Hi there!
We can go about this problem using a summation of torques.
In order to ensure the rod is in equilibrium, we must satisfy the condition:
Στ = 0
Since the rod is "very light", we can disregard its mass.
The equation for torque is:
τ = rFsinθ
In this instance, the torques are the weights of the objects and their distance from the pivot.
As the rod is 40 cm, the pivot is at 20 cm. Also, the torques must sum up to 0, so:
0 = rF1 - rF2
r1F1 = r2F2
0.20(50) = r2(200)
Solve:
10 = r2(200)
r2 = 0.05 m = 5 cm
The 200N weight must be put at a distance of 5 cm from the OTHER SIDE of the pivot in order to balance the rod.