Non-inertial reference frames — pseudo forces: Question bank

Before we start, every symbol used on this page in plain words so nothing is a mystery:

Definition The cast of symbols

$a_{0}$ (read "a-naught") = the acceleration of the frame you are sitting in (bus, lift, car), as measured by someone standing on the ground (an inertial observer).
$m$ = the mass of the object we watch.
$g$ = the acceleration due to gravity ( $\approx 9.8 m/s^{2}$ ), pointing straight down; the real weight of the object is $m g$ , pulling down.
$N$ = the normal force: the upward push a floor or scale exerts on whatever rests on it. A weighing scale literally displays $N$ — this is your "apparent weight".
$T$ = the tension: the pull a taut string exerts along its own length, from the bob toward the pivot. For a hanging pendulum bob, $T$ points up along the string.
$F_{pseudo} = - m a_{0}$ = the fake force we add to keep $F = m a$ true inside the accelerating frame. The minus sign means "point it the opposite way to $a_{0}$ ".

The three scenarios these traps revisit — the bus, the lift, the tilted pendulum — are drawn as free-body diagrams here so you can see every arrow before reasoning about it:

Figure — Non-inertial reference frames — pseudo forces

Notice in the pendulum panel how the tension $T$ splits into two parts: an upward part $T cos θ$ that fights gravity, and a sideways part $T sin θ$ that fights the pseudo force. Balancing each direction gives $T cos θ = m g$ and $T sin θ = m a_{0}$ ; dividing kills $T$ and leaves $tan θ = a_{0} / g$ . That right-triangle of arrows is the "force triangle" referred to below.

True or false — justify

Every item: decide true/false, then give the reason. A bare "true" scores zero here.

A pseudo force acts on every mass in the non-inertial frame, big or small.

True — since

F_{pseudo} = - m a_{0}

scales with

m

, every object feels one; a heavier object feels a proportionally larger fake force, which is exactly why all objects share the same fake acceleration

- a_{0}

A pseudo force can do work on an object.

True — it has a magnitude and direction, so along any displacement in the frame it does

F_{pseudo} \cdot d

; "fictitious" refers to having no source, not to being energetically inert.

The pseudo force obeys Newton's third law like any other force.

False — it has no physical agent pushing, so there is no second body to push back; action–reaction pairs require two real bodies, and here there is only one.

In an inertial frame you may add a pseudo force as long as you are consistent.

False — an inertial frame has

a_{0} = 0

, so

F_{pseudo} = 0

; "adding" one means adding zero, and if you sneak in a nonzero one you double-count and get the wrong answer.

If two frames both accelerate but with the same

a_{0}

, an observer in one needs a pseudo force to describe motion in the other.

False — the relative acceleration between them is zero, so relative to each other they are inertial; no pseudo force is needed between frames sharing the same

a_{0}

The direction you feel pushed is the direction of the pseudo force.

True (in this case) — the pseudo force

- m a_{0}

points opposite to the frame's acceleration, and that is exactly the way your inertia "throws" you; the feeling and the fake force genuinely align.

A pseudo force changes if you switch which inertial frame you measure

a_{0}

from.

False — all inertial frames agree on acceleration (they differ only by constant velocity, see Galilean relativity), so

a_{0}

and hence

F_{pseudo}

is the same no matter which inertial frame you reference.

Spot the error

Each statement has one flawed line. Name the flaw and walk through the vector reasoning that fixes it.

"The bus accelerates forward, so the pseudo force on the passenger points forward."

Wrong direction. Write the rule as a vector:

F_{pseudo} = - m a_{0}

. Here

a_{0}

points forward (call it

+ x

), so

- m a_{0}

carries the minus sign into the

x

-component, flipping it to

- x

— backward. Step by step: forward

a_{0}

→ multiply by

m

(still forward) → multiply by

- 1

(now backward). The reversal is the minus sign, so the fake force points backward, matching the passenger being thrown into the seat.

"Since the ball on the smooth bus floor has a pseudo force on it, in the ground frame it must accelerate too."

Frame mix-up — the pseudo force exists only in the bus frame; in the ground frame there is no horizontal real force, so the ground observer correctly sees the ball stay at rest.

"In the upward-accelerating lift, apparent weight is

N = m (g - a_{0})

because the lift adds to gravity."

Sign error. Set up down as negative. The frame accelerates up (

a_{0} = + a_{0}

), so

F_{pseudo} = - m a_{0}

points down (

- m a_{0}

), stacking on top of the already-downward weight

- m g

. The upward normal

N

must now balance both:

N - m g - m a_{0} = 0

, giving

N = m (g + a_{0})

. The minus-version only appears when

a_{0}

itself points down and the pseudo force flips to up.

"For the pendulum in the car,

tan θ = g / a_{0}

from the force triangle."

Ratio inverted. In the force triangle (see the pendulum panel of the figure), the horizontal leg is

T sin θ

balancing the horizontal pseudo force

m a_{0}

, and the vertical leg is

T cos θ

balancing gravity

m g

. So

T sin θ = m a_{0}

and

T cos θ = m g

. Dividing the first by the second cancels

T

tan θ = \frac{m a _{0}}{m g} = \frac{a _{0}}{g}

. The claimed

g / a_{0}

swaps the legs of the triangle.

