Pseudo Forces

Working inside an accelerating car or lift? Add one imaginary force and carry on.

12 minUpdated 21 October 2026pseudo forcenon-inertial frameapparent weight

Lesson at a glanceShow the key points

In a frame accelerating at a₀ relative to an inertial frame, a(rel) = (F − ma₀)/m.
The extra term −ma₀ is called a pseudo (inertial) force: it acts on every body, opposite to the frame's acceleration, and has no reaction partner.
A pendulum in a car accelerating at a₀ hangs at angle tan⁻¹(a₀/g) from the vertical, toward the back.
In a lift accelerating upward at a, a scale reads m(g + a); downward, m(g − a); in free fall, zero.
Working from an inertial frame never requires pseudo forces.

When a bus brakes, you lurch forward as if shoved, though nothing touched you. Pseudo forces let us describe that feeling honestly and still use F = ma from inside the bus.

Newton’s law from an accelerating frame

Let frame S′ accelerate with constant $\vec a_0$ relative to an inertial frame S, without rotating. A particle P of mass $m$ has acceleration $\vec a$ measured in S′. From the chapter on changing frames,

\vec a_{P,S} = \vec a + \vec a_0 \quad\Rightarrow\quad m\vec a = m\vec a_{P,S} - m\vec a_0

S is inertial, so $m\vec a_{P,S}$ equals the net real force $\vec F$ . Therefore

Motion seen from an accelerating frame(5.3)

\vec a = \frac{\vec F - m\vec a_0}{m}

That is not Newton’s second law: an extra term $-m\vec a_0$ has appeared. Since we are so used to writing “net force = mass × acceleration,” we keep that habit by agreeing to treat $-m\vec a_0$ as if it were one more force on the particle.

Working from an inertial frame, $\vec a_0 = 0$ and no pseudo force is needed. Both approaches give the same physics; choose whichever makes the problem easier.

Interactive

A pendulum hanging in an accelerating car

The car accelerates to the right. From the road (inertial) the bob needs a net force forward to keep up. From inside the car the bob is at rest, so we add a pseudo force −ma₀ to balance the books.

\theta = \tan^{-1}(a_0/g)

14.3°

tension per kg10.11N

Car's acceleration

a_0

2.5m/s²

Analyse from

Worked example

A pendulum in an accelerating car

A small bob hangs from the roof of a car that accelerates at 2.5 m/s² on a level road. At what angle to the vertical does the string settle?

Feeling heavy, feeling light

A bathroom scale measures the normal force it exerts on you, not your mass. In a lift accelerating upward at $a$ , Newton’s second law from the ground frame gives $N - mg = ma$ , so $N = m(g + a)$ : you feel heavier. From inside the lift you would say the pseudo force $ma$ pushes you down and adds to your weight. Same answer.

Interactive

A bathroom scale in a lift

The scale shows the normal force it exerts on you. Accelerate the lift up and you feel heavier; down and you feel lighter; in free fall (a = −g) the scale reads zero.

72.2 kg

true mass 60 kg

feels heavier

N = m(g + a)

708N

scale displays72.2kg

Lift acceleration (up = +)

a

2.0m/s²

Your mass

m

60kg

Worked example

Making the block press lightly

A block of mass M sits on the floor of a box. With what acceleration should the box descend so that the block presses on the floor with only one third of its weight?

What to remember

1In a frame accelerating at $\vec a_0$ , add a pseudo force $-m\vec a_0$ to each body, then use F = ma.
2Pseudo forces have no source body and no reaction partner.
3A hanging bob in a car accelerating at a₀ tilts back by tan⁻¹(a₀/g).
4Apparent weight in a lift: m(g + a); zero in free fall.

Check yourself

In a lift accelerating downward at 3 m/s², a 50 kg person's scale reads (g = 10)…

Which statement about pseudo forces is correct?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchNewton's LawsWalter Lewin · MIT 8.01x Lecture 6 ReadFictitious forceWikipedia ReadCoriolis and other fictitious forcesHyperPhysics ReadMass and weightOpenStax University Physics §5.4

PreviousThe Third Law & Tension Next The Horse-and-Cart Puzzle & Inertia

Pseudo Forces

Working inside an accelerating car or lift? Add one imaginary force and carry on.

