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Torque Rotational Eq

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Physics has been so long, and I don't remember anything except T=rFsin(theta)
In the answer key, it says the 10kg weight is applying a torque of approximately 35Nm. <-how did they calculate this: (35cm to 35Nm?)

F=T/(rsin(theta)) eq manipulation to solve for F
T=35Nm; r=.02m; theta=75`

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