Pendulum problem - What is the magnitude of the torque?

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SUMMARY

The discussion centers on calculating the torque of a pendulum system with a mass of 0.75 kg attached to a 1.25 m long massless rod, positioned at a 30-degree angle from the vertical. The initial calculations incorrectly yielded a torque of -9.2 N, but upon reevaluation using the correct angle and vector resolution, the torque was determined to be 4.6 Nm. The key to resolving the discrepancy was identifying the correct angle of 60 degrees for the vector magnitude acting on the tangent of the rod's arc. Accurate diagram representation is crucial for solving such problems.

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a small mass of 0.75kg is attached to one end of a 1.25m long massless rod, and the other end is hung from a pivot. When the resulting pendulum is 30degrees from the vertical what is the magnitude of the torque about the pivot?

what i did:
-Fsin30 -mg = 0
Fsin30 = -mg
F = -14.7N

t = rFsin30
t = 1.25(-14.7sin30)
t = -9.2N

the answer is supose to be half of this what did i do wrong?
 
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I always always draw out the problem and figure it out geometrically on paper rather than resort to filling in formuals. Check out the attachment. Its easy to see from this that you need to find the vector mag. that acts on the tangent of the rods arc. Get the angle. Its 60. Get the resolved vector magnitude.

cos 60= F/7.36

Torque = 3.68 x 1.25
=4.6 Nm

Was your diagram maybe screwed up?>?
 

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