12 minUpdated 21 October 2026pseudo forcenon-inertial frameapparent weight

Lesson at a glanceShow the key points

In a frame accelerating at a₀ relative to an inertial frame, a(rel) = (F − ma₀)/m.
The extra term −ma₀ is called a pseudo (inertial) force: it acts on every body, opposite to the frame's acceleration, and has no reaction partner.
A pendulum in a car accelerating at a₀ hangs at angle tan⁻¹(a₀/g) from the vertical, toward the back.
In a lift accelerating upward at a, a scale reads m(g + a); downward, m(g − a); in free fall, zero.
Working from an inertial frame never requires pseudo forces.

Newton’s law from an accelerating frame

Let frame S′ accelerate with constant

\vec a_0

relative to an inertial frame S, without rotating. A particle P of mass

m

has acceleration

\vec a

measured in S′. From the chapter on changing frames,

\vec a_{P,S} = \vec a + \vec a_0 \quad\Rightarrow\quad m\vec a = m\vec a_{P,S} - m\vec a_0

S is inertial, so

m\vec a_{P,S}

equals the net real force

\vec F

. Therefore

Motion seen from an accelerating frame(5.3)

\vec a = \frac{\vec F - m\vec a_0}{m}

That is not Newton’s second law: an extra term

-m\vec a_0

has appeared. Since we are so used to writing “net force = mass × acceleration,” we keep that habit by agreeing to treat

-m\vec a_0

as if it were one more force on the particle.

Working from an inertial frame,

\vec a_0 = 0

and no pseudo force is needed. Both approaches give the same physics; choose whichever makes the problem easier.

Interactive

A pendulum hanging in an accelerating car

\theta = \tan^{-1}(a_0/g)

14.3°

tension per kg10.11N

Car's acceleration

a_0

2.5m/s²

Analyse from

Worked example

A pendulum in an accelerating car

A small bob hangs from the roof of a car that accelerates at 2.5 m/s² on a level road. At what angle to the vertical does the string settle?

Feeling heavy, feeling light

A bathroom scale measures the normal force it exerts on you, not your mass. In a lift accelerating upward at

a

, Newton’s second law from the ground frame gives

N - mg = ma

, so

N = m(g + a)

: you feel heavier. From inside the lift you would say the pseudo force

ma

pushes you down and adds to your weight. Same answer.

Interactive

A bathroom scale in a lift

The scale shows the normal force it exerts on you. Accelerate the lift up and you feel heavier; down and you feel lighter; in free fall (a = −g) the scale reads zero.

72.2 kg

true mass 60 kg

feels heavier

N = m(g + a)

708N

scale displays72.2kg

Lift acceleration (up = +)

a

2.0m/s²

Your mass

m

60kg

Worked example

Making the block press lightly

A block of mass M sits on the floor of a box. With what acceleration should the box descend so that the block presses on the floor with only one third of its weight?

What to remember

1In a frame accelerating at $\vec a_0$ , add a pseudo force $-m\vec a_0$ to each body, then use F = ma.
2Pseudo forces have no source body and no reaction partner.
3A hanging bob in a car accelerating at a₀ tilts back by tan⁻¹(a₀/g).
4Apparent weight in a lift: m(g + a); zero in free fall.

Check yourself

In a lift accelerating downward at 3 m/s², a 50 kg person's scale reads (g = 10)…

Which statement about pseudo forces is correct?

Newton’s law from an accelerating frame#

Feeling heavy, feeling light#

Check yourself

Keep exploring

Newton’s law from an accelerating frame#

Feeling heavy, feeling light#

Check yourself

Keep exploring

Newton’s law from an accelerating frame

Feeling heavy, feeling light

Newton’s law from an accelerating frame

Feeling heavy, feeling light