Posted by Joe on .
A wheel of diameter 28.0 cm is constrained to rotate in the xy plane, about the z axis, which passes through its center. A force = (38.0 + 39.0 ) 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 torque about the rotation axis at this instant?
t= i+ j + k(m·N)

physics 
drwls,
Shouldn't there be unit vectors such as i and j associated with your force?
38.0 + 39.0 is just 1.
If the force is applied at 0.14 m (the wheel edge) at the x axis location, only the ycomponent of force will result in a torque. The torque will be in the z direction (k unit vector). The magnitude will be 39 N x 0.14 m = 5.46 N m.