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Ʃτ = 0

ƩF = 0

summing the torques for each piece, i found

Ʃτ = MgLsin(90-ψ) - mgIsin(ψ) = 0

MgLsin(90-ψ) = mgIsin(ψ)

ML/mI = sin(ψ)/sin(90-ψ) = sin(ψ)/cos(-ψ) = tan(ψ)

so ψ = tan

i don't know if this is correct or not...

and I have no idea how I could find the angle without considering torque. any help is appreciated. thanks :)

## Homework Equations

Ʃτ = 0

ƩF = 0

## The Attempt at a Solution

summing the torques for each piece, i found

Ʃτ = MgLsin(90-ψ) - mgIsin(ψ) = 0

MgLsin(90-ψ) = mgIsin(ψ)

ML/mI = sin(ψ)/sin(90-ψ) = sin(ψ)/cos(-ψ) = tan(ψ)

so ψ = tan

^{-1}(ML/mI)i don't know if this is correct or not...

and I have no idea how I could find the angle without considering torque. any help is appreciated. thanks :)