Free body diagrams — systematic drawing technique: Question bank

Intuition How to use this page

These are conceptual traps, not number-crunchers. Cover each answer, say your reasoning out loud before revealing. Why out loud? Silent reading lets your brain nod along to a plausible-looking answer without ever testing it — the "illusion of knowing." Speaking the because forces you to commit to a full chain of logic, which exposes the exact broken link. Example: for "is $N$ always $m g$ ?" don't just think "no" — say "No, because $N$ only equals $m g$ on flat ground with no vertical pull; on a slope it's $m g cos θ$ ." If any word in that sentence made you hesitate, that hesitation is the trap you needed to find. So the scoring rule: if your yes/no was right but your spoken reason was wrong or missing, count it as a miss — the reasoning is the whole point. This deep dive extends the parent Free body diagrams — systematic drawing technique, leaning on Newton's Second Law, Newton's Third Law, Normal Force and Friction, Tension in Strings and Pulleys, Inclined Plane Problems and Resolving Vectors into Components.

Definition Symbol legend — read before the questions

Every trap below reuses the same handful of standard symbols. Keep this list in view:

$m$ — mass of the chosen body, in kilograms.
$g$ — gravitational field strength near Earth, about $9.8 m/s^{2}$ ; it turns mass into weight.
$θ$ (theta) — the incline angle, i.e. the angle between the sloped surface and the horizontal ground. On flat ground $θ = 0^{\circ}$ ; a vertical wall is $θ = 9 0^{\circ}$ .
$N$ — normal force, the push a surface gives perpendicular to itself.
$T$ — tension, the pull a string exerts along its length, away from the body.
$μ_{k}$ (mu-k) — the coefficient of kinetic friction, the fraction linking sliding friction to the normal force: $f_{k} = μ_{k} N$ .
$a$ — acceleration of the body (how fast its velocity changes).
$m g$ — the weight, gravity's downward pull on the body.

Figure — Free body diagrams — systematic drawing technique

True or false — justify

Each begins True or False, and the answer must give the because.

The normal force is always equal to

m g

False — that only holds on a horizontal surface with no vertical push. On an incline

N = m g cos θ

(with

θ

the surface-to-horizontal angle), and if you press down on the block

N

grows, if you pull up it shrinks. See Normal Force and Friction.

The normal force and the weight are a Newton's-third-law action–reaction pair.

False — both act on the same body (the block), so they cannot be a 3rd-law pair. The pair to weight is the block pulling Earth up; the pair to

N

is the block pushing the surface down. See Newton's Third Law.

If the net force on a body is zero, the body must be at rest.

False — zero net force means zero acceleration (

a = 0

), i.e. constant velocity. A body cruising at steady speed has a perfectly balanced FBD.

A free body diagram should include the object's velocity arrow.

False — an FBD shows forces only. Velocity is motion, not a push or pull, so it never appears; drawing it invites confusing speed with force.

The tension is the same throughout a massless string running over a frictionless pulley.

True — an ideal (massless, inextensible) string over a frictionless pulley transmits the same

T

everywhere, because there is no mass to accelerate along the string and no friction to bleed force. See Tension in Strings and Pulleys.

On a frictionless incline, a heavier block slides down faster than a lighter one.

False —

a = g sin θ

has no mass in it; mass cancels just as in free fall, so both slide with the same acceleration.

Friction always opposes the object's motion.

False — friction opposes relative slipping (or its tendency). It can even drive motion: friction from the road is what pushes a walking foot or a driving tyre forward.

The force you apply to a wall equals the force the wall applies to you.

True — that is Newton's 3rd law: the two are equal in magnitude, opposite in direction, and act on different bodies (you and the wall). See Newton's Third Law.

Since a block on a slope stays put, there must be a force pushing it up the slope equal to

m g sin θ

True if it is static — that up-slope force is static friction, a real contact force, not a phantom. It equals

m g sin θ

only while the block does not slip.

A bigger applied force always means a bigger friction force.

False — kinetic friction is roughly constant (

μ_{k} N

) once sliding; pushing harder just increases acceleration, not friction. Static friction does grow with your push, but only up to its maximum.

Spot the error

State what is wrong and how to fix it. (Compare each against the reference FBDs above.)

