Classical-Mechanics · Unit 8 · Video 1 · Interactive Practice

Normal Force Is Not Your Weight

IKey Formulas

Formula	Name	What it says
$\vec{C} = \vec{N} + \vec{f}$	Contact force	Perpendicular normal part $+$ parallel friction part
$N = mg$	Normal force at rest	Holds only when $a = 0$
$N = m(g + a)$	Vertical acceleration	Up taken as positive

Key Insight: The normal force is a constraint force — it obeys no force law of its own. It adjusts to whatever the motion demands, and equals $mg$ only in the special case $a = 0$ .

IIContact Force = Normal + Friction

One contact force resolves into a perpendicular normal part and a parallel friction part.

IIINormal Force in an Elevator

In an elevator the scale reads $N = m(g + a)$ , matching $mg$ only when $a = 0$ .

Elevator acceleration

a

(m/s², up positive) 0

IVThird-Law Pairs Act on Different Bodies

$N$ and $Fg$ both act on the block, so they cannot be a Newton's Third-Law pair.

Show

VQuiz Questions

Problem 1 · Normal Force at Rest

Given: A $2\ \text{kg}$ book lies at rest on a level table, with $g = 10\ \text{m/s}^2$ . Find the normal force $N$ from the table.

✅ Correct! At rest

a = 0

, so

N = mg = 2 \times 10 = 20\ \text{N}

❌ Close. The net force is zero, but that means

N

balances gravity:

N = mg = 20\ \text{N}

, not

0

❌ Not quite. With

a = 0

, Newton's second law gives

N - mg = 0

, so

N = 2 \times 10 = 20\ \text{N}

Show solution

Two forces act on the book: gravity $mg$ down and the normal force $N$ up. Because it is at rest, $a = 0$ .

Newton's second law (up positive):

N - mg = ma = 0 \quad\Rightarrow\quad N = mg = 2 \times 10 = 20\ \text{N}

Here $N = mg$ only because $a = 0$ .

Problem 2 · Accelerating Upward

Given: The same $2\ \text{kg}$ block sits on a scale in an elevator accelerating upward at $a = 5\ \text{m/s}^2$ , with $g = 10\ \text{m/s}^2$ . Find the scale reading $N$ .

✅ Correct!

N = m(g + a) = 2(10 + 5) = 30\ \text{N}

— greater than

mg

because the block accelerates upward.

❌ That is just $mg$ . But

N = mg

holds only when

a = 0

; here

a \neq 0

, so

N = m(g + a) = 30\ \text{N}

❌ Not quite. Up is positive:

N - mg = ma \Rightarrow N = m(g + a) = 2(15) = 30\ \text{N}

Show solution

Take up as positive. The normal force acts up, gravity acts down, and the acceleration is $a = +5\ \text{m/s}^2$ .

N - mg = ma \quad\Rightarrow\quad N = m(g + a) = 2(10 + 5) = 30\ \text{N}

The scale reads $30\ \text{N} > mg = 20\ \text{N}$ : the block feels heavier, yet gravity is unchanged.

Problem 3 · Accelerating Downward

Given: The $2\ \text{kg}$ block on its scale now accelerates downward at $4\ \text{m/s}^2$ , with $g = 10\ \text{m/s}^2$ and up positive.

Newton's second law (up positive) reads:

So the scale reads:

✅ Correct! With

a = -4\ \text{m/s}^2

N = m(g + a) = 2(10 - 4) = 12\ \text{N}

— less than

mg

❌ Check the setup. Up is positive;

N

acts up and

mg

acts down, so

N - mg = ma

❌ Check the arithmetic. The acceleration is

a = -4

, so

N = m(g + a) = 2(10 - 4) = 12\ \text{N}

Show solution

Downward acceleration with up positive means $a = -4\ \text{m/s}^2$ .

Step 1 — Newton's second law (up positive):

N - mg = ma

Step 2 — Solve for $N$ :

N = m(g + a) = 2(10 - 4) = 2(6) = 12\ \text{N}

Since $12\ \text{N} < mg = 20\ \text{N}$ , the block feels lighter. If it fell freely ( $a = -10$ ), then $N = 0$ — weightlessness — even though gravity never switched off.

Problem 4 · The Real Third-Law Partner

Given: A book rests on a table. Which force is the Newton's third-law reaction to the gravitational force the Earth exerts on the book?

✅ Correct! The Earth pulls the book down; the reaction is the book pulling the Earth up — equal, opposite, and on two different bodies.

❌ Common trap. The normal force and gravity both act on the book, so they are not a third-law pair. The reaction to gravity must act on the Earth.

❌ Not quite. A third-law partner acts on the other body — here, the book pulling the Earth up.

Show solution

A Newton's third-law pair consists of forces of the same type acting on two different bodies.

Gravity pair: Earth pulls book down $\;\leftrightarrow\;$ book pulls Earth up.
Contact pair: table pushes book up ( $N$ ) $\;\leftrightarrow\;$ book pushes table down.

The normal force $N$ and the gravitational force $mg$ both act on the book, so they are not a third-law pair — they merely balance while $a = 0$ . The reaction to gravity on the book is the book's gravitational pull on the Earth.

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