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static equilibrium question (1 Viewer)
- Thread starter astroman
- Start date
To me, this is more of an engineering studies question than a physics question, but I'll give it a shot anyway.
Since the stick is in equilibrium, the clockwise torques must equal the counter-clockwise torque.
In the diagram I made, the clockwise torques were the force at 10 cm and the mass (centred at 20 cm) and my counter-clockwise torque was the mass centred at 70 cm. These may be opposite in your diagram, you'll still get the same answer.
Since the stick is in equilibrium, the clockwise torques must equal the counter-clockwise torque.
In the diagram I made, the clockwise torques were the force at 10 cm and the mass (centred at 20 cm) and my counter-clockwise torque was the mass centred at 70 cm. These may be opposite in your diagram, you'll still get the same answer.
Fizzy_Cyst
Well-Known Member
Alternatively, noting the first 40cm on both sides would provide balancing torques and essentially cancel each other out you would be left with:
1 x 0.3 = 20/100m x 9.8 x 0.5
And you should get the exact same answer.
1 x 0.3 = 20/100m x 9.8 x 0.5
And you should get the exact same answer.