Awheel of diameter 23.0 cm is constrained to rotate in the xy plane, about the z axis, which passes through its center. a force =( -31.0 + 40.0) 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 magnitude of the torque about the rotation axis at this instant?

τ=9.2 N.m

Explanation:

The component that contributes to torque is the y-component of the force F, so:

τ = r x F = (0.23 m) * (40 N) = 9.2m

Expressed as a vector equation:

τ = [0.23,0] x [ -31, 40] = [0, 0, 9.2] N.m

prolly a beacuse the momentuem the car has increases while going down hill

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