"I drew an

ma

arrow pointing along the acceleration."

ma

is the result of the forces, not a force. It belongs on the right side of

\sum F = ma

, never as an arrow on the diagram. See Newton's Second Law.

"On the incline I drew the normal force pointing straight up."

Normal force is perpendicular to the surface, so on a slope it tilts away from vertical (see the correct panel of Figure 1). It points straight up only when the surface is horizontal.

"For the hanging block I drew tension pulling down toward the block below."

Tension in a string always pulls away from the body, along the rope. On a hanging block the string goes up, so

T

points up, not down (Figure 2).

"I put both the weight of the block and the block's push on the table on the block's FBD."

The block's push on the table acts on the table, not the block. Only forces on the chosen body belong on its FBD.

"Two blocks joined by a string, so I gave each a different tension."

One ideal string means one tension. Giving them different

T

's breaks the massless-string assumption and over-counts unknowns (Figure 2 shows the shared

T

"The book rests on a table, so the normal force pushes it up and the table's weight also holds it."

The table's weight acts on the table, not the book. The book's FBD has only its own weight down and the table's normal force up.

"On the incline I resolved gravity into

m g cos θ

down-slope and

m g sin θ

into the surface."

Swapped functions. The down-slope (along) component is

m g sin θ

; the into-surface (perpendicular) component is

m g cos θ

. See Resolving Vectors into Components.

"The object moves right, so I added a rightward 'force of motion' arrow."

There is no "force of motion." A body keeps moving with no forward force (inertia); only real pushes/pulls go on the FBD (see the wrong panel of Figure 1).

Why questions

Explain the reasoning, not just the fact.

Why do we draw gravity first in the recipe?

It is a long-range force always present near Earth, independent of contact. Drawing it first guarantees it is never forgotten.

Why is there exactly one set of contact forces per touching surface?

Objects only exchange contact forces where they physically touch. Scanning the boundary for touches finds every contact force and invents none.

Why do we tilt the axes along an incline instead of keeping them horizontal/vertical?

So the acceleration lies entirely along one axis and is zero on the other, killing one unknown and simplifying the equations. See Inclined Plane Problems.

Why does mass cancel in

a = g sin θ

The driving force

m g sin θ

grows with mass, but so does the inertia

m

resisting it. Dividing one by the other removes

m

entirely.

Why must action and reaction never both appear on one FBD?

A 3rd-law pair acts on two different bodies. An FBD isolates one body, so at most one member of any pair belongs on it. See Newton's Third Law.

Why can adding the two equations of a pulley system eliminate the tension?

Tension appears with opposite signs in the two bodies' equations (it pulls each toward the pulley), so summing cancels it, leaving only external forces and total mass.

Why does the normal force adjust its size on its own?

It is a passive constraint force: it becomes whatever is needed to prevent the surfaces from interpenetrating, so it responds to weight, applied pushes, and incline angle.

Edge cases

Boundary and degenerate situations the topic invites.

What is the normal force when the incline angle reaches

9 0^{\circ}

(a vertical wall)?

N = m g cos 9 0^{\circ} = 0

. The surface no longer supports the block against gravity, so the block is effectively in free fall along the wall.

What is the down-slope acceleration when the incline angle is

0^{\circ}

(flat ground)?

a = g sin 0^{\circ} = 0

. On level ground gravity has no along-surface component, so nothing drives the block.

In a pulley system, what happens to the acceleration if the hanging mass

m_{2} \to 0

a = \frac{m _{2}}{m _{1} + m _{2}} g \to 0

. With no hanging weight there is nothing to pull the system, so it stays at rest.

What happens to the acceleration if the table mass

m_{1} \to 0

a = \frac{m _{2}}{m _{1} + m _{2}} g \to g

. With negligible mass on the table, the hanging block is essentially in free fall, dragging a weightless partner.

If a block sits on a table and you pull up on it with force exactly equal to

m g

, what is the normal force?

N = 0

— the upward pull already balances gravity, so the table need not push at all. Pull any harder and the block lifts off (

N

cannot go negative).

A block in free fall (no surface, no string) — how many forces on its FBD?

Exactly one: gravity

m g

down. Nothing touches it, so there are no contact forces at all.

A box pushed at constant velocity across a rough floor — is the FBD balanced?

Yes — constant velocity means zero acceleration, so applied force equals kinetic friction and normal equals weight. Balanced forces, but the box is still moving.