- #1
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Suppose we have a block on an inclined plane.
If we choose the x-y axis to be parallel and perpendicular to the inclined plane, then we have
Fy = N - mgcos30 = 0
But if we choose our trivial x-y coordinate system, where y is parallel to the force of gravity, then we get:
Fy = Ncos30 - mg = 0
Of course, we get two different normal forces (and thus different frictional forces), for the exact same type of problem.
What is going on?
If we choose the x-y axis to be parallel and perpendicular to the inclined plane, then we have
Fy = N - mgcos30 = 0
But if we choose our trivial x-y coordinate system, where y is parallel to the force of gravity, then we get:
Fy = Ncos30 - mg = 0
Of course, we get two different normal forces (and thus different frictional forces), for the exact same type of problem.
What is going on?