- #1

- 1

- 0

Me trying to solve it:

sigma T=0

T1+T2=0

T1 = 0 <--this is the pivot point, i think it equals 0 b/c we what to reach equilibrium

T2 = rFsin(theta)

0+rFsin(theta)

F=-0/(rsin(theta)) ???

- Thread starter supercherrie
- Start date

- #1

- 1

- 0

Me trying to solve it:

sigma T=0

T1+T2=0

T1 = 0 <--this is the pivot point, i think it equals 0 b/c we what to reach equilibrium

T2 = rFsin(theta)

0+rFsin(theta)

F=-0/(rsin(theta)) ???

- #2

- 34,343

- 5,895

It's the '=0' in 'T1+T2=0' that says we're to reach equilibrium. T1 is the torque from the gravitational force. Where is the centre of mass of the stick? How far from the pivot point? What torque does it exert?T1+T2=0

T1 = 0 <--this is the pivot point, i think it equals 0 b/c we what to reach equilibrium

As for the string, it's 30 degrees to the vertical. Be careful about sine versus cosine. (As a check, I always think to myself, what if the specified angle were zero? Would the force be zero or max?)

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