Find Torque of Circle Homework Solution

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SUMMARY

The net torque of a circular object with an inner radius of 12 cm and an outer radius of 20 cm, subjected to three forces of 11 N, 28 N, and 16 N, is calculated incorrectly in the discussion. The force of 28 N acts at an angle of 29 degrees below the horizontal, which affects its contribution to the torque. The correct formula for torque must account for the angle of the force, leading to a revised calculation that includes the sine of the angle. The initial calculation of 2.04 Nm is incorrect due to the misinterpretation of the force's direction.

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Homework Statement



A Circular-shaped object has an inner radius of 12 cm and an outer radius of 20 cm. Three forces (acting perpendicular to the axis of rotation) of magnitudes 11 N, 28N, and 16 N act on the object. The force of magnitude 28 N acts 29 degrees below the horizontal. Find the magnitude of the net torque through the center of the object.

torque.jpg


Homework Equations

The Attempt at a Solution


So I added all the torques together.
-(.12*28)+(.2*11)+(.2*16)=2.04
which came out to 2.04 Nm. Am I right? When I put it, it said it was wrong...
 
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Hi minikk, welcome to PF.
From the diagram I feel that the force 28 N may not be perpendicular to the radius of the smaller circle. That is why they have given the angle of the force below the horizon.
 

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