"A pseudo force is a real force with an unknown source we just haven't found yet."

Category error — it has no source by construction; it is a bookkeeping term born from choosing an accelerating frame, and it vanishes the moment you return to an inertial frame.

"Because centrifugal force is a pseudo force, it appears in the ground frame when a car turns."

Frame error — centrifugal force appears only in the rotating (car) frame; in the ground frame the car turns because of a real inward force (friction/road), with no outward force at all.

Why questions

Answer with the mechanism, not a restatement.

Why does the pseudo force have no reaction partner when every real force does?

Because a reaction partner is the "other body" in an interaction, but the pseudo force arises from the observer's accelerating frame, not from any body — with no second body there is nothing to push back.

Why does

F_{pseudo}

scale with mass, unlike, say, a fixed applied push?

It comes from the term

- m a_{0}

in the rearranged

F = m a

; it must scale with

m

so that dividing by

m

gives every object the same fake acceleration

- a_{0}

, matching what the frame's observer actually sees.

Why is the pendulum-in-a-car problem called an "accelerometer"?

The steady tilt satisfies

tan θ = a_{0} / g

, so measuring

θ

directly reveals

a_{0}

; the pseudo force literally tilts the string in proportion to the frame's acceleration.

Why does adding the pseudo force let a passenger keep using

F = m a

(see Newton's Second Law)?

Starting from the true law in the inertial frame and moving the

- m a_{0}

term to the force side turns "

m a_{frame}

" back into "(all forces including the fake one)", restoring the familiar bookkeeping.

Why does the ground observer never need a pseudo force to explain the same motion?

For them

a_{0} = 0

, so the correction is zero; real forces alone already predict the motion correctly, because the inertial frame is where Newton's laws hold untouched.

Why does the pseudo force point "down" in an upward-accelerating lift but "backward" in a forward bus?

In both cases it is simply

- a_{0}

: the lift accelerates up so the fake force is down; the bus accelerates forward so the fake force is backward. Same rule, different

a_{0}

Edge cases

Push each scenario to its boundary.

What is the pseudo force when the frame's acceleration is exactly zero (

a_{0} = 0

It is exactly zero — the frame is then inertial and Newton's laws need no correction; this is the smooth limit that recovers ordinary mechanics.

In a lift in free fall (

a_{0} = g

downward), what does the scale read and why?

N = m (g - a_{0}) = 0

— the downward pseudo force

m g

exactly cancels real gravity in the falling frame, so nothing presses on the scale; this is weightlessness (see Apparent weight & normal force).

If a lift accelerates downward faster than gravity (

a_{0} > g

), what happens to the apparent weight?

N = m (g - a_{0})

goes negative, which is impossible for a scale pushing up — it means the person leaves the floor and would need the ceiling to hold them; the object rises relative to the lift.

For the accelerating-car pendulum, what does

θ

do as

a_{0} \to 0

and as

a_{0} \to \infty

a_{0} \to 0

tan θ \to 0

so the string hangs vertical (no fake tilt); as

a_{0} \to \infty

tan θ \to \infty

θ \to 9 0^{\circ}

, the string swings toward horizontal.

Is the pseudo-force rule

F_{pseudo} = - m a_{0}

complete for a rotating frame?

No — for a rotating frame the acceleration is not a single constant

a_{0}

; extra velocity- and position-dependent terms appear, namely the centrifugal and Coriolis pseudo forces.

What happens to the ball on a smooth bus floor if the bus accelerates, then holds a constant velocity?

While accelerating, the pseudo force pushes the ball backward; the instant

a_{0} = 0

(constant velocity), the pseudo force vanishes and the ball simply keeps its new velocity — no further slide in the bus frame.

If the bus floor has friction, does the pseudo force disappear?

No — the pseudo force still equals

- m a_{0}

; friction is a separate real force that may hold the ball in place, so in the bus frame friction and pseudo force can balance to give zero net acceleration.

From $- m a_{0}$ to centrifugal and Coriolis

The one-line rule $F_{pseudo} = - m a_{0}$ is the straight-line special case. When the frame rotates, its acceleration is no longer one fixed vector, and two new pseudo-force terms sprout from the same idea. The figure below sketches the bridge:

A frame with straight-line $a_{0}$ → the plain $- m a_{0}$ (this page).
A rotating frame → objects at rest feel an outward centrifugal push (depends on position from the axis).
Objects moving inside a rotating frame feel a sideways Coriolis push (depends on velocity).

Same philosophy every time: whatever extra acceleration the frame injects, subtract $m$ times it and call the result a pseudo force.

Connections

Newton's Second Law — the trap of forgetting why the fake force is added.
Newton's Third Law — the recurring "no reaction pair" trap.
Centrifugal force — pseudo force in a rotating frame (rule extends).
Coriolis force — velocity-dependent pseudo force, beyond $- m a_{0}$ .
Apparent weight & normal force — lift edge cases live here.
Galilean relativity — why all inertial observers agree on $a_{0}$ .