Is the normal force always equal to mg?
Short answer: N adjusts to whatever is needed: N - mg = ma, so N = m(g + a) here.
The trap
The wrong idea
“The normal force always equals mg.”
It sounds reasonable, and a lot of JEE/NEET aspirants carry it into the exam. Here's a one-question check, then the explanation that makes it click.
Quick check
Try it before you peek
You stand in a lift accelerating upward. The floor's normal force on you is:
- Equal to mg
- More than mg
- Less than mg
Show the answer
B. More than mg
N adjusts to whatever is needed: N - mg = ma, so N = m(g + a) here.
The click
The normal force is a contact force: the surface pushes back as hard as needed to stop you sinking into it. It has no fixed formula. You find it from Newton's second law in each situation.
In a lift accelerating upward with acceleration a, two forces act on you: your weight mg downward and the normal force N upward. Taking up as positive, N − mg = ma, so N = m(g + a), which is more than mg. That is why you feel heavier: a weighing scale reads N, not mg.
If the lift accelerates downward, N = m(g − a), less than mg, and you feel lighter. In free fall (a = g), N = 0, which is apparent weightlessness. Moving at constant velocity, up or down, gives a = 0 and N = mg.
N differs from mg in many other cases too. On a smooth incline of angle θ, N = mg cos θ. Push down on a book lying on a table and N increases; pull up on it with a string and N decreases. At the bottom of a dip in a road, N is more than mg because a net upward force is needed for the circular motion. Always draw the free-body diagram and solve for N; never assume it.
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