Non-Uniform Bar Center of Gravity
Calculating the center of gravity for a suspended bar
This simulation calculates the center of gravity (d) for a non-uniform bar suspended by two strings at angles of 36.9° and 53.1° from the vertical. The bar is 2 meters long and has weight W.
36.9°
53.1°
2.0
Center of Gravity (d)
0.72 m
Distance from left end
Left Tension (T₁)
0.60W
Relative to weight
Right Tension (T₂)
0.80W
Relative to weight
Physics Solution
For equilibrium, the sum of vertical forces must equal the weight (W), and the sum of horizontal forces must be zero. The torques about any point must also balance.
The center of gravity (d) is calculated using the torque equilibrium condition about the left end:
d = (L × T₂ × sinθ₂) / (T₁ × sinθ₁ + T₂ × sinθ₂)
With θ₁=36.9° and θ₂=53.1°, the center of gravity is located 0.72m from the left end.
