 \text{ m}$ from the left end. Pivot Y is at $0.30 \text{ m}$ from the left end. $$d_W = 1.50 \text{ m} - 0.30 \text{ m} = 1.20 \text{ m}$$ This weight creates a clockwise moment about Y.
• Distance of the reaction force at X ($R_X$) from Y: The total length of the plank is $3.00 \text{ m}$. The distance from the left end to Y is $0.30 \text{ m}$. The distance from the right end to X is $0.50 \text{ m}$. The distance between Y and X is: $$d_X = L - (\text{distance from left end to Y}) - (\text{distance from right end to X})$$ $$d_X = 3.00 \text{ m} - 0.30 \text{ m} - 0.50 \text{ m} = 2.20 \text{ m}$$ This reaction force creates an anti-clockwise moment about Y.

Sum of clockwise moments = Sum of anti-clockwise moments

Substitute the known values:

Comparing this result with the given options:
A. $40.5 \text{ N}$
B. $24.0 \text{ N}$
C. $20.0 \text{ N}$
D. $16.2 \text{ N}$

The calculated reaction at X matches option B.

The final answer is $\boxed{\text{B. } 24.0 \text{ N}}$.