Calculate Torque of 24 cm Wheel in XY Plane w/ Force (-35i + 39j) N

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SUMMARY

The torque of a wheel with a diameter of 24 cm, subjected to a force of F = (-35.0 i + 39.0 j) N, can be calculated using the cross-product definition of torque. The position vector at the edge of the wheel, located on the x-axis, is crucial for this calculation. The torque vector is determined by the formula τ = r × F, where r is the position vector and F is the force vector. The resulting torque about the z-axis is expressed in terms of its components.

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A wheel of diameter 24.0 cm is constrained to rotate in the xy plane, about the z axis, which passes through its center. A force F = (-35.0 i + 39.0 j) N acts at a point on the edge of the wheel that lies exactly on the x-axis at a particular instant. What is the torque about the rotation axis at this instant?

i already got T=0i+0j mN which i still don't exactly know why, but i can't figure out what k is.
 
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Eeh??
Your answer is wrong in so much.
Do you know about the cross-product definition of the torque?
 
1. What is the position vector of the point at which the force is being applied?
2. What is the definition of torque?

If you can answer these two points, then you can answer your question.
 

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