The normal force is not always mg

01
The tempting model
Setting N = mg on an incline, in an elevator, or under an applied force.
02
Why it feels right
On a level floor with no vertical acceleration, N = mg. Reusing that special case can feel automatic.
03
Where it breaks · how to catch it
You wrote N = mg without checking the perpendicular force balance.
04
The correct model
N is found from Newton's second law perpendicular to the surface. On a slope N = mg·cos θ; in an accelerating lift N = m(g ± a).
05
Repair · see it, then test it
worked example A 3.0 kg block rests on a 25° incline. Find the normal force.
Tempting move
N = mg = 29.4 N.
Correct move
N = mg·cos 25° ≈ 26.6 N.
Test the repair NL-07 · transfer

A 50 kg rider is in an elevator accelerating downward at 2.0 m/s². What normal force does the floor exert?
include N

Make a fresh prediction before referring to the worked example.
Then practise it on the ladder
R2Choose axes on an inclineadds: Axes parallel and perpendicular to the surface isolate the constrained direction.R3Find the normal force on an inclineadds: The normal force balances only the perpendicular component of weight.