A uniform heavy beam of length 10 m and mass 50 kg is supported...

Let $N_A$ and $N_B$ be the normal reaction forces exerted by pillars A and B, respectively.

The forces acting on the beam are:
1. Weight of the uniform beam, $W_{\text{beam}} = 50 \times 10 = 500\text{ N}$, acting at its center of mass, which is at a distance of $5\text{ m}$ from pillar A.
2. Weight of the block, $W_{\text{block}} = 30 \times 10 = 300\text{ N}$, acting at a distance of $2\text{ m}$ from pillar A.
3. Normal reaction $N_B$ acting upwards at a distance of $10\text{ m}$ from pillar A.

Since the beam is in rotational equilibrium, the net torque about any point must be zero. Taking the torque about point A:

$\sum \tau_A = 0$

The clockwise torques due to the weights are balanced by the counter-clockwise torque due to $N_B$:

$N_B \times 10 = (W_{\text{beam}} \times 5) + (W_{\text{block}} \times 2)$

$10 N_B = (500 \times 5) + (300 \times 2)$

$10 N_B = 2500 + 600$

$10 N_B = 3100$

$N_B = 310\text{ N}$

Therefore, the normal reaction force exerted by pillar B is $310\text{ N}$.

Answer: $310\text{ N}$